RD Sharma Class 12 Ex 24.2 Solutions Chapter 24 Scalar or Dot Product

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter24
Exercise24.2
CategoryRD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 24.2 Solutions Chapter 24 Scalar or Dot Product

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

Question 1. Find \vec{a}.\vec{b} when

(i) \vec{a} = \hat{i}-2 \hat{j}+ \hat{k} and \vec{b} = 4 \hat{i} -4\hat{j} +7\hat{k}

Solution:

 = (\hat{i}-2 \hat{j}+ \hat{k})(\hat{i}-2 \hat{j}+ \hat{k})

= (1)(4) + (-2)(-4) + (1)(7)

= 4 + 8 + 7

= 19

(ii) \vec{a} = \hat{j}+2 \hat{k} and \vec{b} = 2\hat{i}+\hat{k}

Solution:

(\hat{j}+2 \hat{k} )(2\hat{i}+\hat{k})

= (0)(2) + (1)(0) + (2)(1)

= 2

(iii)  \vec{a} = \hat{j}-\hat{k}  and  \vec{b} = 2\hat{i}+3\hat{j}-2 \hat{k}

Solution:

(\hat{j}-\hat{k})(2\hat{i}+3\hat{j}-2 \hat{k} )

= (0)(2) + (1)(3) + (-1)(-2)

= 0 + 3 + 2

= 5

Question 2. For what value of λ are the vector \vec{a}  and \vec{b}  perpendicular to each other? where:

(i) \vec{a} =  λ\hat{i}+2\hat{j}+\hat{k}  and \vec{b} =4\hat{i}-9\hat{j}+2\hat{k}

Solution:

\vec{a}  and \vec{b}  are perpendicular to each other

So \vec{a} . \vec{b} = 0

(λ\hat{i}+2\hat{j}+\hat{k})(4\hat{i}-9\hat{j}+2\hat{k}) = 0         

⇒ λ(4) + (2)(-9) + (1)(2) = 0

⇒ 4λ – 18 + 2 = 0

⇒ 4λ = 16

⇒ λ = 4

(ii) \vec{a} = λ\hat{i}+2\hat{j}+\hat{k}  and \vec{b} =5\hat{i}-9\hat{j}+2\hat{k}

Solution:

\vec{a}  and \vec{b}  are perpendicular to each other

so \vec{a} . \vec{b}  = 0 

⇒ (λ\hat{i}+2\hat{j}+\hat{k})(5\hat{i}-9\hat{j}+2\hat{k})

⇒ λ(5) + (2)(-9) + (1)(2) = 0

⇒ 5λ – 18 + 2 = 0

⇒ 5λ = 16

⇒ λ = 16/5

(iii) \vec{a} = 2\hat{i}+3\hat{j}+4\hat{k}  and \vec{b} =3\hat{i}+2\hat{j}-λ\hat{k}

Solution:

\vec{a}  and \vec{b}  are perpendicular to each other

so \vec{a} . \vec{b}   = 0  

 (2\hat{i}+3\hat{j}+4\hat{k})(3\hat{i}+2\hat{j}-λ\hat{k})   =0

⇒ (2)(3) + (3)(2) – (4)λ = 0

⇒ 6 + 6 – 4λ = 0

⇒ 4λ = 12

⇒ λ = 3

(iv) \vec{a} = λ\hat{i}+3\hat{j}+2\hat{k}  and \vec{b} =\hat{i}-\hat{j}+3\hat{k}

Solution:

\vec{a}  and \vec{b}  are perpendicular to each other

so \vec{a} . \vec{b}=0 

⇒ (λ\hat{i}+3\hat{j}+2\hat{k})(\hat{i}-\hat{j}+3\hat{k})=0  

⇒ λ(1) + (3)(-1) + (2)(3) = 0

⇒ λ – 3 + 6 = 0

⇒ λ = 3

Question 3. If \vec{a}  and  \vec{b}  are two vectors such that |\vec{a}  |=4, |\vec{b}  | = 3 and \vec{a}.\vec{b} = 6. Find the angle between  \vec{a} and  \vec{b}

Solution:

Let the angle be θ 

cos θ = \frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}

= 6 /(4×3) = 1/2

Therefore, θ = cos-1(1/2)

= π/3

Question 4. If \vec{a} = \hat{i}-\hat{j}  and  \vec{b} =-\hat{j}+2\hat{k}, find (\vec{a}-2\vec{b}). (\vec{a}+\vec{b})

Solution:

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

= \hat{i}-\hat{j}+2\hat{j}-4\hat{k}

= \hat{i}+\hat{j}-4\hat{k}         

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

\hat{i}-\hat{j}-\hat{j}+2\hat{k}

\hat{i}-2\hat{j}+2\hat{k}

Now, (\vec{a}-2\vec{b}).(\vec{a}+\vec{b})

(\hat{i}+\hat{j}-4\hat{k})(\hat{i}-2\hat{j}+2\hat{k})

= (1)(1) + (1)(-2) + (-4)(2)

= 1 – 2 – 8

= -9

Therefore, (\vec{a}-2\vec{b}).(\vec{a}+\vec{b}) = -9

Question 5. Find the angle between the vectors \vec{a}  and \vec{b}  where :

(i) \vec{a} = \hat{i}-\hat{j}  and \vec{b} = \hat{j}+\hat{k}

Solution:

 Let the angle be θ between \vec{a}  and \vec{b}

cos θ = \frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}

Now, \vec{a} . \vec{b}

(\hat{i}-\hat{j})(\hat{j}+\hat{k})

= (1)(0) + (-1)(1) + (0)(1)

= 0 – 1 + 0 = -1

|\vec{a}  |= |\hat{i}-\hat{j}  |

\sqrt{(1)^2+(-1)^2}

= √2

|\vec{b}| = |\hat{j}+\hat{k}|

\sqrt{(1)^2+(1)^2}

= √2

Now, cos θ = -1/(√2×√2)

= -1/2

θ = cos-1(-1/2)

= 2π/3

(ii) \vec{a} =3\hat{i}-2\hat{j}-6\hat{k} and \vec{b} =4\hat{i}-\hat{j}+8\hat{k}

Solution:

Let the angle be θ between  \vec{a} and \vec{b}

Now,  \vec{a} . \vec{b}

=(3\hat{i}-2\hat{j}-6\hat{k})(4\hat{i}-\hat{j}+8\hat{k})

=(3)(4) + (-2)(-1) + (-6)(8)

= 12 + 2 – 48

= -34

|\vec{a}| = |3\hat{i}-2\hat{j}-6\hat{k}|

\sqrt{(3)^2+(-2)^2+(-6)^2}

= √49 = 7

|\vec{b}| = |4\hat{i}-\hat{j}+8\hat{k}|

\sqrt{(4)^2+(-1)^2+(8)^2}

= √81 = 9

cos θ = \frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}

Now, cos θ = -34/(7×9)

= -34/63

θ = cos-1(-34/63)

(iii) \vec{a} =2\hat{i}-\hat{j}+2\hat{k} and \vec{b} =4\hat{i}+4\hat{j}-2\hat{k}

Solution:

Let the angle be θ between \vec{a} and \vec{b}

Now,  \vec{a} . \vec{b}

=(2\hat{i}-\hat{j}+2\hat{k})(4\hat{i}+4\hat{j}-2\hat{k})

= (2)(4) + (-1)(4) + (2)(-2)

= 8 – 4 – 4 = 0

|\vec{a}| = |2\hat{i}-\hat{j}+2\hat{k}|

\sqrt{(2)^2+(-1)^2+(2)^2}

= √9 = 3

|\vec{b}| = |4\hat{i}+4\hat{j}-2\hat{k}|

= \sqrt{(4)^2+(4)^2+(-2)^2}

= √36 = 6

Now, cos θ = \frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}          

cos θ = 0/(3×6) = 0

θ = cos-1(0)

