RD Sharma Class 12 Ex 25.1 Solutions Chapter 25 Vector or Cross Product

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter25
Exercise25.1
CategoryRD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 25.1 Solutions Chapter 25 Vector or Cross Product

Question 1: Ten cards numbered 1 through 10 are placed in a box, mixed

Question 1. If \vec{a}= \hat{i}+3\hat{j}-2\hat{k}   and \vec{b}= -\hat{i}+3\hat{k} , find |\vec{a} \times \vec{b}|

Solution:

Given, \vec{a}= \hat{i}+3\hat{j}-2\hat{k}    and \vec{b}= -\hat{i}+3\hat{k}    .

=> \vec{a} \times \vec{b}   \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \end{vmatrix}

=> \vec{a} \times \vec{b}   \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 1 & 3 & -2\\ -1 & 0 & 3 \end{vmatrix}

=> \vec{a} \times \vec{b}   \hat{i}[(3)(3)-(0)(-2)]-\hat{j}[(1)(3)-(-1)(-2)]+\hat{k}[(1)(0)-(-1)(3)]

=> \vec{a} \times \vec{b}   \hat{i}[9-0]-\hat{j}[3-2]+\hat{k}[0-(-3)]

=> \vec{a} \times \vec{b}   9\hat{i}-\hat{j}+3\hat{k}

Now, |\vec{a} \times \vec{b}|    

=> |\vec{a} \times \vec{b}|   \sqrt{(9)^2+(-1)^2+(3)^2}

=> |\vec{a} \times \vec{b}|   \sqrt{81+1+9}

=> |\vec{a} \times \vec{b}|   = √91

Question 2(i). If \vec{a}= 3\hat{i}+4\hat{j}    and \vec{b}= \hat{i}+\hat{j}+\hat{k}   , find the value of |\vec{a} \times \vec{b}|    

Solution:

Given, \vec{a}= 3\hat{i}+4\hat{j}   and \vec{b}= \hat{i}+\hat{j}+\hat{k}   

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \end{vmatrix}

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 3 & 4 & 0\\ 1 & 1 & 1 \end{vmatrix}

=> \vec{a} \times \vec{b}  \hat{i}[(4)(1)-(1)(0)]-\hat{j}[(3)(1)-(1)(0)]+\hat{k}[(3)(1)-(1)(4)]

=> \vec{a} \times \vec{b}  \hat{i}[4-0]-\hat{j}[3-0]+\hat{k}[3-4]

=> \vec{a} \times \vec{b}  4\hat{i}-3\hat{j}-\hat{k}

Now, |\vec{a} \times \vec{b}|   

=> |\vec{a} \times \vec{b}|  \sqrt{(4)^2+(-3)^2+(-1)^2}

=> |\vec{a} \times \vec{b}|  \sqrt{16+1+9}

=> |\vec{a} \times \vec{b}|  \sqrt{26}

Question 2(ii). If \vec{a}= 2\hat{i}+\hat{j}   and \vec{b}= \hat{i}+\hat{j}+\hat{k}  , find the magnitude of |\vec{a} \times \vec{b}|   

Solution:

Given, \vec{a}= 2\hat{i}+\hat{j}   and \vec{b}= \hat{i}+\hat{j}+\hat{k}

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \end{vmatrix}

=> \vec{a} \times \vec{b}   \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2 & 1 & 0\\ 1 & 1 & 1 \end{vmatrix}

=> \vec{a} \times \vec{b}  \hat{i}[(1)(1)-(1)(0)]-\hat{j}[(2)(1)-(1)(0)]+\hat{k}[(2)(1)-(1)(1)]

=> \vec{a} \times \vec{b}  \hat{i}[1-0]-\hat{j}[2-0]+\hat{k}[2-1]

=> \vec{a} \times \vec{b}  \hat{i}-2\hat{j}+\hat{k}

Now, |\vec{a} \times \vec{b}|   

=> |\vec{a} \times \vec{b}|  \sqrt{(1)^2+(-2)^2+(1)^2}

=> |\vec{a} \times \vec{b}|  \sqrt{1+4+1}

=> |\vec{a} \times \vec{b}|  = √6

Question 3(i). Find a unit vector perpendicular to both the vectors 4\hat{i}-\hat{j}+3\hat{k}   and -2\hat{i}+\hat{j}-2\hat{k}   

Solution:

Given 4\hat{i}-\hat{j}+3\hat{k}    and -2\hat{i}+\hat{j}-2\hat{k}   

A vector perpendicular to 2 vectors is given by \vec{a} \times \vec{b}   

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \end{vmatrix}

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 4 & -1 & 3\\ -2 & 1 & -2 \end{vmatrix}

=> \vec{a} \times \vec{b}    \hat{i}[(-1)(-2)-(1)(3)]-\hat{j}[(4)(-2)-(-2)(3)]+\hat{k}[(4)(1)-(-2)(-1)]

=> \vec{a} \times \vec{b}    \hat{i}[2-3]-\hat{j}[-8+6]+\hat{k}[4-2]

=> \vec{a} \times \vec{b}  -\hat{i}+2\hat{j}+2\hat{k}

Unit vector is given by 

=> \hat{p}   = \dfrac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}

=> |\vec{a} \times \vec{b}|  \sqrt{(-1)^2+(2)^2+(2)^2}

=> |\vec{a} \times \vec{b}|  = 3

=> Unit vector is,

=> \hat{p}  \dfrac{1}{3}(-\hat{i}+2\hat{j}+2\hat{k})

Question 3(ii). Find a unit vector perpendicular to the plane containing the vectors \vec{a}=2\hat{i}+\hat{j}+\hat{k}   and \vec{b}=\hat{i}+2\hat{j}+\hat{k}   .

Solution:

 Given, \vec{a}=2\hat{i}+\hat{j}+\hat{k}   and \vec{b}=\hat{i}+2\hat{j}+\hat{k}   

A vector perpendicular to 2 vectors is given by \vec{a} \times \vec{b}

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \end{vmatrix}

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2 & 1 & 1\\ 1 & 2 & 1 \end{vmatrix}

=> \vec{a} \times \vec{b}  \hat{i}[(1)(1)-(2)(1)]-\hat{j}[(2)(1)-(1)(1)]+\hat{k}[(2)(2)-(1)(1)]

=> \vec{a} \times \vec{b}  \hat{i}[1-2]-\hat{j}[2-1]+\hat{k}[4-1]

=> \vec{a} \times \vec{b}  -\hat{i}-\hat{j}+3\hat{k}

Unit vector is given by 

=> \hat{p}  \dfrac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}

=> |\vec{a} \times \vec{b}|  \sqrt{(-1)^2+(-1)^2+(3)^2}

=> |\vec{a} \times \vec{b}|  \sqrt{11}

=> Unit vector is,

=> \hat{p}  \dfrac{1}{\sqrt{11}}(-\hat{i}-\hat{j}+3\hat{k})

Question 4. Find the magnitude of vector \vec{a} = (3\hat{k}+4\hat{j})\times(\hat{i}+\hat{j}-\hat{k})   

Solution:

Given \vec{a} = (3\hat{k}+4\hat{j})\times(\hat{i}+\hat{j}-\hat{k})   

=> \vec{a} = (3\hat{k}+4\hat{j})\times(\hat{i}+\hat{j}-\hat{k})      

=> \vec{a}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 0 & 4 & 3\\ 1 & 1 & -1 \end{vmatrix}

=> \vec{a}  \hat{i}[(4)(-1)-(1)(3)]-\hat{j}[(0)(-1)-(1)(3)]+\hat{k}[(0)(1)-(1)(4)]

=> \vec{a}  \hat{i}[-4-3]-\hat{j}[0-3]+\hat{k}[0-4]

=> \vec{a}  -7\hat{i}+3\hat{j}-4\hat{k}

Unit vector is,

=> |\vec{a}|  \sqrt{(-7)^2+(3)^2+(-4)^2}

=> |\vec{a}|  \sqrt{49+9+16}

=> |\vec{a}|  = √74

Question 5. If \vec{a}=4\hat{i}+3\hat{j}+\hat{k}    and \vec{b}=\hat{i}-2\hat{k}   , then find |2\hat{b}\times\vec{a}|   

Solution:

Given, \vec{a}=4\hat{i}+3\hat{j}+\hat{k}    and \vec{b}=\hat{i}-2\hat{k}   

=> \hat{b}  \dfrac{\vec{b}}{|\vec{b}|}

=> \hat{b}  \dfrac{(\hat{i}-2\hat{k})}{\sqrt{1^2+(-2)^2}}

=> \hat{b}  \dfrac{(\hat{i}-2\hat{k})}{\sqrt{5}}

=> 2\hat{b}  \dfrac{2}{\sqrt{5}}{(\hat{i}-2\hat{k})}

=> 2\hat{b}\times\vec{a}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ \dfrac{2}{\sqrt{5}} & 0 &\dfrac{-4}{\sqrt{5}}\\ 4 & 3 & 1 \end{vmatrix}

=> 2\hat{b}\times\vec{a}   \hat{i}[(0)(1)-(3)(-\dfrac{4}{\sqrt{5}})]-\hat{j}[(\dfrac{2}{\sqrt{5}})(1)-(4)(\dfrac{-4}{\sqrt{5}})]+\hat{k}[(\dfrac{2}{\sqrt{5}})(3)-(4)(0)]