θ = π/2

(iv) \vec{a} =2\hat{i}-3\hat{j}+\hat{k} and \vec{b} =\hat{i}+\hat{j}-2\hat{k}

Solution:

Let the angle be θ between \vec{a} and \vec{b}

Now, \vec{a} . \vec{b}

=(2\hat{i}-3\hat{j}+\hat{k})(\hat{i}+\hat{j}-2\hat{k})

= (2)(1) + (-3)(1) + (1)(-2)

= 2 – 3 – 2

= -3

|\vec{a}| = |2\hat{i}-3\hat{j}+\hat{k}|

= \sqrt{(2)^2+(-3)^2+(-1)^2}

= √14 

|\vec{b}| =|\hat{i}+\hat{j}-2\hat{k}|

= \sqrt{(1)^2+(1)^2+(-2)^2}

= √6

cos θ =  \frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}

Now, cos θ = -3/(√14×√6)

= -3/√84

θ = cos-1(-3/√84)

(v) \vec{a} =\hat{i}+2\hat{j}-\hat{k} and \vec{b} =\hat{i}-\hat{j}+\hat{k}

Solution:

Let the angle be θ between \vec{a} and \vec{b}

Now, \vec{a} . \vec{b}

=(\hat{i}+2\hat{j}-\hat{k})(\hat{i}-\hat{j}+\hat{k})

= (1)(1) + (2)(-1) + (-1)(1)

= 1 – 2 – 1

= -2

|\vec{a}| = |\hat{i}+2\hat{j}-\hat{k}|

= \sqrt{(1)^2+(2)^2+(-1)^2}

= √6

|\vec{b}| = |\hat{i}-\hat{j}+\hat{k}|

= \sqrt{(1)^2+(-1)^2+(1)^2}

= √3 

cos θ = \frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}

Now, cos θ = -2/(√6×√3)

= -2/√18

= -2/3√2

θ = cos-1(-√2 /3)

Question 6. Find the angles which the vectors \vec{a} =\hat{i}-\hat{j}+\sqrt2\hat{k} makes with the coordinate axes.

Solution:

Components along x, y and z axis are  \hat{i},\hat{j} and \hat{k} respectively.

Let the angle between  \vec{a} and  \hat{i} be θ1

Now, \vec{a} . \hat{i}

(\hat{i}-\hat{j}-\sqrt2\hat{k})(\hat{i}-0\hat{j}+0\hat{k})

= (1)(1) + (-1)(0) + (√2)(0) 

= 1

|\vec{a}| = |\hat{i}-\hat{j}+\sqrt2\hat{k}|

\sqrt{(1)^2+(-1)^2+(√2)^2}

= √4 = 2

|\hat{i}| = |\hat{i}+0\hat{j}+0\hat{k}|

= √1 = 1

cos θ1 = \frac{\vec{a}.\hat{i}}{|\vec{a}||\hat{i}|}

Now, cos θ1 = 1/(2×1)

= 1/2

θ1 = cos-1(1/2) = π/3

Let the angle between \vec{a} and  \hat{j} be θ2

Now, \vec{a} . \hat{j}

= (\hat{i}-\hat{j}+\sqrt2\hat{k})(0\hat{i}+\hat{j}+0\hat{k})

= (1)(0) + (-1)(1) + (√2)(0)  

= -1

|\hat{j}| = |0\hat{i}+\hat{j}+0\hat{k}|

= √1 = 1

cos θ2 = \frac{\vec{a}.\hat{j}}{|\vec{a}||\hat{j}|}

Now, cos θ2 = -1/(2×1)

= -1/2

θ2 = cos-1(-1/2) = 2π/3

 Let the angle between \vec{a} and \hat{k} be θ3

Now, \vec{a} . \hat{k}

= (\hat{i}-\hat{j}+\sqrt2\hat{k})(0\hat{i}+0\hat{j}+\hat{k})

= (1)(0) + (-1)(0) + (√2)(1)  

= √2

|\hat{k}| = |0\hat{i}+0\hat{j}+\hat{k}|

= √1 = 1

cos θ3 =  \frac{\vec{a}.\hat{k}}{|\vec{a}||\hat{k}|}

= 1/(√2)

= cos-1(1/√2) = π/4

Question 7(i). Dot product of a vector with \hat{i}+\hat{j}-3\hat{k}, \hat{i}+3\hat{j}-2\hat{k} and 2\hat{i}+\hat{j}+4\hat{k} are 0, 5 and 8respectively. Find the vector.

Solution:

Let \vec{a} =\hat{i}+\hat{j}-3\hat{k}, \vec{b} =\hat{i}+3\hat{j}-2\hat{k} and  \vec{c}=2\hat{i}+\hat{j}+4\hat{k} be three given vectors.

Let \vec{r}=x\hat{i}+y\hat{j}+j\hat{k} be a vector such that its dot products with  \vec{a},  \vec{b}, and \vec{c} are 0, 5 and 8 respectively. Then, 

 \vec{r}. \vec{a} = 0

⇒ (x\hat{i}+y\hat{j}+j\hat{k})(\hat{i}+\hat{j}-3\hat{k})       = 0

⇒ x + y – 3z = 0        ….(1)

 \vec{r}. \vec{b} = 5

 (x\hat{i}+y\hat{j}+j\hat{k})(\hat{i}+3\hat{j}-2\hat{k})       = 5

⇒ x + 3y – 2z = 5     …..(2)

 \vec{r}. \vec{c} = 8

⇒ (x\hat{i}+y\hat{j}+j\hat{k})(2\hat{i}+\hat{j}+4\hat{k})       = 8

⇒ 2x + y + 4z = 8    …..(3)

Solving 1,2 and 3 we get x = 1, y = 2 and z = 1,

Hence, the required vector is \vec{r}=\hat{i}+2\hat{j}+\hat{k}

Question 8. If  \vec{a} and  \vec{b} are unit vectors inclined at an angle θ then prove that 

(i) cos θ/2 = 1/2|\hat{a}+\hat{b}|

Solution:

|\hat{a}| = |\hat{b}| = 1

|\hat{a}+\hat{b}|2 =(\hat{a}+\hat{b})2 

(\hat{a})^2+(\hat{b})^2+2\hat{a}.\hat{b}

= 1 + 1 + 2\hat{a}.\hat{b}

= 2 + 2|\hat{a}||\hat{b}|cos θ 

= 2(1 + (1)(1)cos θ)

= 2(2cos2 θ/2)

|\hat{a}+\hat{b}|= 4cos2 θ/2

\hat{a}+\hat{b} = 2 cos θ/2

cos θ/2 = 1/2|\hat{a}+\hat{b}|

(ii) tan θ/2 = \frac{|\hat{a}-\hat{b}|}{|\hat{a}+\hat{b}|}

Solution:

|\hat{a}| = |\hat{b}| = 1

\frac{|\hat{a}-\hat{b}|^2}{|\hat{a}+\hat{b}|^2}\frac{(\hat{a}-\hat{b})^2}{(\hat{a}+\hat{b})^2}

=\frac{(\hat{a})^2+(\hat{b})^2-2\hat{a}.\hat{b}}{(\hat{a})^2+(\hat{b})^2+2\hat{a}.\hat{b}}

\frac{2-2|\hat{a}||\hat{b}|cos θ }{2+2|\hat{a}||\hat{b}|cos θ }

=\frac{2(1-cos θ)}{2(1+cos θ)}

\frac{2sin^2 θ/2}{2cos^2 θ/2}

= tan2 θ/2

Therefore, tan θ/2 =\frac{|\hat{a}-\hat{b}|}{|\hat{a}+\hat{b}|}

Question 9. If the sum of two unit vectors is a unit vector prove that the magnitude of their difference is √3.

Solution:

Let \hat{a} and \hat{b} be two unit vectors

Then,  |\hat{a}| = |\hat{b}| = 1

According to question:

|\hat{a}+\hat{b}| = 1

Taking square on both sides

|\hat{a}+\hat{b}|^2 = (1)^2

(\hat{a})^2+(\hat{b})^2+2\hat{a}.\hat{b} = 1

⇒ (1)2+(1)2+2\hat{a}.\hat{b} = 1

⇒ 2+ 2\hat{a}.\hat{b} = 1

⇒ 2\hat{a}.\hat{b}= -1

⇒ \hat{a}.\hat{b} =-1/2

Now, |\hat{a}-\hat{b}|^2 = (\hat{a}-\hat{b})^2

(\hat{a})^2 + (\hat{b})^2 - 2\hat{a}.\hat{b}

= (1)2 + (1)2  – 2 (-1/2)

= 2 + 1 = 3

Therefore, |\hat{a}-\hat{b}|^2 = 3 

|\hat{a}-\hat{b}|=√3

Question 10. If \vec{a},\vec{b},\vec{c}  are three mutually perpendicular unit vectors, then prove that |\vec{a}+\vec{b}+\vec{c} | =√3.

Solution:

Given \vec{a},\vec{b},\vec{c}   are mutually perpendicular so,

\vec{a}.\vec{b}=\vec{b}.\vec{c}=\vec{c}.\vec{a} = 0
|\vec{a}| = |\vec{b}| = |\vec{c}|=1

Now, 

|\vec{a}+\vec{b}+\vec{c}|^2   = (\vec{a}+\vec{b}+\vec{c})^2

=(\vec{a})^2+(\vec{b})^2+(\vec{c})^2+2\vec{a}\vec{b}+2\vec{b}\vec{c}+2\vec{c}\vec{a}

= (1)2 + (1)2 +(1)2 + 0

= 3

 |\vec{a}+\vec{b}+\vec{c}|   = √3

Question 11. If |\vec{a}+\vec{b}| = 60, |\vec{a}-\vec{b}| = 40 and |\vec{b}|= 46, find |\vec{a}|

Solution:

Given |\vec{a}+\vec{b}|=60, |\vec{a}-\vec{b}| = 40 and  |\vec{b}|= 46

We know that, 

(a + b)+ (a – b)2 = 2(a2 + b2)

⇒ |\vec{a}+\vec{b}|^2+|\vec{a}-\vec{b}|^2 = 2(|\vec{a}|^2+|\vec{b}|^2)

⇒ 602 + 402 = 2(|\vec{a}| 2 + 492)

⇒ 3600 + 1600 = 2|\vec{a}|^2   + 2401 

⇒ 2|\vec{a}|   = 968

⇒ |\vec{a}|   = √484 =22

Question 12. Show that the vector \hat{i}+\hat{j}+\hat{k} is equally inclined with the coordinate axes. 

Solution:

Let \vec{a} = \hat{i}+\hat{j}+\hat{k}

|\vec{a}| =√(1+1+1) = √3

Let θ1, θ2, θbe the angle between the coordinate axes and the \vec{a}

cos θ1 = \frac{\vec{a}.\hat{i}}{|\vec{a}||\hat{i}|}

= 1/√3

cos θ2 = \frac{\vec{a}.\hat{j}}{|\vec{a}||\hat{j}|}

= 1/√3

cos θ3 = \frac{\vec{a}.\hat{k}}{|\vec{a}||\hat{k}|}

= 1/√3

Since, cos θ= cos θ= cos θ3 

Therefore, Given vector is equally inclined with coordinate axis.

Question 13. Show that the vectors \vec{a}=\frac{1}{7}(2\hat{i}+3\hat{j}+6\hat{k}), \vec{b}=\frac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k}),\vec{c}=\frac{1}{7}(6\hat{i}+2\hat{j}-3\hat{k}) are mutually perpendicular unit vectors.

Solution: 

Given, \vec{a}=\frac{1}{7}(2\hat{i}+3\hat{j}+6\hat{k})

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

|\vec{a}|   = (1/7)√(2+ 3+ 62) = (1/7)(√49) = 1

|\vec{b}|   = (1/7)√(3+ (-6)+ 22) = (1/7)(√49) = 1

|\vec{c}|   = (1/7)√(6+ 2+ (-3)2) = (1/7)(√49) = 1

Now, \vec{a}.\vec{b} =   1/49[3 × 2 – 3 × 6 + 6 × 2]

= 1/49[6 – 18 + 12] = 0 

\vec{b}.\vec{c} =   1/49[3 × 6 – 6 × 2 – 2 × 3]

= 1/49[18 – 12 – 6] = 0

Since, \vec{a}.\vec{b} = \vec{b}.\vec{c}=0 they are mutually perpendicular unit vectors.

Question 14. For any two vectors \vec{a}  and \vec{b}  , Show that (\vec{a}+\vec{b}).(\vec{a}-\vec{b})=0\Leftrightarrow|\vec{a}|=|\vec{b}|.

Solution:

To prove (\vec{a}+\vec{b}).(\vec{a}-\vec{b})=0\Leftrightarrow|\vec{a}|=|\vec{b}|

(\vec{a}+\vec{b}).(\vec{a}-\vec{b})=0

|\vec{a}|^2-|\vec{b}|^2=0

|\vec{a}|=|\vec{b}|    

Hence Proved

Question 15. If \vec{a}=2\hat{i}-\hat{j}+\hat{k} \vec{b}=\hat{i}+\hat{j}-2\hat{k} and \vec{c}=\hat{i}+3\hat{j}-\hat{k} , find such that \vec{a} is perpendicular to λ\vec{b}+\vec{c} .

Solution:

Given: \vec{a}=2\hat{i}-\hat{j}+\hat{k}

\vec{b}=\hat{i}+\hat{j}-2\hat{k}
\vec{c}=\hat{i}+3\hat{j}-\hat{k}

According to question

\vec{a}(λ\vec{b}+\vec{c})=0

⇒ (2\hat{i}-\hat{j}+\hat{k})[λ(\hat{i}+\hat{j}-2\hat{k})+(\hat{i}+3\hat{j}-\hat{k})]=0

⇒ (2\hat{i}-\hat{j}+\hat{k})(λ\hat{i}+λ\hat{j}-2λ\hat{k}+\hat{i}+3\hat{j}-\hat{k})=0

⇒ 2(λ+1) – (λ+3) -2λ-1 = 0

⇒ 2λ + 2 -λ – 3 – 2λ – 1 = 0

⇒ -λ = 2

⇒ λ = -2

Question 16. If \vec{p}=5\hat{i}+λ\hat{j}-3\hat{k} and \vec{q}=\hat{i}+3\hat{j}-5\hat{k} , then find the value of λ so that \vec{p}+\vec{q}  and \vec{p}-\vec{q}  are perpendicular vectors.

Solution:

Given, \vec{p}=5\hat{i}+λ\hat{j}-3\hat{k}   

\vec{q}=\hat{i}+3\hat{j}-5\hat{k}

According to question

(\vec{p}+\vec{q})(\vec{p}-\vec{q})=0

|\vec{p}|^2-|\vec{q}|^2=0

⇒ |\vec{p}|^2=|\vec{q}|^2

⇒ \sqrt{5^2+λ^2+(-3)^2}=\sqrt{1^2+3^2+(-5)^2}

⇒ 25 + λ+ 9 = 1 + 9 + 25

⇒ λ2 = 1

⇒ λ = 1

Question 17. If \vec{α}=3\hat{i}+4\hat{j}+5\hat{k}   and \vec{β}=2\hat{i}+\hat{j}-4\hat{k}  , then express \vec{β}    in the form of \vec{β}=\vec{β_1}+\vec{β_2}   where \vec{β_1}   is parallel to \vec{α}   and \vec{β_2}   is perpendicular to \vec{α}   .  