=> 2\hat{b}\times\vec{a}  \hat{i}[0+\dfrac{12}{\sqrt{5}}]-\hat{j}[\dfrac{2}{\sqrt{5}}+\dfrac{16}{\sqrt{5}}]+\hat{k}[\dfrac{6}{\sqrt{5}}-0]

=> 2\hat{b}\times\vec{a}  \dfrac{12}{\sqrt{5}}\hat{i}-\dfrac{18}{\sqrt{5}}\hat{j}+\dfrac{6}{\sqrt{5}}\hat{k}

Now, |2\hat{b}\times\vec{a}|   

=> |2\hat{b}\times\vec{a}|  \sqrt{(\dfrac{12}{\sqrt{5}})^2+(\dfrac{-18}{\sqrt{5}})^2+(\dfrac{6}{\sqrt{5}})^2}

=> |2\hat{b}\times\vec{a}|  \sqrt{\dfrac{144}{5} + \dfrac{324}{5}+\dfrac{36}{5}}

=> |2\hat{b}\times\vec{a}|  \sqrt{\dfrac{504}{5}}

Question 6. If \vec{a}=3\hat{i}-\hat{j}-2\hat{k}   and \vec{b}=2\hat{i}+3\hat{j}+\hat{k}  , find (\vec{a}+2\vec{b})\times(2\vec{a}-\vec{b})   

Solution:

Given, \vec{a}=3\hat{i}-\hat{j}-2\hat{k}    and \vec{b}=2\hat{i}+3\hat{j}+\hat{k}   

=> 2\vec{a}  2(3\hat{i}-\hat{j}-2\hat{k})

=> 2\vec{a}  6\hat{i}-2\hat{j}-4\hat{k}

=> 2\vec{b}  2(2\hat{i}+3\hat{j}+\hat{k})

=> 2\vec{b}  4\hat{i}+6\hat{j}+2\hat{k}

=> \vec{a}+2\vec{b}  (3\hat{i}-\hat{j}-2\hat{k})+(4\hat{i}+6\hat{j}+2\hat{k})

=> \vec{a}+2\vec{b}  (3+4)\hat{i}+(-1+6)\hat{j}+(-2+2)\hat{k}

=> \vec{a}+2\vec{b}  7\hat{i}+5\hat{j}

=> 2\vec{a}-\vec{b}  (6\hat{i}-2\hat{j}-4\hat{k})-(2\hat{i}+3\hat{j}+\hat{k})

=> 2\vec{a}-\vec{b}  (6-2)\hat{i}+(-2-3)\hat{j}+(-4-1)\hat{k}

=> 2\vec{a}-\vec{b}  4\hat{i}-5\hat{j}-5\hat{k}

=> (\vec{a}+2\vec{b})\times(2\vec{a}-\vec{b})  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 7 & 5 & 0\\ 4 & -5 & -5 \end{vmatrix}

=> (\vec{a}+2\vec{b})\times(2\vec{a}-\vec{b})  \hat{i}[(5)(-5)-(-5)(0)]-\hat{j}[(7)(-5)-(4)(0)]+\hat{k}[(7)(-5)-(4)(5)]

=> (\vec{a}+2\vec{b})\times(2\vec{a}-\vec{b})  \hat{i}[-25-0]-\hat{j}[-35-0]+\hat{k}[-35-20]

=> (\vec{a}+2\vec{b})\times(2\vec{a}-\vec{b})  -25\hat{i}+35\hat{j}-55\hat{k}

Question 7(i). Find a vector of magnitude 49, which is perpendicular to both the vectors 2\hat{i}+3\hat{j}+6\hat{k}    and 3\hat{i}-6\hat{j}+2\hat{k}   

Solution:

Given, \vec{a} =2\hat{i}+3\hat{j}+6\hat{k}    and \vec{b}=3\hat{i}-6\hat{j}+2\hat{k}   

A vector perpendicular to 2 vectors is given by \vec{a} \times \vec{b}

=> \vec{a} \times \vec{b}   = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \end{vmatrix}

=>  \vec{a} \times \vec{b}   = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2 & 3 & 6\\ 3 & -6 & 2 \end{vmatrix}

=>  \vec{a} \times \vec{b}   = \hat{i}[(3)(2)-(-6)(6)]-\hat{j}[(2)(2)-(3)(6)]+\hat{k}[(2)(-6)-(3)(3)]

=>  \vec{a} \times \vec{b}   = \hat{i}[6+36]-\hat{j}[4-18]+\hat{k}[-12-9]

=>  \vec{a} \times \vec{b}  42\hat{i}+14\hat{j}-21\hat{k}

Magnitude of vector is given by,

=> |\vec{a} \times \vec{b}|  \sqrt{(42)^2+(14)^2+(-21)^2}

=> |\vec{a} \times \vec{b}|  \sqrt{1764+196+441}

=> |\vec{a} \times \vec{b}|  \sqrt{2401}

=> |\vec{a} \times \vec{b}|  49

=> Vector is, 42\hat{i}+14\hat{j}-21\hat{k}

Question 7(ii). Find the vector whose length is 3 and which is perpendicular to the vector 3\hat{i}+\hat{j}-4\hat{k}   and 6\hat{i}+5\hat{j}-2\hat{k}   

Solution:

Given, 3\hat{i}+\hat{j}-4\hat{k}    and 6\hat{i}+5\hat{j}-2\hat{k}   

A vector perpendicular to 2 vectors is given by   \vec{a} \times \vec{b}

=>  \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \end{vmatrix}

=>  \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 3 & 1 & -4\\ 6 & 5 & -2 \end{vmatrix}

=>  \vec{a} \times \vec{b}  \hat{i}[(1)(-2)-(5)(-4)]-\hat{j}[(3)(-2)-(6)(-4)]+\hat{k}[(3)(5)-(6)(1)]

=>  \vec{a} \times \vec{b}  \hat{i}[-2+20]-\hat{j}[-6+24]+\hat{k}[15-6]

=>  \vec{a} \times \vec{b}  18\hat{i}-18\hat{j}+9\hat{k}

Magnitude of vector is given by,

=> |\vec{a} \times \vec{b}|  \sqrt{(18)^2+(-18)^2+(9)^2}

=> |\vec{a} \times \vec{b}|  \sqrt{324+324+81}

=> |\vec{a} \times \vec{b}|  \sqrt{729}

=> |\vec{a} \times \vec{b}|  = 27

=> Unit vector is,

=> \hat{p}  \dfrac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}

=> \hat{p}  \dfrac{1}{27}(18\hat{i}-18\hat{j}+9\hat{k})

Required vector is, 

=> 3\times\dfrac{1}{27}(18\hat{i}-18\hat{j}+9\hat{k})=2\hat{i}-2\hat{j}+\hat{k}      

Question 8(i). Find the parallelogram determined by the vectors: 2\hat{i}    and 3\hat{j}   

Solution:

Given that, \vec{a}=2\hat{i}  and \vec{b} =3\hat{j}   

=> Area of the parallelogram is |\vec{a} \times \vec{b}|

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \end{vmatrix}

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2 & 0 & 0\\ 0 & 3 & 0 \end{vmatrix}

=> \vec{a} \times \vec{b}  \hat{i}[(0)(0)-(3)(0)]-\hat{j}[(2)(0)-(0)(0)]+\hat{k}[(2)(3)-(0)(0)]

=> \vec{a} \times \vec{b}  \hat{i}[0-0]-\hat{j}[0-0]+\hat{k}[6-0]

=> \vec{a} \times \vec{b}  6\hat{k}

Thus the area of parallelogram is,

=> |\vec{a} \times \vec{b}|  \sqrt{(0)^2+(0)^2+(6)^2}

=> |\vec{a} \times \vec{b}|  6

=> Area = 6 square units.

Question 8(ii). Find the parallelogram determined by the vectors: 2\hat{i}+\hat{j}+3\hat{k}    and \hat{i}-\hat{j}  .

Solution:

Given that, \vec{a}=2\hat{i}+\hat{j}+3\hat{k}   and \vec{b}=\hat{i}-\hat{j}

=> Area of the parallelogram is |\vec{a} \times \vec{b}|

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \end{vmatrix}

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2 & 1 & 3\\ 1 & -1 & 0 \end{vmatrix}

=> \vec{a} \times \vec{b}  \hat{i}[(1)(0)-(-1)(3)]-\hat{j}[(2)(0)-(1)(3)]+\hat{k}[(2)(-1)-(1)(1)]

=> \vec{a} \times \vec{b}  \hat{i}[0+3]-\hat{j}[0-3]+\hat{k}[-2-1]

=> \vec{a} \times \vec{b}  3\hat{i}+3\hat{j}-3\hat{k}

Thus, the area of parallelogram is,

=> |\vec{a} \times \vec{b}|  \sqrt{(3)^2+(3)^2+(-3)^2}

=> |\vec{a} \times \vec{b}|  \sqrt{9+9+9}

=> Area = 3\sqrt{3}

Question 8(iii). Find the area of the parallelogram determined by the vectors: 3\hat{i}+\hat{j}-2\hat{k}    and \hat{i}-3\hat{j}+4\hat{k}   

Solution:

Given that, \vec{a} = 3\hat{i}+\hat{j}-2\hat{k}   and \vec{b} = \hat{i}-3\hat{j}+4\hat{k}

=> Area of the parallelogram is |\vec{a} \times \vec{b}|

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \end{vmatrix}

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 3 & 1 & -2\\ 1 & -3 & 4 \end{vmatrix}

=> \vec{a} \times \vec{b}  \hat{i}[(1)(4)-(-3)(-2)]-\hat{j}[(3)(4)-(1)(-2)]+\hat{k}[(3)(-3)-(1)(1)]

=> \vec{a} \times \vec{b}  \hat{i}[4-6]-\hat{j}[12+2]+\hat{k}[-9-1]

=> \vec{a} \times \vec{b}  -2\hat{i}-14\hat{j}-10\hat{k}

Thus the area of parallelogram is,

=> |\vec{a} \times \vec{b}|    \sqrt{(-2)^2+(-14)^2+(-10)^2}

=> |\vec{a} \times \vec{b}|    \sqrt{4+196+100}

=> Area = 10\sqrt{3}

Question 8(iv). Find the area of the parallelogram determined by the vectors: \hat{i}-3\hat{j}+\hat{k}   and \hat{i}+\hat{j}+\hat{k}   

Solution:

Given that, \vec{a} = \hat{i}-3\hat{j}+\hat{k}   and \vec{b} = \hat{i}+\hat{j}+\hat{k}

=> Area of the parallelogram is |\vec{a} \times \vec{b}|

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \end{vmatrix}

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 1 & -3 & 1\\ 1 & 1 & 1 \end{vmatrix}

=> \vec{a} \times \vec{b}  \hat{i}[(-3)(1)-(1)(1)]-\hat{j}[(1)(1)-(1)(1)]+\hat{k}[(1)(1)-(1)(-3)]

=> \vec{a} \times \vec{b}  \hat{i}[-3-1]-\hat{j}[1-1]+\hat{k}[1+3]

=> \vec{a} \times \vec{b}  -4\hat{i}+4\hat{k}

Thus the area of parallelogram is,

=> |\vec{a} \times \vec{b}|  \sqrt{(-4)^2+(0)^2+(4)^2}

=> |\vec{a} \times \vec{b}|  \sqrt{16+16}

=> Area = 4\sqrt{2}

Question 9(i). Find the area of the parallelogram whose diagonals are: 4\hat{i}-\hat{j}-3\hat{k}   and -2\hat{i}+\hat{j}-2\hat{k}   

Solution:

Given, \vec{a}=4\hat{i}-\hat{j}-3\hat{k}   and \vec{b}=-2\hat{i}+\hat{j}-2\hat{k}

=> Area of the parallelogram is \dfrac{1}{2}|\vec{a} \times \vec{b}|

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \end{vmatrix}

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 4 & -1 & -3\\ -2 & 1 & -2 \end{vmatrix}

=> \vec{a} \times \vec{b}  \hat{i}[(-1)(-2)-(1)(-3)]-\hat{j}[(4)(-2)-(-2)(-3)]+\hat{k}[(4)(1)-(-2)(-1)]

=> \vec{a} \times \vec{b}  \hat{i}[2+3]-\hat{j}[-8-6]+\hat{k}[4-2]

=> \vec{a} \times \vec{b}  5\hat{i}+14\hat{j}+2\hat{k}

Thus the area of parallelogram is,

=> \dfrac{1}{2}|\vec{a} \times \vec{b}|  \dfrac{1}{2}\sqrt{(5)^2+(14)^2+(2)^2}

=> \dfrac{1}{2}|\vec{a} \times \vec{b}|  \dfrac{1}{2}\sqrt{225}

=> Area = 15/2 = 7.5 square units

Question 9(ii). Find the area of the parallelogram whose diagonals are: 2\hat{i}+\hat{k}    and \hat{i}+\hat{j}+\hat{k}   

Solution:

Given, \vec{a}=2\hat{i}+\hat{k}   and \vec{b}=\hat{i}+\hat{j}+\hat{k}

=> Area of the parallelogram is \dfrac{1}{2}|\vec{a} \times \vec{b}|

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \end{vmatrix}

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2 & 0 & 1\\ 1 & 1 & 1 \end{vmatrix}

=> \vec{a} \times \vec{b}  \hat{i}[(0)(1)-(1)(1)]-\hat{j}[(2)(1)-(1)(1)]+\hat{k}[(2)(1)-(1)(0)]

=> \vec{a} \times \vec{b}  \hat{i}[0-1]-\hat{j}[2-1]+\hat{k}[2-0]

=> \vec{a} \times \vec{b}  -\hat{i}-\hat{j}+2\hat{k}

Thus the area of parallelogram is,

=> \dfrac{1}{2}|\vec{a} \times \vec{b}|  \dfrac{1}{2}\sqrt{(-1)^2+(-1)^2+(2)^2}

=> \dfrac{1}{2}|\vec{a} \times \vec{b}|  \dfrac{1}{2}\sqrt{6}

=> Area = \dfrac{\sqrt{6}}{2}

Question 9(iii). Find the area of the parallelogram whose diagonals are: 3\hat{i}+4\hat{j}   and \hat{i}+\hat{j}+\hat{k}   

Solution:

Given, \vec{a}=3\hat{i}+4\hat{j}   and \vec{b}=\hat{i}+\hat{j}+\hat{k}

=> Area of the parallelogram is \dfrac{1}{2}|\vec{a} \times \vec{b}|

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \end{vmatrix}

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 3 & 4 & 0\\ 1 & 1 & 1 \end{vmatrix}

=> \vec{a} \times \vec{b}  \hat{i}[(4)(1)-(1)(0)]-\hat{j}[(3)(1)-(1)(0)]+\hat{k}[(3)(1)-(1)(4)]

=> \vec{a} \times \vec{b}  \hat{i}[4-0]-\hat{j}[3-0]+\hat{k}[3-4]

=> \vec{a} \times \vec{b}  4\hat{i}-3\hat{j}-\hat{k}

Thus the area of parallelogram is,

=> \dfrac{1}{2}|\vec{a} \times \vec{b}|  \dfrac{1}{2}\sqrt{(4)^2+(-3)^2+(-1)^2}

=> \dfrac{1}{2}|\vec{a} \times \vec{b}|  \dfrac{1}{2}\sqrt{26}

=> Area = \dfrac{\sqrt{26}}{2}

Question 9(iv). Find the area of the parallelogram whose diagonals are: 2\hat{i}+3\hat{j}+6\hat{k}    and 3\hat{i}-6\hat{j}+2\hat{k}   

Solution:

Given, \vec{a}=2\hat{i}+3\hat{j}+6\hat{k}  and \vec{b}=3\hat{i}-6\hat{j}+2\hat{k}

=> Area of the parallelogram is \dfrac{1}{2}|\vec{a} \times \vec{b}|

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \end{vmatrix}

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2 & 3 & 6\\ 3 & -6 & 2 \end{vmatrix}

=> \vec{a} \times \vec{b}  \hat{i}[(3)(2)-(-6)(6)]-\hat{j}[(2)(2)-(3)(6)]+\hat{k}[(2)(-6)-(3)(3)]

=> \vec{a} \times \vec{b}  \hat{i}[6+36]-\hat{j}[4-18]+\hat{k}[-12-9]

=> \vec{a} \times \vec{b}  42\hat{i}+14\hat{j}-21\hat{k}

Thus the area of parallelogram is,

=> \dfrac{1}{2}|\vec{a} \times \vec{b}|  \dfrac{1}{2}\sqrt{(42)^2+(14)^2+(-21)^2}

=> \dfrac{1}{2}|\vec{a} \times \vec{b}|  \dfrac{1}{2}\sqrt{2401}

=> Area = \dfrac{49}{2}

=> Area = 24.5

Question 10. If  \vec{a}=2\hat{i}+5\hat{j}-7\hat{k}  \vec{b}=-3\hat{i}+4\hat{j}+\hat{k}    and \vec{c}=\hat{i}-2\hat{j}-3\hat{k}  , compute (\vec{a}\times\vec{b})\times\vec{c}    and \vec{a}\times(\vec{b}\times\vec{c})    and verify these are not equal.

Solution:

Given \vec{a}=2\hat{i}+5\hat{j}-7\hat{k}  \vec{b}=-3\hat{i}+4\hat{j}+\hat{k}   and \vec{c}=\hat{i}-2\hat{j}-3\hat{k}

=> \vec{a} \times \vec{b}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2 & 5 & -7\\ -3 & -4 & 1 \end{vmatrix}

=> \vec{a} \times \vec{b}  \hat{i}[(5)(1)-(4)(-7)]-\hat{j}[(2)(1)-(-3)(-7)]+\hat{k}[(2)(4)-(-3)(5)]

=> \vec{a} \times \vec{b}  \hat{i}[5+28]-\hat{j}[2-21]+\hat{k}[8+15]

=> \vec{a} \times \vec{b}  33\hat{i}+19\hat{j}+23\hat{k}

=> (\vec{a}\times\vec{b})\times\vec{c}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 33 & 19 & 23\\ 1 & -2 & -3 \end{vmatrix}

=> (\vec{a}\times\vec{b})\times\vec{c}   = \hat{i}[(19)(-3)-(-2)(23)]-\hat{j}[(33)(-3)-(1)(23)]+\hat{k}[(33)(-2)-(1)(19)]

=> (\vec{a}\times\vec{b})\times\vec{c}   = \hat{i}[-57+46]-\hat{j}[-99-23]+\hat{k}[-66-19]

=> (\vec{a}\times\vec{b})\times\vec{c}  -11\hat{i}+122\hat{j}-85\hat{k}

=> \vec{b} \times \vec{c}  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ -3 & 4 & 1\\ 1 & -2 & -3 \end{vmatrix}

=> \vec{b} \times \vec{c}  \hat{i}[(4)(-3)-(-2)(1)]-\hat{j}[(-3)(-3)-(1)(1)]+\hat{k}[(-3)(-2)-(1)(4)]

=> \vec{b} \times \vec{c}  \hat{i}[-12+2]-\hat{j}[9-1]+\hat{k}[6-4]

=> \vec{b} \times \vec{c}  -10\hat{i}-8\hat{j}+2\hat{k}

=> \vec{a}\times(\vec{b}\times\vec{c})  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2 & 5 & -7\\ -10 & -8 & 2 \end{vmatrix}

=> \vec{a}\times(\vec{b}\times\vec{c})  \hat{i}[(5)(2)-(-8)(-7)]-\hat{j}[(2)(2)-(-10)(-7)]+\hat{k}[(2)(-8)-(-10)(5)]

=> \vec{a}\times(\vec{b}\times\vec{c})  \hat{i}[10-56]-\hat{j}[4-70]+\hat{k}[-16+50]

=> \vec{a}\times(\vec{b}\times\vec{c})  -46\hat{i}+66\hat{j}+34\hat{k}

=> (\vec{a}\times\vec{b})\times\vec{c}   is not equal to \vec{a}\times(\vec{b}\times\vec{c})

=> Hence verified.