Solution:

Given, \vec{α}=3\hat{i}+4\hat{j}+5\hat{k}

\vec{β}=2\hat{i}+\hat{j}-4\hat{k}

According to question

\vec{β_1} = λ\vec{α}    also \vec{β_2}.\vec{α}    = 0

Now,

\vec{β_1} = λ(3\hat{i}+4\hat{j}+5\hat{k})

⇒ \vec{β_1} = 3λ\hat{i}+4λ\hat{j}+5λ\hat{k}

\vec{β_2}=\vec{β}-\vec{β_1}

⇒ \vec{β_2}=2\hat{i}+\hat{j}-4\hat{k}-(3λ\hat{i}+4λ\hat{j}+5λ\hat{k})

⇒ \vec{β_2}=(2-3λ)\hat{i}+(1-4λ)\hat{j}-(4+5λ)\hat{k}    

Now, 

\vec{β_2}.\vec{α}=0

⇒ [(2-3λ)\hat{i}+(1-4λ)\hat{j}-(4+5λ)\hat{k}](3\hat{i}+4\hat{j}+5\hat{k})=0

⇒ 3(2-3λ)+4(1-4λ)-5(4+5λ) = 0

⇒ 6-9λ+4-16λ-20-25λ = 0

⇒ -10 -50λ = 0

⇒ λ = -1/5

\vec{β_1} = -3/5\hat{i}-4/5\hat{j}-1\hat{k}
\vec{β_2} = 13/5\hat{i}+9/5\hat{j}-3\hat{k}

Question 18. If either \vec{a}=\vec{0}   or \vec{b}=\vec{0}   , then \vec{a}.\vec{b}=0   . But, The converse need not be true. Justify your answer with an example.

Solution:

Given, 

\vec{a}=\vec{0}    or \vec{b}=\vec{0}    then \vec{a}.\vec{b}=0

Suppose \vec{a}=2\hat{i}+\hat{j}+\hat{k}

\vec{b}=\hat{i}-\hat{j}-\hat{k}
\vec{a}.\vec{b}=0

But,

|\vec{a}|   = √(2)2+(1)2+(1)2

= √4+1+1

= √6 ≠ 0

|\vec{b}|   = √(1)2+(1)2+(1)2

= √3 ≠ 0

Hence Proved

Question 19. Show that the vectors \vec{a}=3\hat{i}-2\hat{j}+\hat{k}, \vec{b}=\hat{i}-3\hat{j}+5\hat{k},\vec{c}=2\hat{i}+\hat{j}-4\hat{k}   form a right-angled triangle.

Solution:

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

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

To prove given vectors form a right angle triangle

|\vec{a}|  = √(32+(-2)2+12) = √14

|\vec{b}|  = √(12+(-3)2+52) = √35

|\vec{c}|  = √(22+12+(-4)2) = √21

|\vec{a}|^2+|\vec{c}|^2 = (\sqrt14)^2+(\sqrt21)^2

= 14 + 21 = 35

Since, |\vec{b}|^2=|\vec{a}|^2+|\vec{c}|^2       (Pythagoras Theorem)

Hence, \vec{a},\vec{b}  and \vec{c}   form a right angled triangle.

Question 20. If \vec{a}=2\hat{i}+2\hat{j}+3\hat{k}   \vec{b}=-\hat{i}+2\hat{j}+\hat{k}   and \vec{c}=3\hat{i}+\hat{j}   are such that \vec{a}+λ\vec{b}   is perpendicular to \vec{c}   , then find the value of λ.

Solution:

Given:

\vec{a}=2\hat{i}+2\hat{j}+3\hat{k}
\vec{b}=-\hat{i}+2\hat{j}+\hat{k}
\vec{c}=3\hat{i}+\hat{j}

Now, (\vec{a}+λ\vec{b}).\vec{c}=0

⇒ [2\hat{i}+2\hat{j}+3\hat{k}+λ(-\hat{i}+2\hat{j}+\hat{k})](3\hat{i}+\hat{j})=0

⇒ [(2-λ)\hat{i}+(2+2λ)\hat{j}+(3+λ)\hat{k}](3\hat{i}+\hat{j})=0

⇒ (2 – λ)3 + (2 + 2λ) + 0 = 0

⇒ 6 – 3λ + 2 + 2λ =0

⇒ λ = 8

Question 21.  Find the angles of a triangle whose vertices are A (0, -1. -2), B (3, 1, 4), and C(5, 7, 1).

Solution:

Given that angle of a triangle whose vertices are A (0, -1. -2), B (3, 1, 4) and C(5, 7, 1).

\vec{AB}=(\vec{B}-\vec{A}) =3\hat{i}+2\hat{j}+6\hat{k}
\vec{BC}=(\vec{C}-\vec{B}) =2\hat{i}+6\hat{j}-3\hat{k}
\vec{AC}=(\vec{C}-\vec{A}) =5\hat{i}+8\hat{j}+3\hat{k}
|\vec{AB}|=\sqrt{3^2+2^2+6^2}=7
|\vec{BC}|=\sqrt{2^2+6^2+(-3)^2}=7
|\vec{AC}|=\sqrt{5^2+8^2+3^2}

= √98 = 7√2

Now, 

\vec{AB}.\vec{BC}   = (3 × 2 + 2 × 6 – 6 × 3) = 0

Thus, we can say AB is perpendicular to BC.

Hence, AB = BC = 7, ∠A =∠C and ∠B = 90°

∠A + ∠B + ∠C = 180°

2∠A = 180° – 90°

∠A = 45°

∠C = 45°  

∠B = 90°

Question 22. Find the magnitude of two vectors \vec{a}   and \vec{b}  , having the same magnitude and such that the angle between them is 60° and their scalar product is 1/2.

Solution:

We know \vec{a}.\vec{b}=|\vec{a}||\vec{b}|cos  θ

⇒ 1/ 2 = |\vec{a}||\vec{a}|cos θ

⇒ 1/2 = |\vec{a}|^2   (1/2)

⇒ |\vec{a}| = 1

or

⇒ |\vec{a}| = |\vec{b}| = 1

Question 23. Show that the points whose position vector are \vec{a} =4\hat{i}-3\hat{j}+\hat{k},\vec{b} =2\hat{i}-4\hat{j}+5\hat{k}, \vec{c} =\hat{i}-\hat{j}   form a right triangle.

Solution:

Given that positions vectors

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

Now,

\vec{AB} = (\vec{b}-\vec{a})

⇒ \vec{AB} =-2\hat{i}-\hat{j}+4\hat{k}

\vec{BC} = (\vec{c}-\vec{b})

⇒ \vec{BC} =-\hat{i}+3\hat{j}-5\hat{k}

\vec{CA} = (\vec{a}-\vec{c})

⇒ \vec{CA} =3\hat{i}-2\hat{j}+\hat{k}

Now, \vec{AB}.\vec{BC} =(-2\hat{i}-\hat{j}+4\hat{k})(-\hat{i}+3\hat{j}-5\hat{k})         

= 2 – 3 – 20 = -21

\vec{BC}.\vec{CA} =(-\hat{i}+3\hat{j}-5\hat{k})(3\hat{i}-2\hat{j}+\hat{k})

= -3 – 6 – 5 = -14

\vec{AB}.\vec{CA} =(-2\hat{i}-\hat{j}+4\hat{k})(3\hat{i}-2\hat{j}+\hat{k})

= -6 + 2 + 4 = 0

So, AB is perpendicular to CA or the given position vectors form a right-angled triangle.

Question 24. If the vertices A, B, C of △ABC have position vectors (1, 2, 3), (-1, 0, 0), (0, 1, 2) respectively, what is the magnitude of ∠ABC?