Question 11. If |\vec{a}|=2   |\vec{b}|=5    and |\vec{a}\times\vec{b}|=8   , find \vec{a}.\vec{b}    

Solution:

We know that,

=> \vec{a}\times\vec{b} = |\vec{a}||\vec{b}|\sin\theta\hat{n}

=> |\vec{a}\times\vec{b}|  =  |\vec{a}||\vec{b}||\sin\theta||\hat{n}|

We know that |\hat{n}|  is 1, as \hat{n}  is a unit vector

=> 8 = 2\times5\times\sin\theta\times1

=> 10\sin\theta  = 8

=> \sin\theta  = \dfrac{4}{5}

Also,

=> \vec{a}.\vec{b} = |\vec{a}||\vec{b}|\cos\theta

And \sin^2\theta + \cos^2\theta = 1

=> \cos\theta  = \sqrt{1-\sin^2\theta}

=> \cos\theta  = \sqrt{1-(\dfrac{4}{5})^2}

=> \cos\theta  = \sqrt{\dfrac{9}{16}}

=> \cos\theta  = \dfrac{3}{5}

=> \vec{a}.\vec{b}  = 2\times5\times\dfrac{3}{5}

=> \vec{a}.\vec{b}  = 6

Question 12. Given \vec{a} = \dfrac{1}{7}(2\hat{i}+3\hat{j}+6\hat{k})   \vec{b} = \dfrac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k})   \vec{c} = \dfrac{1}{7}(6\hat{i}+2\hat{j}-3\hat{k})   \hat{i}   \hat{j}   , \hat{k}    being a right-handed orthogonal system of unit vectors in space, show that \vec{a}   \vec{b}    and \vec{c}    is also another system.

Solution:

To show that \vec{a} \vec{b}  and \vec{c}  is a right-handed orthogonal system of unit vectors, we need to prove:

(1) |\vec{a}|=|\vec{b}|=|\vec{c}|=1

=> |\vec{a}| = \dfrac{1}{7}\sqrt{2^2+3^2+6^2}

=> |\vec{a}| = \dfrac{1}{7}\sqrt{4+9+36}

=> |\vec{a}| = \dfrac{1}{7}\sqrt{49}

=> |\vec{a}| = \dfrac{1}{7}\times7

=> |\vec{a}| = 1

=> |\vec{b}| = \dfrac{1}{7}\sqrt{3^2+(-6)^2+2^2}

=> |\vec{b}| = \dfrac{1}{7}\sqrt{9+36+4}

=> |\vec{b}| = \dfrac{1}{7}\sqrt{49}

=> |\vec{b}| = \dfrac{1}{7}\times7

=> |\vec{b}| = 1

=> |\vec{c}| = \dfrac{1}{7}\sqrt{6^2+2^2+(-3)^2}

=> |\vec{c}| = \dfrac{1}{7}\sqrt{36+4+9}

=> |\vec{c}| = \dfrac{1}{7}\sqrt{49}

=> |\vec{c}| = \dfrac{1}{7}\times7

=> |\vec{c}| = 1

(2) \vec{a}\times\vec{b} = \vec{c}

=> \vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_1 & a_2 & a_3\\b_1 & b_ 2& b_3\end{vmatrix}

=>\vec{a}\times\vec{b}= \dfrac{1}{7}\times\dfrac{1}{7}\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2 & 3 & 6\\3 & -6& 2\end{vmatrix}

=>\vec{a}\times\vec{b} = \dfrac{1}{49}(\hat{i}[6+36]-\hat{j}[4-18]+\hat{k}[-12-9])

=>\vec{a}\times\vec{b} = \dfrac{1}{49}(42\hat{i}+14\hat{j}-21\hat{k})

=> \vec{a}\times\vec{b} = \dfrac{1}{7}(6\hat{i}+2\hat{j}-3\hat{k})=\vec{c}

(3) \vec{b}\times\vec{c} =\vec{a}

=>\vec{b}\times\vec{c} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\b_1 & b_2 & b_3\\c_1 & c_ 2& c_3\end{vmatrix}

=>\vec{b}\times\vec{c}= \dfrac{1}{7}\times\dfrac{1}{7}\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\3 & -6 & 2\\6 & 2& -3\end{vmatrix}

=>\vec{b}\times\vec{c} = \dfrac{1}{49}(\hat{i}[18-4]-\hat{j}[-9-12]+\hat{k}[6+36])

=>\vec{b}\times\vec{c} = \dfrac{1}{49}(14\hat{i}+21\hat{j}+42\hat{k})

=>\vec{b}\times\vec{c} = \dfrac{1}{7}(2\hat{i}+3\hat{j}+6\hat{k})=\vec{a}

(4) \vec{c}\times\vec{a} = \vec{b}

=>\vec{c}\times\vec{a} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\c_1 & c_2 & c_3\\a_1 & a_ 2& a_3\end{vmatrix}

=> \vec{c}\times\vec{a}= \dfrac{1}{7}\times\dfrac{1}{7}\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\6 & 2 & -3\\2 & 3& 6\end{vmatrix}

=>\vec{c}\times\vec{a} = \dfrac{1}{49}(\hat{i}[12+9]-\hat{j}[36+6]+\hat{k}[18-4])

=>\vec{c}\times\vec{a} = \dfrac{1}{49}(21\hat{i}-42\hat{j}+14\hat{k})

=>\vec{c}\times\vec{a} = \dfrac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k})=\vec{b}

Hence proved.

Question 13. If |\vec{a}|=13 |\vec{b}|=5   and \vec{a}.\vec{b}=60 , find |\vec{a}\times\vec{b}|
Solution:

We know that,
=> \vec{a}.\vec{b}  = |\vec{a}||\vec{b}|\cos\theta
=>|\vec{a}.\vec{b}|  =  |\vec{a}||\vec{b}||\sin\theta|
=> 60 = 13\times5\times\cos\theta
=> 65\cos\theta  = 60
=> \cos\theta  = \dfrac{12}{13}
Also,
=>| \vec{a}\times\vec{b}| = |\vec{a}||\vec{b}||\cos\theta||\hat{n}|
And \sin^2\theta + \cos^2\theta = 1
=> \sin\theta  = \sqrt{1-\cos^2\theta}
=> \sin\theta  = \sqrt{1-(\dfrac{12}{13})^2}
=> \sin\theta  = \sqrt{\dfrac{25}{169}}
=> \sin\theta  = \dfrac{5}{13}
=> |\vec{a}\times\vec{b}| = 13\times5\times\dfrac{5}{13}
=>| \vec{a}\times\vec{b}|  = 25
Question 14. Find the angle between 2 vectors \vec{a}   and  \vec{b}    , if  |\vec{a}\times\vec{b}| = \vec{a}.\vec{b}
Solution:
Given |\vec{a}\times\vec{b}| = \vec{a}.\vec{b}
=>|\vec{a}||\vec{b}|\sin\theta|\hat{n}| = |\vec{a}||\vec{b}|\cos\theta
=> |\vec{a}||\vec{b}|\sin\theta = |\vec{a}||\vec{b}|\cos\theta   , as \hat{n}  is a unit vector.
=> \sin\theta = \cos\theta
=> \tan\theta = 1
=> \theta = \dfrac{\pi}{4}
Question 15. If \vec{a}\times\vec{b} = \vec{b}\times\vec{c} \neq \vec{0}   , then show that \vec{a}+\vec{c}=m\vec{b}   , where m is any scalar.
Solution:
Given that \vec{a}\times\vec{b} = \vec{b}\times\vec{c} \neq \vec{0}
=> \vec{a}\times\vec{b} - \vec{b}\times\vec{c} = \vec{0}
=> \vec{a}\times\vec{b} -[-(\vec{c}\times\vec{a})] = \vec{0}
=> \vec{a}\times\vec{b} + \vec{c}\times\vec{b} = \vec{0}
Using distributive property,
=> (\vec{a}+\vec{c})\times\vec{b}=\vec{0}
If two vectors are parallel, then their cross-product is 0 vector.
=> (\vec{a}+\vec{c})    and \vec{b}    are parallel vectors.
=> (\vec{a}+\vec{c}) = m\vec{b}
Hence proved.
 Question 16. If |\vec{a}|=2   |\vec{b}|=7   and \vec{a}\times\vec{b} = 3\hat{i}+2\hat{j}+6\hat{k}   , find the angle between \vec{a}    and \vec{b}
Solution:
Given that, |\vec{a}|=2   |\vec{b}|=7    and \vec{a}\times\vec{b} = 3\hat{i}+2\hat{j}+6\hat{k}
We know that,
=> \vec{a}\times\vec{b} =|\vec{a}||\vec{b}|\sin\theta\hat{n}
=> |\vec{a}\times\vec{b}| =|\vec{a}||\vec{b}|\sin\theta\|\hat{n}|
=> |\vec{a}\times\vec{b}| =|\vec{a}||\vec{b}|\sin\theta
=> \sqrt{3^2+2^2+6^2} = 2\times7\times\sin\theta
=> \sqrt{9+4+36} = 14\sin\theta
=> 7 = 14\sin\theta
=> \sin\theta = \dfrac{1}{2}
=> \theta = \dfrac{\pi}{6}
Question 17. What inference can you draw if \vec{a}\times\vec{b}=\vec{0}   and \vec{a}.\vec{b}=0
Solution:
Given, \vec{a}\times\vec{b}=\vec{0}   and \vec{a}.\vec{b}=0
=>\vec{a}\times\vec{b} = \vec{0}
=> |\vec{a}|\vec{b}|\sin\theta\hat{n} = \vec{0}
Either of the following conditions is true,
1. |\vec{a}| = 0
2. |\vec{b}| =0
3. |\vec{a}| = |\vec{b}| = 0
4. \vec{a}   is parallel to \vec{b}
=> \vec{a}.\vec{b}=0
=> |\vec{a}||\vec{b}|\cos\theta = 0
Either of the following conditions is true,
1. |\vec{a}| = 0
2. |\vec{b}| =0
3. |\vec{a}| = |\vec{b}| = 0
4. \vec{a}   is perpendicular to \vec{b}
Since both these conditions are true, that implies atleast one of the following conditions is true,
1. |\vec{a}| = 0
2. |\vec{b}| =0
3. |\vec{a}| = |\vec{b}| = 0
Question 18. If \vec{a}   \vec{b}   and \vec{c}   are 3 unit vectors such that \vec{a}\times\vec{b} = \vec{c}   \vec{b}\times\vec{c}=\vec{a}   and \vec{c}\times\vec{a} = \vec{b}   . Show that \vec{a}   , \vec{b}   and \vec{c}    form an orthogonal right handed triad of unit vectors.
Solution:
Given, \vec{a}\times\vec{b} = \vec{c}   \vec{b}\times\vec{c}=\vec{a}   and \vec{c}\times\vec{a} = \vec{b}
As,
=> \vec{c} = \vec{a}\times\vec{b}
=> \vec{c}    is perpendicular to both \vec{a}   and \vec{b}   .
Similarly,
=> \vec{a}    is perpendicular to both \vec{b}   and \vec{c}
=> \vec{b}    is perpendicular to both \vec{a}   and \vec{c}
=> \vec{a}   \vec{b}   and \vec{c}    are mutually perpendicular.
As, \vec{a}   \vec{b}   and \vec{c}    are also unit vectors,
=> \vec{a}   \vec{b}   and \vec{c}    form an orthogonal right-handed triad of unit vectors
Hence proved.
Question 19. Find a unit vector perpendicular to the plane ABC, where the coordinates of A, B, and C are A(3, -1, 2), B(1, -1, 3), and C(4, -3, 1).
Solution:
Given A(3, -1, 2), B(1, -1, 3) and C(4, -3, 1).
Let,
=> \vec{a} = A = 3\hat{i}- \hat{j} +2\hat{k}
=> \vec{b} = B = \hat{i} -\hat{j} + 3\hat{k}
=> \vec{c} = C = 4\hat{i}-3\hat{j}+\hat{k}
Plane ABC has two vectors \vec{AB}    and \vec{AC}
=> \vec{AB} = \vec{b} - \vec{a}
=> \vec{AB} = (\hat{i}-\hat{j}-3\hat{k})-(3\hat{i}-\hat{j}+2\hat{k})
=> \vec{AB}= (1-3)\hat{i}+ (-1+1)\hat{j} +(-3-2)\hat{k}
=> \vec{AB} = -2\hat{i}-5\hat{k}
=> \vec{AC} = \vec{c} - \vec{a}
=> \vec{AC} = (4\hat{i}-3\hat{j}+\hat{k})-(3\hat{i}-\hat{j}+2\hat{k})
=> \vec{AC} = (4-3)\hat{i}+ (-3+1)\hat{j} +(1-2)\hat{k}
=> \vec{AC} = \hat{i}-2\hat{i}-\hat{k}
A vector perpendicular to both \vec{AB}   and \vec{AC}   is given by,
=> \vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_1 & a_2 & a_3\\b_1 & b_ 2& b_3\end{vmatrix}
=>  \vec{AB}\times\vec{AC}= \dfrac{1}{7}\times\dfrac{1}{7}\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\-2 & 0 & -5\\1 & -2& -1\end{vmatrix}
=> \vec{AB}\times\vec{AC} = \hat{i}[(0)(-1)-(-2)(-5)] -\hat{j}[(-2)(-1)-(1)(-5)] +\hat{k}[(-2)(-2)-(1)(0)]
=> \vec{AB}\times\vec{AC} = \hat{i}[0-10]-\hat{j}[2+5]+\hat{k}[4-0]
=> \vec{AB}\times\vec{AC} = -10\hat{j}-7\hat{j}+4\hat{k}
To find the unit vector,
=> \hat{p} = \dfrac{\vec{AB}\times\vec{AC}}{|\vec{AB}\times\vec{AC}|}
=> \hat{p} = \dfrac{1}{\sqrt{(-10^2)+(-7)^2+4^2}}(-10\hat{j}-7\hat{j}+4\hat{k})
=> \hat{p} = \dfrac{1}{\sqrt{100+49+16}}(-10\hat{j}-7\hat{j}+4\hat{k})
=> \hat{p} = \dfrac{1}{\sqrt{165}}(-10\hat{j}-7\hat{j}+4\hat{k})
Question 20. If a, b and c are the lengths of sides BC, CA and AB of a triangle ABC, prove that \vec{BC} +\vec{CA} +\vec{AB} = \vec{0}   and deduce that \dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}