Solution:

Given the vertices of △ABC are A(1, 2, 3), B(-1, 0, 0), C(0, 1, 2)

Now, \vec{AB} = (-1-1)\hat{i}+(0-2)\hat{j}+(0-3)\hat{k}

-2\hat{i}-2\hat{j}-3\hat{k}

Or, \vec{BA}=2\hat{i}+2\hat{j}+3\hat{k}

\vec{BC}=\hat{i}+\hat{j}+2\hat{k}

We know that \vec{BA}.\vec{BC} = |\vec{BA}||\vec{BC}|

 \vec{BA}.\vec{BC} =   (2 × 1) + (2 × 1) + (3 × 2)

= 2 + 2 + 6 = 10

Now, |\vec{BA}|=\sqrt{4+4+9}   = √17                          

|\vec{BC}|=\sqrt{1+1+4}   = √6

Therefore,

 cos θ = = \frac{\vec{BA}.\vec{BC}}{|\vec{BA}||\vec{BC}|}

 cos θ = 10/ √(17×6)

θ = cos-1(10/√102)                 

Question 25. If A, B, C have position vectors (0, 1, 1), (3, 1, 5), (0, 3, 3) respectively, show that △ABC is right-angled at C. 

Solution:

Given, position vectors A(0, 1, 1), B(3, 1, 5), C(0, 3, 3)

Now, \vec{AC} = (0-0)\hat{i}+(3-1)\hat{j}+(3-1)\hat{k}

=2\hat{j}+2\hat{k}
\vec{BC}=(0-3)\hat{i}+(3-1)\hat{j}+(3-5)\hat{k}

-3\hat{i}+2\hat{j}-2\hat{k}

\vec{AC}.\vec{BC}=(2\hat{j}+2\hat{k}).(-3\hat{i}+2\hat{j}-2\hat{k})

= 2 × 2 – 2 × 2 = 0

Thus, \vec{AC}   and \vec{BC}   are perpendicular hence △ABC is right-angled at C

Question 26. Find the projection of \vec{b}+\vec{c}   on \vec{a}  , where \vec{a}=2\hat{i}-2\hat{j}+\hat{k}, \vec{b}=\hat{i}+2\hat{j}-2\hat{k}    and \vec{c}=2\hat{i}-\hat{j}+4\hat{k}   .

Solution:

Given: 

\vec{a}=2\hat{i}-2\hat{j}+\hat{k}
|\vec{a}|=\sqrt{4+4+1} = 3
\vec{b}=\hat{i}+2\hat{j}-2\hat{k}
 \vec{c}=2\hat{i}-\hat{j}+4\hat{k}

To find the projection of \vec{b}+\vec{c}    on \vec{a}

\vec{b}+\vec{c} = 3\hat{i}+\hat{j}+2\hat{k}

Now, Projection of \vec{b}+\vec{c}   [\frac{(\vec{b}+\vec{c})\vec{a}}{|\vec{a}|^2}]\vec{a}

[\frac{6-2+2}{3^2}](2\hat{i}-2\hat{j}+\hat{k})

= 6/9 × 3

= 2

Question 27. If \vec{a}=5\hat{i}-\hat{j}-3\hat{k}   and \vec{b}=\hat{i}+3\hat{j}-5\hat{k}  , then show that the vectors \vec{a}+\vec{b}   and \vec{a}-\vec{b}   are orthogonal.

Solution:

Given: \vec{a}=5\hat{i}-\hat{j}-3\hat{k}

\vec{b}=\hat{i}+3\hat{j}-5\hat{k}

To prove 

(\vec{a}+\vec{b})(\vec{a}-\vec{b})=0

Taking LHS

|\vec{a}|^2-|\vec{b}|^2

\sqrt{5^2+1+3^2}-\sqrt{1+3^2+5^2}

= √35 – √35

= 0 

Thus, the given vectors \vec{a}+\vec{b}   and \vec{a}-\vec{b}   are orthogonal.

Question 28. A unit vector \vec{a}   makes angle π/2 and π/3 with \hat{i}   and \hat{j}   respectively and an acute angle θ with \hat{k}  . Find the angle θ and components of \vec{a}  .

Solution:

Let us assume \vec{a}=a_1i + a_2j+a_3k

We know that 

a12+ a22 + a32 = 1  ….(1)

So, 

\vec{a}i=a_1
|\vec{a}||i|cos\frac{π}{4}=a_1

(1)(1)(1/√2) = a1

a= 1/√2

Again we take 

\vec{a}j=a_2
|\vec{a}||j|cos\frac{π}{3}=a_2

(1)(1)(1/2) = a2

a= 1/2

Put all these values in eq(1) to find the value of a3

(1/√2)2 + (1/2)2 + a32 = 1  ….(1)

a32 = 1/4

a3 = 1/2

Now we find the value of θ 

\vec{a}k=a_3
|\vec{a}||k|cos\theta=a_3

(1)(1)cosθ = 1/2

cosθ = 1/2

cosθ = π/3

and components of \vec{a}

\vec{a}=\frac{1}{\sqrt{2}}i + \frac{1}{2}j+\frac{1}{2}k

Question 29. If two vectors \vec{a}   and \vec{b}   are such that |\vec{a}|   = 2, |\vec{b}|   = 1, and \vec{a}.\vec{b}   =1. Find the value of (3\vec{a}-5\vec{b}).(2\vec{a}+7\vec{b}).

Solution:

Given, (3\vec{a}-5\vec{b}).(2\vec{a}+7\vec{b})

=6|\vec{a}|^2+21\vec{a}.\vec{b}-10\vec{a}.\vec{b}-35|\vec{b}|^2

=6|\vec{a}|^2+11\vec{a}.\vec{b}-35|\vec{b}|^2        

= 6(2)2 + 11(1) – 35(1)2 

= 24 + 11 – 35

= 35 – 35 = 0

Question 30. If \vec{a}   is a unit vector, then find |\vec{x}|   in each of the following:

(i) (\vec{x}-\vec{a})(\vec{x}+\vec{a})=8

Solution:

Given, (\vec{x}-\vec{a})(\vec{x}+\vec{a})=8       

|\vec{x}|^2-\vec{x}.\vec{a}+\vec{a}\vec{x}+|\vec{a}|^2=8

⇒ |\vec{x}|^2-|\vec{a}|^2=8

⇒ |\vec{x}|^2-1=8

⇒ |\vec{x}|^2=9

|\vec{x}|=3

(ii) (\vec{x}-\vec{a})(\vec{x}+\vec{a})=12

Solution:

Given, (\vec{x}-\vec{a})(\vec{x}+\vec{a})=12

⇒ |\vec{x}|^2-\vec{x}.\vec{a}+\vec{a}\vec{x}+|\vec{a}|^2=12

⇒ |\vec{x}|^2-|\vec{a}|^2=12

⇒ |\vec{x}|^2-1=12

⇒ |\vec{x}|^2=13

⇒ |\vec{x}|   =√13

Question 31.  Find |\vec{a}|   and |\vec{b}|   , if  

(i) (\vec{a}+\vec{b})(\vec{a}-\vec{b})   = 12 and |\vec{a}| = 2 |\vec{b}|

Solution:

Given, (\vec{a}+\vec{b})(\vec{a}-\vec{b})   = 12

⇒  |\vec{a}|^2-\vec{a}.\vec{b}+\vec{a}\vec{b}-|\vec{b}|^2=12

⇒  |\vec{a}|^2-|\vec{b}|^2    = 12

⇒ 4|\vec{b}|^2-|\vec{b}|^2    = 12

⇒ 3|\vec{b}|^2   = 12

⇒ |\vec{b}|   = 2

So, 

|\vec{a}|   = 4

(ii) (\vec{a}+\vec{b})(\vec{a}-\vec{b})   = 8 and  |\vec{a}|   = 8|\vec{b}|    

Solution:

Given, (\vec{a}+\vec{b})(\vec{a}-\vec{b})   = 8

⇒ |\vec{a}|^2-\vec{a}.\vec{b}+\vec{a}\vec{b}-|\vec{b}|^2=8

⇒ |\vec{a}|^2-|\vec{b}|^2=8

 64|\vec{b}|^2-|\vec{b}|^2=8

⇒ 63|\vec{b}|^2=8

⇒ |\vec{b}|   = √(8/63)

So, 

|\vec{a}|   = 8√(8/63)

(iii) (\vec{a}+\vec{b})(\vec{a}-\vec{b})   = 3 and  |\vec{a}|    = 2|\vec{b}|

Solution:

Given, (\vec{a}+\vec{b})(\vec{a}-\vec{b})= 3

⇒ |\vec{a}|^2-\vec{a}.\vec{b}+\vec{a}\vec{b}-|\vec{b}|^2=3

⇒ |\vec{a}|^2-|\vec{b}|^2 = 3

⇒ 4|\vec{b}|^2-|\vec{b}|^2=3

⇒ 3|\vec{b}|^2   = 3

⇒ |\vec{b}|   = 1

So, 

|\vec{a}|   = 2

Question 32. Find |\vec{a}-\vec{b}|   , if  

(i) |\vec{a}| = 2,  |\vec{b}| = 5   and \vec{a}.\vec{b} = 8

Solution:

We have, |\vec{a}-\vec{b}|

 |\vec{a}-\vec{b}|^2 =|\vec{a}|^2-2\vec{a}.\vec{b}+|\vec{b}|^2

 |\vec{a}-\vec{b}|^2   = 2– 2 × 8 + 52

⇒ |\vec{a}-\vec{b}|^2   = 4 – 16 + 25

⇒ |\vec{a}-\vec{b}|^2   = 13

 |\vec{a}-\vec{b}|   = √13

(ii) |\vec{a}|    = 3,  |\vec{b}|    = 4 and \vec{a}.\vec{b}    = 1

Solution:

We have, |\vec{a}-\vec{b}|

⇒  |\vec{a}-\vec{b}|^2 =|\vec{a}|^2-2\vec{a}.\vec{b}+|\vec{b}|^2

 |\vec{a}-\vec{b}|^2    = 3– 2 × 1 + 42

⇒ |\vec{a}-\vec{b}|^2    = 9 – 2 + 16

⇒ |\vec{a}-\vec{b}|^2   = 23

 |\vec{a}-\vec{b}|    = √23

(iii) |\vec{a}|=2,  |\vec{b}| = 3    and \vec{a}.\vec{b}   = 4

Solution:

We have,  |\vec{a}-\vec{b}|

⇒ |\vec{a}-\vec{b}|^2 =|\vec{a}|^2-2\vec{a}.\vec{b}+|\vec{b}|^2

 |\vec{a}-\vec{b}|^2   = 2– 2 × 4 + 32

⇒ |\vec{a}-\vec{b}|^2   = 4 – 8 + 9

 |\vec{a}-\vec{b}|^2   = 5

⇒ |\vec{a}-\vec{b}|    = √5

Question 33. Find the angle between the two vectors \vec{a}   and \vec{b} , if

(i) |\vec{a}|   =√3, |\vec{b}|   = 2 and \vec{a}.\vec{b}   = √6

Solution:

We know, \vec{a}\vec{b} = |\vec{a}||\vec{b}|cos\theta

⇒ √6 = 2√3 cos θ

⇒ cos θ = 1/√2 

⇒ θ = cos-1(1/√2)

⇒ θ = π/4

(ii) |\vec{a}|   = 3, |\vec{b}|   = 3 and \vec{a}.\vec{b}  = 1

Solution:

We know, \vec{a}\vec{b} = |\vec{a}||\vec{b}|cos\theta

⇒ 1 = 3×3 cos θ

⇒ cos θ = 1/9

⇒ θ = cos-1(1/9)

Question 34. Express the vector \vec{a}=5\hat{i}-2\hat{j}+5\hat{k}   as the sum of two vectors such that one is parallel to the vector \vec{b}=3\hat{i}+\hat{k}   and other is perpendicular to \vec{b}         

Solution: 

Given, \vec{a}=5\hat{i}-2\hat{j}+5\hat{k}

\vec{b}=3\hat{i}+\hat{k}

Let the two vectors be \vec{a_1}, \vec{a_2}

Now, \vec{a}=\vec{a_1}+\vec{a_2}           ….(1)

Assuming \vec{a_1}  is parallel to \vec{b}

Then, \vec{a_1}=λ\vec{b}             ……(2)

\vec{a_2}  is perpendicular to \vec{b}

Then, \vec{a_2}.\vec{b}=0      ……(3)

From eq(1) 

\vec{a_1}=λ\vec{b}

⇒ \vec{a_2} = \vec{a}-λ\vec{b}

⇒ \vec{a_2} =5\hat{i}-2\hat{j}+5\hat{k} -λ(3\hat{i}+\hat{k})

⇒ \vec{a_2} =(5-3λ)\hat{i}-2\hat{j}+(5-λ)\hat{k}

From eq(3)

 \vec{a_2}.\vec{b}=0

⇒ ((5-3λ)\hat{i}-2\hat{j}+(5-λ)\hat{k})(3\hat{i}+\hat{k})

⇒ (5-3λ)3+(5-λ)=0

⇒ 15-9λ+5-λ=0

⇒ -10λ = -20

⇒ λ=2

From eq(2)

\vec{a_1}=6\hat{i}+2\hat{k}
\vec{a_2}=-\hat{i}-2\hat{j}+3\hat{k}
\vec{a}=(6\hat{i}+2\hat{k})+(-\hat{i}-2\hat{j}+3\hat{k})

Question 35. If \vec{a} and \vec{b} are two vectors of the same magnitude inclined at an angle of 30° such that \vec{a}.\vec{b} = 3, find |\vec{a}|, |\vec{b}|.

Solution:

Given that two vectors of the same magnitude inclined at an angle of 30°, and \vec{a}.\vec{b} = 3

To find |\vec{a}|, |\vec{b}|

We know, \vec{a}\vec{b} = |\vec{a}||\vec{b}|cos\theta  

⇒ 3 = |\vec{a}||\vec{a}|cos θ    

⇒ 3 = |\vec{a}|^2 cos 30°

⇒ 3 = |\vec{a}|^2   (√3/2)

 |\vec{a}|^2   = 6/√3

|\vec{a}|=|\vec{b}|=\sqrt{2\sqrt{3}}

Question 36. Express 2\hat{i}-\hat{j}+3\hat{k}   as the sum of a vector parallel and a vector perpendicular to 2\hat{i}+4\hat{j}-2\hat{k}.

Solution:

Assuming \vec{a}=2\hat{i}-\hat{j}+3\hat{k}

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

Let the two vectors be \vec{a_1}, \vec{a_2}

Now, \vec{a}=\vec{a_1}+\vec{a_2}         

or \vec{a_2}=\vec{a}-\vec{a_1}          ….(1)

Assuming \vec{a_1}   is parallel to \vec{b}

then, \vec{a_1}=λ\vec{b}             …(2)

\vec{a_2}   is perpendicular to \vec{b}

then,\vec{a_2}.\vec{b}=0     ……(3)

Putting eq(2) in eq(1), we get

\vec{a_1}=λ\vec{b}

⇒ \vec{a_2} = \vec{a}-λ\vec{b}

⇒ \vec{a_2} =2\hat{i}-\hat{j}+3\hat{k} -λ(2\hat{i}+4\vec{j}-2\hat{k})

⇒ \vec{a_2} =(2-2λ)\hat{i}-(1+4λ)\hat{j}+(3+2λ)\hat{k}

From eq(3)

\vec{a_2}.\vec{b}=0

((2-2λ)\hat{i}-(1+4λ)\hat{j}+(3+2λ)\hat{k})(2\hat{i}+4\hat{j}-2\hat{k})=0

⇒ (2 – 2λ)2 – (1 + 4λ)4 – (3 + 2λ)2 = 0

⇒ 4 – 4λ – 4 – 16λ – 6 – 4λ = 0

⇒ 24λ = -6

⇒ λ = -6/24

From eq(2)

\vec{a_1}=-1/4(2\hat{i}+4\hat{j}-2\hat{k})
\vec{a_1}=-1/2\hat{i}-1\hat{j}+1/2\hat{k}
\vec{a_2}=(2+1/2)\hat{i}-0\hat{j}+(3-1/2)\hat{k}
\vec{a_2}=5/2\hat{i}+5/2\hat{k}
\vec{a}=\vec{a_1}+\vec{a_2}
\vec{a}=(-1/2\hat{i}-1\hat{j}+1/2\hat{k})+(5/2\hat{i}+5/2\hat{k})

Question 37. Decompose the vector 6\hat{i}-3\hat{j}-6\hat{k}   into vectors which are parallel and perpendicular to the vector \hat{i}+\hat{j}+\hat{k}.

Solution:

Let \vec{a}=6\hat{i}-3\hat{j}-6\hat{k}   and \vec{m}=\hat{i}+\hat{j}+\hat{k}

Let \vec{b}   be a vector parallel to \hat{i}+\hat{j}+\hat{k}.

Therefore, \vec{b} =λ(\hat{i}+\hat{j}+\hat{k})

\vec{a}   to be decomposed into two vectors

\vec{a}=\vec{b}+\vec{c}

⇒ \vec{c}=\vec{a}-\vec{b}

⇒ \vec{c}=(6-λ)\hat{i}+(-3-λ)\hat{j}+(-6-λ)\vec{k}

Now, \vec{c}   is perpendicular to \vec{m}

or \vec{c}.\vec{m}=0

((6-λ)\hat{i}+(-3-λ)\hat{j}+(-6-λ)\vec{k})(\hat{i}+\hat{j}+\hat{k})=0

⇒ 6 – λ – 3 – λ – 6 – λ = 0

⇒ λ = -1

Therefore, the required vectors are \vec{b} =-\hat{i}-\hat{j}-\hat{k}  and \vec{c}=7\hat{i}-2\hat{j}-5\vec{k}

Question 38. Let \vec{a}=5\hat{i}-\hat{j}+7\hat{k}   and \vec{b}=\hat{i}-\hat{j}+λ\hat{k} . Find λ such that \vec{a}+\vec{b}  is orthogonal to \vec{a}-\vec{b}

Solution:

Given, \vec{a}=5\hat{i}-\hat{j}+7\hat{k}

\vec{b}=\hat{i}-\hat{j}+λ\hat{k}

According to question

(\vec{a}+\vec{b})(\vec{a}-\vec{b})=0

|\vec{a}|^2-|\vec{b}|^2=0

⇒ |\vec{a}|^2=|\vec{b}|^2

⇒ \sqrt{5^2+(-1)^2+7^2}=\sqrt{1^2+(-1)^2+λ^2}

⇒ 25 + 1 + 49 = 1 + 1 + λ2

⇒ λ= 73

⇒ λ = √73

Question 39. If \vec{a}.\vec{a}=0  and \vec{a}.\vec{b}=0 , what can you conclude about the vector \vec{b} ?

Solution:

Given, \vec{a}.\vec{a}=0 ,\vec{a}.\vec{b}=0

|\vec{a}| = 0

Now, \vec{a}.\vec{b}=0

We conclude that \vec{a}=0 or \vec{b}=0  or θ = 90°

Thus, \vec{b}  can be any arbitrary vector.

Question 40. If \vec{c}  is perpendicular to both \vec{a}  and \vec{b} , then prove that it is perpendicular to both \vec{a}+\vec{b}  and \vec{a}-\vec{b}

Solution:

Given \vec{c}  is perpendicular to both \vec{a}  and \vec{b}

\vec{c}.\vec{a}=0      ….(1)

\vec{c}.\vec{b}=0      ….(2)

To prove \vec{c}.(\vec{a}+\vec{b})=0  and \vec{c}.(\vec{a}-\vec{b})=0

Now,\vec{c}.(\vec{a}+\vec{b})

 \vec{c}.\vec{a}+\vec{c}.\vec{b}=0          [From eq(1) and (2)]

Again, \vec{c}.(\vec{a}-\vec{b})

 \vec{c}.\vec{a}-\vec{c}.\vec{b}=0          [From eq(1) and (2)] 

Hence Proved

Question 41. If |\vec{a}|= a  and |\vec{b}|= b , prove that (\frac{\vec{a}}{a^2}-\frac{\vec{b}}{b^2})^2= (\frac{\vec{a}-\vec{b}}{ab})^2

Solution:

Given, |\vec{a}|  = a  and  |\vec{b}|  = b,

To prove  

(\frac{\vec{a}}{a^2}-\frac{\vec{b}}{b^2})^2= (\frac{\vec{a}-\vec{b}}{ab})^2.

Taking LHS

(\frac{\vec{a}}{a^2}-\frac{\vec{b}}{b^2})^2

=\frac{\vec{a}.\vec{a}}{a^4}+\frac{\vec{b}.\vec{b}}{b^4}-2\frac{\vec{a}.\vec{b}}{a^2b^2}

=\frac{a^2}{a^4}+\frac{b^2}{b^4}-2\frac{\vec{a}.\vec{b}}{a^2b^2}

\frac{1}{a^2}+\frac{1}{b^2}-2\frac{\vec{a}.\vec{b}}{a^2b^2}

Taking RHS

(\frac{\vec{a}-\vec{b}}{ab})^2  \vec{d}.\vec{a}=0

=(\frac{a^2+b^2-2\vec{a}\vec{b}}{a^2b^2})

=\frac{1}{a^2}+\frac{1}{b^2}-2\frac{\vec{a}.\vec{b}}{a^2b^2}

LHS = RHS 

Hence Proved

Question 42. If \vec{a},\vec{b},\vec{c}  are three non- coplanar vectors such that \vec{d}.\vec{a}=\vec{d}.\vec{b}=\vec{d}.\vec{c}  =0 then show that \vec{d}  is the null vector.

Solution:

Given that \vec{d}.\vec{a}=0

So either \vec{d} = 0  or \vec{d}⊥\vec{a}=0

Similarly, \vec{d}.\vec{b} = 0

Either \vec{d}= 0   or \vec{d}⊥\vec{b}=0

Also, \vec{d}.\vec{c} =0

So \vec{d}  = 0   or \vec{d}⊥\vec{c}=0

But \vec{d} can’t be perpendicular to \vec{a},\vec{b}  and \vec{c}  because \vec{a},\vec{b},\vec{c}  are non-coplanar.

So \vec{d}  = 0 or \vec{d}   is a null vector  

Question 43. If a vector \vec{a}   is perpendicular to two non- collinear vectors \vec{b}   and \vec{c}   , then is \vec{a}   perpendicular to every vector in the plane of \vec{b}   and \vec{c}          

Solution:

Given that \vec{a}  is perpendicular to \vec{b}  and \vec{c}         

\vec{a}.\vec{b}=0 , \vec{a}.\vec{c}=0

Let \vec{r}    be any vector in the plane of  \vec{b}  and \vec{c}  and \vec{r}   is the linear combination of  \vec{b}   and \vec{c}        

\vec{r} =x\vec{b}+y\vec{c}                      [x, y are scalars]

Now, \vec{a}.\vec{r}

⇒ \vec{a}.\vec{r} =\vec{a}(x\vec{b}+y\vec{c})

⇒ \vec{a}.\vec{r} =x(\vec{a}.\vec{b})+y(\vec{a}.\vec{c})

⇒ \vec{a}.\vec{r} =x0+y0

 \vec{a}.\vec{r} = 0

Therefore,  \vec{a} is perpendicular to \vec{r}  i.e. \vec{a}  is perpendicular to every vector.