Solution:
Given that |\vec{BC}|=a   |\vec{CA}|=b   and |\vec{AB}| = c
From triangle law of vector addition, we have
=> \vec{AB} + \vec{BC} = \vec{AC}
=> \vec{AB} + \vec{BC} = -\vec{CA}
=> \vec{AB} + \vec{BC} + \vec{CA} = \vec{0}
=> \vec{BC} +\vec{CA} +\vec{AB} = \vec{0}
=> \vec{a} + \vec{b} + \vec{c} = \vec{0}
=> \vec{a}\times(\vec{a}+\vec{b}+\vec{c}) = \vec{a}\times\vec{0}
=> \vec{a}\times\vec{a} + \vec{a}\times\vec{b} + \vec{a}\times\vec{c} = \vec{0}
=> \vec{0} + \vec{a}\times\vec{b} + \vec{a}\times\vec{c} = \vec{0}
=> \vec{a}\times\vec{b} + \vec{a}\times\vec{c} = \vec{0}
=> \vec{a}\times\vec{b} = -\vec{a}\times\vec{c}
=> \vec{a}\times\vec{b} = \vec{c}\times\vec{a}
=> |\vec{a}||\vec{b}|\sin C = |\vec{c}||\vec{a}|\sin B
=> |\vec{b}|\sin C = |\vec{c}|\sin B
=> b\sin C = c\sin B
=> \dfrac{b}{\sin B} = \dfrac{c}{\sin C}
Similarly,
=> \vec{b}\times(\vec{a}+\vec{b}+\vec{c}) = \vec{b}\times\vec{0}
=> \vec{b}\times\vec{a} + \vec{b}\times\vec{b} + \vec{b}\times\vec{c} = \vec{0}
=> \vec{0} + \vec{b}\times\vec{a} + \vec{b}\times\vec{c} = \vec{0}
=> \vec{b}\times\vec{a} + \vec{b}\times\vec{c} = \vec{0}
=> \vec{b}\times\vec{a} = -\vec{b}\times\vec{c}
=> \vec{b}\times\vec{a} = \vec{c}\times\vec{b}
=> |\vec{b}||\vec{a}|\sin C = |\vec{c}||\vec{b}|\sin A
=> |\vec{a}|\sin C = |\vec{c}|\sin A
=> a\sin C = c\sin A
=> \dfrac{a}{\sin A} = \dfrac{c}{\sin C}
=> \dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{sin C}
Hence proved.
Question 21. If \vec{a} = \hat{i}-2\hat{j}+3\hat{k}  and \vec{b}=2\hat{i}+3\hat{j}-5\hat{k}  , then find \vec{a}\times\vec{b}  . Verify that \vec{a}  and \vec{a}\times\vec{b}   are perpendicular to each other.
Solution:
Given, \vec{a} = \hat{i}-2\hat{j}+3\hat{k}  and \vec{b}=2\hat{i}+3\hat{j}-5\hat{k}
=> \vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_1 & a_2 & a_3\\b_1 & b_2 & b_3\end{vmatrix}
=> \vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & -2 & 3\\2 & 3 & -5\end{vmatrix}
=> \vec{a}\times\vec{b} = \hat{i}[(-2)(-5)-(3)(3)]-\hat{j}[(1)(-5)-(2)(3)]+\hat{k}[(1)(3)-(2)(-2)]
=> \vec{a}\times\vec{b} = \hat{i}[10-9]-\hat{j}[-5-6]+\hat{k}[3+4]
=> \vec{a}\times\vec{b} = \hat{i}+11\hat{j}+7\hat{k}
Two vectors are perpendicular if their dot product is zero.
=> (\vec{a}\times\vec{b}).\vec{a} =  (\hat{i}-2\hat{j}+3\hat{k}).(\hat{i}+11\hat{j}+7\hat{k})
=> (\vec{a}\times\vec{b}).\vec{a} = \hat{i}.\hat{i}-2\hat{j}.11\hat{j}+3\hat{k}.7\hat{k}
=> (\vec{a}\times\vec{b}).\vec{a} = 1-22+21
=> (\vec{a}\times\vec{b}).\vec{a} =0
Hence proved.
Question 22. If \vec{p}   and \vec{q}   are unit vectors forming an angle of 30\degree  , find the area of the parallelogram having \vec{a}=\vec{p}+2\vec{q}  and \vec{b}=2\vec{p}+\vec{q}   as its diagonals.
Solution:
Given \vec{p}  and \vec{q}  forming an angle of 30\degree  .
Area of a parallelogram having diagonals \vec{a}   and \vec{b}   is \dfrac{1}{2}|\vec{a}\times\vec{b}|
=> \vec{p}\times\vec{q} = |\vec{p}||\vec{q}|\sin 30\degree \hat{n}
=> \vec{p}\times\vec{q} = 1\times1\times\dfrac{1}{2}\times \hat{n}
=> \vec{p}\times\vec{q} = \dfrac{1}{2} \hat{n}
Thus area is,
=> Area = \dfrac{1}{2}|(\vec{p}+2\vec{q})\times( 2\vec{p}+\vec{q})|
=> Area = \dfrac{1}{2}|\vec{p}\times( 2\vec{p}+\vec{q})+2\vec{q}\times( 2\vec{p}+\vec{q})|
=> Area = \dfrac{1}{2}|\vec{p}\times\vec{q}+4(\vec{q}\times\vec{p})|
=> Area = \dfrac{1}{2}|\vec{p}\times\vec{q}+4(-\vec{p}\times\vec{q})|
=> Area = \dfrac{1}{2}|-3(\vec{p}\times\vec{q})|
=> Area = \dfrac{3}{2}|(\vec{p}\times\vec{q})|
=> Area = \dfrac{3}{2}|\dfrac{1}{2} \hat{n}|
=> Area = \dfrac{3}{2}\times\dfrac{3}{2}\times1
=> Area = \dfrac{3}{4}   square units
Question 23. For any two vectors \vec{a}   and \vec{b}   , prove that |\vec{a}\times\vec{b}|^2 = \begin{vmatrix} \vec{a}.\vec{a}& \vec{a}.\vec{b}\\\vec{b}.\vec{a} & \vec{b}.\vec{b}\end{vmatrix}   
Solution:
We know that,
=> \vec{a}\times\vec{b}= |\vec{a}||\vec{b}|\sin\theta\hat{n}
=> |\vec{a}\times\vec{b}|= |\vec{a}||\vec{b}|\sin\theta|\hat{n}|
=> |\vec{a}\times\vec{b}|= |\vec{a}||\vec{b}|\sin\theta
=> |\vec{a}\times\vec{b}|^2= |\vec{a}|^2|\vec{b}|^2\sin^2\theta
=> |\vec{a}\times\vec{b}|^2= |\vec{a}|^2|\vec{b}|^2(1-\cos^2\theta)
=> |\vec{a}\times\vec{b}|^2= |\vec{a}|^2|\vec{b}|^2 -(|\vec{a}|^2|\vec{b}|^2\cos^2\theta)
=> |\vec{a}\times\vec{b}|^2= |\vec{a}|^2|\vec{b}|^2 -(|\vec{a}|.|\vec{b}|)^2
=> |\vec{a}\times\vec{b}|^2= (\vec{a}.\vec{a})(\vec{b}.\vec{b})-(\vec{a}.\vec{b})(\vec{b}.\vec{a})
=> |\vec{a}\times\vec{b}|^2 = \begin{vmatrix} \vec{a}.\vec{a}& \vec{a}.\vec{b}\\\vec{b}.\vec{a} & \vec{b}.\vec{b}\end{vmatrix}
Hence proved.
Question 24. Define \vec{a}\times\vec{b}   and prove that |\vec{a}\times\vec{b}|=(\vec{a}.\vec{b})\tan \theta  , where \theta   is the angle between \vec{a}  and \vec{b}
Solution:
Definition of \vec{a}\times\vec{b}  : Let \vec{a}   and \vec{b}   be 2 non-zero, non-parallel vectors. Then \vec{a}\times\vec{b}  , is defined as a vector with the magnitude of |\vec{a}||\vec{b}|\sin\theta  , and which is perpendicular to both the vectors \vec{a}   and \vec{b}  .
We know that,
=> \vec{a}\times\vec{b} = |\vec{a}||\vec{b}|\sin\theta\hat{n}
=> |\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta|\hat{n}|
=> |\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta    ……………..(eq.1)
And as,
=> \vec{a}.\vec{b} = |\vec{a}||\vec{b}|\cos\theta
=> |\vec{a}||\vec{b}| = \dfrac{\vec{a}.\vec{b}}{\cos\theta}
Substituting in (eq.1),
=> |\vec{a}\times\vec{b}| = \dfrac{\vec{a}.\vec{b}}{\cos\theta} \sin\theta
=> |\vec{a}\times\vec{b}| = (\vec{a}.\vec{b})\tan\theta
Question 25. If |\vec{a}|=\sqrt{26}  |\vec{b}|= 7   and |\vec{a}\times\vec{b}|=35  , find \vec{a}.\vec{b}
Solution:

We know that,
=> \vec{a}\times\vec{b} = |\vec{a}||\vec{b}|\sin\theta\hat{n}
=> |\vec{a}\times\vec{b} |= |\vec{a}||\vec{b}|\sin\theta\|hat{n}|
=> 35 = \sqrt{26}\times7|\sin\theta|\times1
=> \sin\theta = \dfrac{35}{7\sqrt{26}}
=> \sin\theta = \dfrac{5}{\sqrt{26}}
As \cos^2\theta + \sin^2\theta =1  ,
=> \cos\theta = \sqrt{1-\sin^2\theta}
=> \cos\theta = \sqrt{1-(\dfrac{5}{\sqrt{26}})^2}
=> \cos\theta = \sqrt{1-\dfrac{25}{26}}
=> \cos\theta = \dfrac{1}{\sqrt{26}}
Thus,
=> \vec{a}.\vec{b} = |\vec{a}||\vec{b}|\cos\theta
=> \vec{a}.\vec{b} = 7\sqrt{26}\times\dfrac{1}{\sqrt{26}}
=> \vec{a}.\vec{b} = 7
Question 26. Find the area of the triangle formed by O, A, B when \vec{OA} = \hat{i}+2\hat{j}+3\hat{k}  \vec{OB} = -3\hat{i}-2\hat{j}+\hat{k}
Solution:
The area of a triangle whose adjacent sides are given by \vec{a}   and \vec{b}   is \dfrac{1}{2}|\vec{a}\times\vec{b}|
=> \vec{OA}\times\vec{OB} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_1 & a_2 & a_3\\b_1 & b_2 & b_3\end{vmatrix}
=> \vec{OA}\times\vec{OB} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & 2 & 3\\-3 & -2 & 1\end{vmatrix}
=> \vec{OA}\times\vec{OB} = \hat{i}[2+6]-\hat{j}[1+9]+\hat{k}[-2+6]
=> \vec{OA}\times\vec{OB} = 8\hat{i}-10\hat{j}+4\hat{k}
=> Area = \dfrac{1}{2} |\vec{OA}\times\vec{OB}|
=> Area = \dfrac{1}{2}\sqrt{8^2+(-10^2)+4^2}
=> Area = \dfrac{1}{2}\sqrt{64+100+16}
=> Area = \dfrac{1}{2}\times6\sqrt{45}
=> Area = 3\sqrt{5}   square units.
Question 27. Let \vec{a}=\hat{i}+4\hat{j}+2\hat{k}  \vec{b}=3\hat{i}-2\hat{j}+7\hat{k}   and \vec{c} = 2\hat{i}-\hat{j}+4\hat{k}  . Find a vector \vec{d}   which is perpendicular to both \vec{a}   and \vec{b}  and \vec{c}.\vec{d} = 15
Solution:
Given that \vec{d}   is perpendicular to both \vec{a}   and \vec{b}  .
=> \vec{d}.\vec{a} =0   ……….(1)
=> \vec{d}.\vec{b} =0   ……….(2)
Also,
=> \vec{c}.\vec{d} = 15   …….(3)
Let \vec{d} = d_1\hat{i}+d_2\hat{j}+d_3\hat{k}
From eq(1),
=> d1 + 4d2 + 2d3 = 0 
From eq(2),
=> 3d1 – 2d2 + 7d3 = 0
From eq(3),
=> 2d1 – d2 + 4d3 = 15 
On solving the 3 equations we get,
d= 160/3, d= -5/3, and d= -70/3, 
=> \vec{d} = \dfrac {1}{3}(160\hat{i}-5\hat{j}-70\hat{k})
Question 28. Find a unit vector perpendicular to each of the vectors \vec{a}+\vec{b}   and \vec{a}-\vec{b}  , where \vec{a}=3\hat{i}+2\hat{j}+2\hat{k}   and \vec{b} = \hat{i}+2\hat{j}-2\hat{k}  .
Solution:
Given that, \vec{a}=3\hat{i}+2\hat{j}+2\hat{k}   and \vec{b} = \hat{i}+2\hat{j}-2\hat{k}
Let \vec{c} = \vec{a}+\vec{b}
=> \vec{c} = (3\hat{i}+2\hat{j}+2\hat{k})+ (\hat{i}+2\hat{j}-2\hat{k})
=> \vec{c} = \hat{i}[3+1] +\hat{j}[2+2] +\hat{k}[2-2]
=> \vec{c} = 4\hat{i} +4\hat{j}
Let \vec{d} = \vec{a}-\vec{b}
=> \vec{d} = (3\hat{i}+2\hat{j}+2\hat{k})-(\hat{i}+2\hat{j}-2\hat{k})
=> \vec{d} = \hat{i}[3-1] +\hat{j}[2-2] +\hat{k}[2+2]
=> \vec{d} = 2\hat{i} +4\hat{k}
A vector perpendicular to both \vec{c}   and \vec{d}   is,
=> \vec{c}\times\vec{d} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\c_1 & c_2 & c_3\\d_1 & d_2 & d_3\end{vmatrix}
=> \vec{c}\times\vec{d} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\4 & 4 & 0\\2 & 0 & 4\end{vmatrix}
=> \vec{c}\times\vec{d} = \hat{i}[16-0]-\hat{j}[16-0]+\hat{k}[0-8]
=> \vec{c}\times\vec{d} = 16\hat{i}-16\hat{j}-8\hat{k}
To find the unit vector,
=> \hat{p} = \dfrac{\vec{c}\times\vec{d}}{|\vec{c}\times\vec{d}|}
=> \hat{p} = \dfrac{1}{\sqrt{16^2+(-16)^2+(-8)^2}}(16\hat{i}-16\hat{j}-8\hat{k})
=> \hat{p} = \dfrac{1}{\sqrt{256+256+64}}(16\hat{i}-16\hat{j}-8\hat{k})
=> \hat{p} = \dfrac{1}{24}(16\hat{i}-16\hat{j}-8\hat{k})
=> \hat{p} = \dfrac{1}{3}(2\hat{i}-2\hat{j}-\hat{k})
Question 29. Using vectors, find the area of the triangle with the vertices A(2, 3, 5), B(3, 5, 8), and C(2, 7, 8).
Solution:
Given, A(2, 3, 5), B(3, 5, 8), and C(2, 7, 8)
Let,
=> \vec{a} = A = 2\hat{i}+3\hat{j}+5\hat{k}
=> \vec{b} = B = 3\hat{i}+5\hat{j}+8\hat{k}
=> \vec{c} = C = 2\hat{2}+7\hat{j}+8\hat{k}
Then,
=> \vec{AB} = \vec{b}-\vec{a}
=> \vec{AB} = (3\hat{i}+5\hat{j}+8\hat{k})-(2\hat{i}+3\hat{j}+5\hat{k})
=> \vec{AB} = \hat{i}[3-2]+\hat{j}[5-3]+\hat{k}[8-5]
=> \vec{AB} = \hat{i}+2\hat{j}+3\hat{k}
=> \vec{AC} = \vec{c}-\vec{a}
=> \vec{AC} = (2\hat{2}+7\hat{j}+8\hat{k})-(2\hat{i}+3\hat{j}+5\hat{k})
=> \vec{AC} = \hat{i}[2-2]+\hat{j}[7-3]+\hat{k}[8-5]
=> \vec{AC} = 4\hat{j}+3\hat{k}
The area of a triangle whose adjacent sides are given by \vec{a}   and \vec{b}   is \dfrac{1}{2}|\vec{a}\times\vec{b}|
=> \vec{AB}\times\vec{AC}= \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & 2& 3\\0 & 4 & 3\end{vmatrix}
=> \vec{AB}\times\vec{AC} = \hat{i}[6-12]-\hat{j}[3-0]+\hat{k}[4-0]
=> \vec{AB}\times\vec{AC} = -6\hat{i}-3\hat{j}+4\hat{k}
=> Area = \dfrac{1}{2}|\vec{AB}\times\vec{AC}|
=> Area = \dfrac{1}{2}\sqrt{(-6)^2+(-3)^2+4^2}
=> Area = √61/2
Question 30. If \vec{a}=2\hat{i}-3\hat{j}+\hat{k}  \vec{b}=-\hat{i}+\hat{k}  \vec{c}=2\hat{j}-\hat{k}   are three vectors, find the area of the parallelogram having diagonals (\vec{a}+\vec{b})   and (\vec{b}+\vec{c})  .
Solution:
Given, \vec{a}=2\hat{i}-3\hat{j}+\hat{k}  \vec{b}=-\hat{i}+\hat{k}  \vec{c}=2\hat{j}-\hat{k}
Let,
=> \vec{c} = (\vec{a}+\vec{b})
=> \vec{c} = (2\hat{i}-3\hat{j}+\hat{k})+(-\hat{i}+\hat{k})
=> \vec{c} = (2-1)\hat{i}+(-3)\hat{j}+(1+1)\hat{k}
=> \vec{c} = \hat{i}-3\hat{j}+2\hat{k}
=> \vec{d} = (\vec{b}+\vec{c})
=> \vec{d} = (-\hat{i}+\hat{k})+(2\hat{j}-\hat{k})
=> \vec{d} = -\hat{i}+2\hat{j}
The area of the parallelogram having diagonals \vec{c}   and \vec{d}   is \dfrac{1}{2}|\vec{c}\times\vec{d}|
=> \vec{c}\times\vec{d} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & -3& 2\\-1 & 2 & 0\end{vmatrix}
=> \vec{c}\times\vec{d} = \hat{i}[0-4] -\hat{j}[0+2]+\hat{k}[2-3]
=> \vec{c}\times\vec{d} = -4\hat{i}-2\hat{j}-\hat{k}
=> Area = \dfrac{1}{2}|\vec{c}\times\vec{d}|
=> Area = \dfrac{1}{2}\sqrt{(-4)^2+(-2)^2+(-1)^2}
=> Area = \dfrac{1}{2}\sqrt{21}
=> Area = √21/2
Question 31. The two adjacent sides of a parallelogram are 2\hat{i}-4\hat{j}+5\hat{k}  and \hat{i}-2\hat{j}-3\hat{k}  . Find the unit vector parallel to one of its diagonals. Also, find its area.
Solution:
Given a parallelogram ABCD and its 2 sides AB and BC.
By triangle law of addition,
=> \vec{AC} = \vec{AB}+\vec{BC}
=> \vec{AC} = (2\hat{i}-4\hat{j}+5\hat{k})+(\hat{i}-2\hat{j}-3\hat{k})
=> \vec{AC} = \hat{i}[2+1] +\hat{j}[-4-2]+\hat{k}[5-3]
=> \vec{AC} = 3\hat{i}-6\hat{j}+2\hat{k}
Unit vector is,
=> \hat{p} = \dfrac{\vec{AC}}{|\vec{AC}|}
=> \hat{p} = \dfrac{1}{\sqrt{3^2+(-6)^2+2^2}}(3\hat{i}-6\hat{j}+2\hat{k})
=> \hat{p} = \dfrac{1}{\sqrt{49}}(3\hat{i}-6\hat{j}+2\hat{k})
=> \hat{p} = \dfrac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k})
Area of a parallelogram whose adjacent sides are given is |\vec{a}\times\vec{b}|