Question 44. If \vec{a}+\vec{b}+\vec{c}=\vec{0} , how that the angle θ between the vectors \vec{b}  and \vec{c}  is given by cos θ = \frac{|\vec{a}|^2-|\vec{b}|^2-|\vec{c}|^2}{2|\vec{b}||\vec{c}|}

Solution:

Given that \vec{a}+\vec{b}+\vec{c}=\vec{0}          

 \vec{a}=-(\vec{b}+\vec{c})

 (\vec{a})^2=(\vec{b}+\vec{c})^2

⇒ \vec{a}.\vec{a}=(\vec{b}+\vec{c})(\vec{b}+\vec{c})

⇒ |\vec{a}|^2=|\vec{b}|^2+|\vec{b}||\vec{c}|+|\vec{b}||\vec{c}|+|\vec{c}|^2

⇒ |\vec{a}|^2-|\vec{b}|^2-|\vec{c}|^2=2|\vec{b}||\vec{c}|

⇒ |\vec{b}||\vec{c}|=(|\vec{a}|^2-|\vec{b}|^2-|\vec{c}|^2)/2

⇒ cos θ = \frac{|\vec{a}|^2-|\vec{b}|^2-|\vec{c}|^2}{2|\vec{b}||\vec{c}|}.

Question 45. Let \vec{u},\vec{v}  and \vec{w}   be vector such \vec{u}+\vec{v}+\vec{w}=\vec{0} |\vec{u}|  = 3, |\vec{v}|  = 4 and |\vec{w}|  = 5, then find \vec{u}.\vec{v}+\vec{v}.\vec{w}+\vec{w}.\vec{u}

Solution:

Given that \vec{u},\vec{v} and \vec{w}   are vectors such that \vec{u}+\vec{v}+\vec{w}=\vec{0} |\vec{u}|  = 3, |\vec{v}|  = 4 and |\vec{w}| =5,  

To find \vec{u}.\vec{v}+\vec{v}.\vec{w}+\vec{w}.\vec{u}

Taking  

\vec{u}+\vec{v}+\vec{w}=\vec{0}

Squaring on both side, we get 

⇒ (\vec{u}+\vec{v}+\vec{w})^2=\vec{0}

⇒ |\vec{u}|^2+|\vec{v}|^2+|\vec{w}|^2+2(\vec{u}.\vec{v}+\vec{v}.\vec{w}+\vec{w}.\vec{u})=0

⇒ 2(\vec{u}.\vec{v}+\vec{v}.\vec{w}+\vec{w}.\vec{u})=-(|\vec{u}|^2+|\vec{v}|^2+|\vec{w}|^2)

⇒ 2(\vec{u}.\vec{v}+\vec{v}.\vec{w}+\vec{w}.\vec{u})=-(3^2+4^2+5^2)

⇒ 2(\vec{u}.\vec{v}+\vec{v}.\vec{w}+\vec{w}.\vec{u})=-50

Therefore, \vec{u}.\vec{v}+\vec{v}.\vec{w}+\vec{w}.\vec{u}=-25

Question 46. Let \vec{a}=x^2\hat{i}+2\hat{j}-2\hat{k}, \vec{b}=\hat{i}-\hat{j}+\hat{k},   and \vec{c}=x^2\hat{i}+5\hat{j}-4\hat{k}   be three vectors. Find the values of x for which the angle between \vec{a}   and \vec{b}   is acute and the angle between \vec{b}   and \vec{c}   is obtuse.

Solution:

Given \vec{a}=x^2\hat{i}+2\hat{j}-2\hat{k}, \vec{b}=\hat{i}-\hat{j}+\hat{k}, \vec{c}=x^2\hat{i}+5\hat{j}-4\hat{k}

Case I: When angle between \vec{a}   and \vec{b}   is acute:-

\vec{a}.\vec{b}   >0

 (x^2\hat{i}+2\hat{j}-2\hat{k})(\hat{i}-\hat{j}+\hat{k})>0″ height=”32″ width=”341″></p><p>⇒ x<sup>2 </sup>– 2 – 2 > 0</p><p>⇒ x<sup>2 </sup>> 4</p><p>x ∈ (2, -2)</p><p>Case II: When angle between <img loading=and \vec{c}   is obtuse:-

\vec{b}.\vec{c}<0

⇒ (\hat{i}-\hat{j}+\hat{k})(x^2\hat{i}+5\hat{j}-4\hat{k})<0

⇒ x– 5 – 4 < 0

⇒ x< 9

x ∈ (3, -3)

Therefore, x ∈ (-3, -2)∪(2, 3)

Question 47. Find the value of x and y if the vectors \vec{a}=3\hat{i}+x\hat{j}-\hat{k}   and \vec{b}=2\hat{i}+\hat{j}+y\hat{k}  are mutually perpendicular vectors of equal magnitude.

Solution:

Given  \vec{a}=3\hat{i}+x\hat{j}-\hat{k}, \vec{b}=2\hat{i}+\hat{j}+y\hat{k}   are mutually perpendicular vectors of equal magnitude.

|\vec{a}|^2=|\vec{b}|^2

⇒ 3+ x+ (-1)2 = 2+ 1+ y2

⇒ x2+10 = y2+5

⇒ x– y+ 5 = 0    ….(1)

Now,  \vec{a}.\vec{b} = 0

⇒ 6 + x – y = 0

⇒ y = x + 6      …..(2)

From eq(1)

x– (x + 6)+ 5 = 0

⇒ x2 – (x2 + 36 – 12x) + 5 = 0

⇒ -12x – 31 = 0

⇒ x = -31/12

Now, y = -31/12 + 6

y = 41/12

Question 48. If \vec{a}   and \vec{b}   are two non-coplanar unit vectors such that |\vec{a}+\vec{b}|  =√3 , find (2\vec{a}-5\vec{b}).(3\vec{a}+\vec{b}).

Solution:

Given that \vec{a}   and \vec{b}    are two non-coplanar unit vectors such that |\vec{a}+\vec{b}|  =√3  

To find (2\vec{a}-5\vec{b}).(3\vec{a}+\vec{b}).

Now,  

|\vec{a}+\vec{b}|^2  =3
⇒|\vec{a}|^2+|\vec{b}|^2+2\vec{a}.\vec{b}=3
⇒ 1+1+2\vec{a}.\vec{b}=3
⇒ \vec{a}.\vec{b}=1/2

Now, (2\vec{a}-5\vec{b}).(3\vec{a}+\vec{b})

6|\vec{a}|^2-13\vec{a}.\vec{b}-5|\vec{b}|^2

= 6 – 13(1/2) – 5

= 1 – 13/2

= -11/2

Question 49. If \vec{a},\vec{b}   are two vectors such that |\vec{a}+\vec{b}  | = |\vec{b}|  , then prove that \vec{a}+2\vec{b}   is perpendicular to \vec{a}.   

Solution:

To prove 

(\vec{a}+2\vec{b})\vec{a}= 0

Now, 

\vec{a}+\vec{b} = \vec{b}

Squaring on both side, we get

|\vec{a}+\vec{b}|^2=|\vec{b}|^2

⇒ (\vec{a}+\vec{b})(\vec{a}+\vec{b}) = \vec{b}.\vec{b}

⇒ \vec{a}\vec{a}+\vec{a}\vec{b}+\vec{b}\vec{a}+\vec{b}\vec{b} = \vec{b}.\vec{b}

⇒ \vec{a}\vec{a}+2\vec{a}\vec{b} = 0

⇒ \vec{a}(\vec{a}+2\vec{b}) = 0

Therefore, \vec{a}+2\vec{b}  is perpendicular to \vec{a}

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