=> |\vec{AB}\times\vec{BC}| = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2 & -4 & 5\\1 & -2 & -3\end{vmatrix}
=> |\vec{AB}\times\vec{BC}| = \hat{i}[12+10]-\hat{j}[-6-5]+\hat{k}[-4+4]
=> |\vec{AB}\times\vec{BC}| = 22\hat{i}+11\hat{j}
Thus area is,
=> Area = |22\hat{i}+11\hat{j}|
=> Area = \sqrt{22^2+11^2}
=> Area = \sqrt{605}
=> Area = 11 √5 square units
Question 32. If either \vec{a}=0  or \vec{b}=0  , then \vec{a}\times\vec{b}=\vec{0}  . Is the converse true? Justify with example.
Solution:
Let us take two parallel non-zero vectors \vec{a}   and \vec{b}
=> \vec{a}\times\vec{b} = \vec{0}
For example,
\vec{a} = \hat{i}   and \vec{b}=2\hat{i}
=> \vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & 0 & 0\\2 & 0 & 0\end{vmatrix}
=> \vec{a}\times\vec{b} = 0
But,
=> |\vec{a}| = \sqrt{1^2} =1
=> |\vec{b}| = \sqrt{2^2} =2
Hence the converse may not be true.
Question 33. If \vec{a} = a_1\hat{i}+a_2\hat{j}+a_3\hat{k}  \vec{b} = b_1\hat{i}+b_2\hat{j}+b_3\hat{k}  and \vec{c} = c_1\hat{i}+c_2\hat{j}+c_3\hat{k}  , then verify that \vec{a}\times(\vec{b}\times\vec{c})=\vec{a}\times\vec{b}+\vec{a}\times\vec{c}  .
Solution:
Given, \vec{a} = a_1\hat{i}+a_2\hat{j}+a_3\hat{k}  \vec{b} = b_1\hat{i}+b_2\hat{j}+b_3\hat{k}  and \vec{c} = c_1\hat{i}+c_2\hat{j}+c_3\hat{k}
=> (\vec{b}+\vec{c}) = (b_1+c_1)\hat{i}+(b_2+c_2)\hat{j}+(b_3+c_3)\hat{k}
=> \vec{a}\times(\vec{b}\times\vec{c}) = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_1 & a_2 & a_3\\(b_1+c_1) & (b_2+c_2) & (b_3+c_3)\end{vmatrix}
=> \vec{a}\times(\vec{b}\times\vec{c}) = \hat{i}[a_2(b_3+c_3)-a_3(b_2+c_2)]-\hat{j}[a_1(b_3+c_3)-a_3(b_1+c_1)]+\hat{k}[a_1(b_2+c_2)-a_2(b_1+c_1)]
=> \vec{a}\times(\vec{b}\times\vec{c}) = \hat{i}[a_2b_3+a_2c_3-a_3b_2-a_3c_2]+\hat{j}[-a_1b_3-a_1c_3+a_3b_1+a_3c_1]+\hat{k}[a_1b_2+a_1c_2-a_2b_1-a_2c_1]   …..eq(1)
Now,
=> \vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_1 & a_2 & a_3\\b_1 & b_2 & b_3\end{vmatrix}
=> \vec{a}\times\vec{b} = \hat{i}[a_2b_3-b_2a_3]-\hat{j}[a_1b_3-b_1a_3]+\hat{k}[a_1b_2-b_1a_2]
And,
=> \vec{a}\times\vec{c} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_1 & a_2 & a_3\\c_1 & c_2 & c_3\end{vmatrix}
=> \vec{a}\times\vec{c} = \hat{i}[a_2b_3-c_2a_3]-\hat{j}[a_1c_3-c_1a_3]+\hat{k}[a_1c_2-c_1a_2]
Thus, 
=> \vec{a}\times\vec{b}+\vec{a}\times\vec{c} = (\hat{i}[a_2b_3-b_2a_3]-\hat{j}[a_1b_3-b_1a_3]+\hat{k}[a_1b_2-b_1a_2]) + (\hat{i}[a_2b_3-c_2a_3]-\hat{j}[a_1c_3-c_1a_3]+\hat{k}[a_1c_2-c_1a_2])
=> \vec{a}\times\vec{b}+\vec{a}\times\vec{c} = \hat{i}[a_2b_3+a_2c_3-a_3b_2-a_3c_2]+\hat{j}[-a_1b_3-a_1c_3+a_3b_1+a_3c_1]+\hat{k}[a_1b_2+a_1c_2-a_2b_1-a_2c_1]   …eq(2)
Thus eq(1) = eq(2)
Hence proved.
Question 34(i). Using vectors find the area of the triangle with the vertices A(1, 1, 2), B(2, 3, 5), and C(1, 5, 5).
Solution:
Given, A(1, 1, 2), B(2, 3, 5), and C(1, 5, 5)
=> \vec{a} = A = \hat{i}+\hat{j}+2\hat{k}
=> \vec{b} = B = 2\hat{i}+3\hat{j}+5\hat{k}
=> \vec{c} = C = \hat{i}+5\hat{j}+5\hat{k}
Now 2 sides of the triangle are given by,
=> \vec{AB} = \vec{b}-\vec{a}
=> \vec{AB} = (2\hat{i}+3\hat{j}+5\hat{k})-(\hat{i}+\hat{j}+2\hat{k})
=> \vec{AB} = \hat{i}[2-1] +\hat{j}[3-1]+\hat{j}[5-2]
=> \vec{AB} = \hat{i}+2\hat{j}+3\hat{k}
=> \vec{AC} = \vec{c}-\vec{a}
=> \vec{AC} = (\hat{i}+5\hat{j}+5\hat{k})-(\hat{i}+\hat{j}+2\hat{k})
=> \vec{AC} = \hat{i}[1-1] +\hat{j}[5-1]+\hat{j}[5-2]
=> \vec{AC} = 4\hat{j}+3\hat{k}
Area of the triangle whose adjacent sides are given is \dfrac{1}{2}|\vec{a}\times\vec{b}|
=> \vec{AB}\times\vec{AC} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & 2 & 3\\0 & 4 & 3\end{vmatrix}
=> \vec{AB}\times\vec{AC} = \hat{i}[6-12]-\hat{j}[3-0]+\hat{k}[4-0]
=> \vec{AB}\times\vec{AC} = -6\hat{i}-3\hat{j}+4\hat{k}
Thus area of the triangle is,
=> Area = \dfrac{1}{2}\sqrt{(-6)^2+(-3)^2+4^2}
=> Area = \dfrac{1}{2}\sqrt{36+9+16}
=> Area = √61/2
Question 34(ii). Using vectors find the area of the triangle with the vertices A(1, 2, 3), B(2, -1, 4), and C(4, 5, -1).
Solution:
Given, A(1, 2, 3), B(2, -1, 4), and C(4, 5, -1)
=> \vec{a} = A = \hat{i}+2\hat{j}+3\hat{k}
=> \vec{b} = B = 2\hat{i}-1\hat{j}+4\hat{k}
=> \vec{c} = C = 4\hat{i}+5\hat{j}-1\hat{k}
Now 2 sides of the triangle are given by,
=> \vec{AB} = \vec{b}-\vec{a}
=> \vec{AB} = (2\hat{i}-1\hat{j}+4\hat{k})-(4\hat{i}+5\hat{j}-1\hat{k})
=> \vec{AB} = \hat{i}[2-1] +\hat{j}[-1-2]+\hat{j}[4-3]
=> \vec{AB} = \hat{i}-3\hat{j}+\hat{k}
=> \vec{AC} = \vec{c}-\vec{a}
=> \vec{AC} = (4\hat{i}+5\hat{j}-1\hat{k})-(\hat{i}+2\hat{j}+3\hat{k})
=> \vec{AC} = \hat{i}[4-1] +\hat{j}[5-2]+\hat{j}[-1-3]
=> \vec{AC} = 3\hat{i}+3\hat{j}-4\hat{k}
Area of the triangle whose adjacent sides are given is \dfrac{1}{2}|\vec{a}\times\vec{b}|
=> \vec{AB}\times\vec{AC} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & -3 & 1\\3 & 3 & -4\end{vmatrix}
=> \vec{AB}\times\vec{AC} = \hat{i}[12-3]-\hat{j}[-4-3]+\hat{k}[3+9]
=> \vec{AB}\times\vec{AC} = 9\hat{i}+7\hat{j}+12\hat{k}
Thus area of the triangle is,
=> Area = \dfrac{1}{2}\sqrt{(9)^2+(7)^2+12^2}
=> Area = \dfrac{1}{2}\sqrt{81+49+144}
=> Area = √274/2
Question 35. Find all the vectors of magnitude 10\sqrt{3}   that are perpendicular to the plane of \hat{i}+2\hat{j}+\hat{k}  and -\hat{i}+3\hat{j}+4\hat{k}  .
Solution:
Given, \vec{a} = \hat{i}+2\hat{j}+\hat{k}  and \vec{b}=\hat{i}+3\hat{j}+4\hat{k}
A vector perpendicular to both \vec{a}   and \vec{b}   is,
=> \vec{a}\times\vec{b} =  \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & 2 & 1\\-1 & 3 & 4\end{vmatrix}
=> \vec{a}\times\vec{b} = \hat{i}[8-3]-\hat{j}[4+1]+\hat{k}[3+2]
=> \vec{a}\times\vec{b} = 5\hat{i}-5\hat{j}+5\hat{k}
Unit vector is,
=> \hat{p} = \dfrac{\vec{a}\times\vec{b}}{|\vec{a}\times\vec{b}|}
=> \hat{p} = \dfrac{1}{\sqrt{5^2+(-5)^2+5^2}}(5\hat{i}-5\hat{j}+5\hat{k})
=> \hat{p} = \dfrac{1}{5\sqrt{3}}(5\hat{i}-5\hat{j}+5\hat{k})
=> \hat{p} = \dfrac{1}{\sqrt{3}}(\hat{i}-\hat{j}+\hat{k})
Now vectors of magnitude 10\sqrt{3}   are given by,
=> 10\sqrt{3}\hat{p} = \pm10\sqrt{3}\times \dfrac{1}{\sqrt{3}}(\hat{i}-\hat{j}+\hat{k})
=> Required vectors, \pm10(\hat{i}-\hat{j}+\hat{k})
Question 36. The adjacent sides of a parallelogram are 2\hat{i}-4\hat{j}-5\hat{k}  and 2\hat{i}+2\hat{j}+3\hat{k}  . Find the 2 unit vectors parallel to its diagonals. Also, find its area of the parallelogram.
Solution:
Given, \vec{AB}=2\hat{i}-4\hat{j}-5\hat{k}   and \vec{BC} = 2\hat{i}+2\hat{j}+3\hat{k}
=> \vec{AC} = \vec{AB} + \vec{BC}
=> \vec{AC} = (2\hat{i}-4\hat{j}-5\hat{k})+(2\hat{i}+2\hat{j}+3\hat{k})
=> \vec{AC} = 4\hat{i}-\hat{j}-2\hat{k}
Unit vector is,
=> \hat{p} = \dfrac{\vec{AC}}{|\vec{AC}|}
=> \hat{p} = \dfrac{1}{\sqrt{4^2+(-1)^2+(-2)^2}}(4\hat{i}-\hat{j}-2\hat{k})
=> \hat{p} = \dfrac{1}{\sqrt{21}}(4\hat{i}-\hat{j}-2\hat{k})
Area is given by |\vec{AB}\times\vec{BC}|  ,

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