RD Sharma Class 12 Ex 23.6 Solutions Chapter 23 Algebra of Vectors

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter23
Exercise23.6
CategoryRD Sharma Solutions

RD Sharma Class 12 Ex 23.6 Solutions Chapter 23 Algebra of Vectors

Question 1: Find the magnitude of the vector \vec{a} = 2\hat{i}+3\hat{j}-6\hat{k}  .

Solution:

Magnitude of a vector x\hat{i}+y\hat{j}+z\hat{k} = \sqrt{x^2+y^2+z^2}

=> |\vec{a}| = \sqrt{2^2+3^2+(-6)^2}

=> |\vec{a}| = \sqrt{4+9+36}

=> |\vec{a}| = \sqrt{49}

=> |\vec{a}| = 7

Question 2: Find the unit vector in the direction of 3\hat{i}+4\hat{j}-12\hat{k}  .

Solution:

We know that unit vector of a vector \vec{a}   is given by,

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

=> \hat{p} = \dfrac{1}{\sqrt{3^2+4^2+(-12)^2}}(3\hat{i}+4\hat{j}-12\hat{k})

=> \hat{p} = \dfrac{1}{\sqrt{9+16+144}}(3\hat{i}+4\hat{j}-12\hat{k})

=> \hat{p} = \dfrac{1}{\sqrt{169}}(3\hat{i}+4\hat{j}-12\hat{k})

=> \hat{p} = \dfrac{1}{13}(3\hat{i}+4\hat{j}-12\hat{k})

Question 3: Find a unit vector in the direction of the resultant of the vectors \hat{i}-\hat{j}+3\hat{k}2\hat{i}+\hat{j}-2\hat{k}   and \hat{i}+2\hat{j}-2\hat{k}.

Solution:

Let,

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

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

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

Let \vec{d}   be the resultant,

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

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

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

Unit vector is,

=> \hat{p} = \dfrac{\vec{d}}{|\vec{d}|}

=> \hat{p} = \dfrac{1}{\sqrt{4^2+2^2+(-1)^2}}(4\hat{i}+2\hat{j}-\hat{k})

=> \hat{p} = \dfrac{1}{\sqrt{16+4+1}}(4\hat{i}+2\hat{j}-\hat{k})

=> \hat{p} = \dfrac{1}{\sqrt{21}}(4\hat{i}+2\hat{j}-\hat{k})

Question 4: The adjacent sides of a parallelogram are represented by the vectors \vec{a}=\hat{i}+\hat{j}-\hat{k}   and \vec{b}=-2\hat{i}+\hat{j}+2\hat{k}. Find the unit vectors parallel to the diagonals of the parallelogram.

Solution:

Let PQRS be the parallelogram.

Given that, PQ = \hat{i}+\hat{j}-\hat{k} and QR = -2\hat{i}+\hat{j}+2\hat{k}.

Thus, the diagonals are: PR and SQ.

=> \vec{PR} = \vec{PQ} + \vec{QR}

=> \vec{PR} = \vec{a} + \vec{b}

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

=> \vec{PR} = -\hat{i} + 2\hat{j} +\hat{k}

=> \vec{SQ} = \vec{PQ} - \vec{PS}

=> \vec{SQ} = \vec{a} - \vec{b}

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

=> \vec{SQ} = 3\hat{i}-3\hat{k}

Thus the unit vectors in the direction of the diagonals are:

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

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

=> \hat{PR} = \dfrac{1}{\sqrt{6}}(-\hat{i}+2\hat{j}+\hat{k})

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

=> \hat{SQ} = \dfrac{1}{\sqrt{3^2+(-3)^2}}( 3\hat{i}-3\hat{k})

=> \hat{SQ} = \dfrac{1}{3\sqrt{2}}(3\hat{i}-3\hat{k})

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

Solution:

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

Let,

=> \vec{d} = 3\vec{a}-2\vec{b}+4\vec{c}

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

=> \vec{d} = (9\hat{i}-3\hat{j}-12\hat{k})+ (4\hat{i}-8\hat{j}+6\hat{k})+(4\hat{i}+8\hat{j}-4\hat{k})

=> \vec{d} = 17\hat{i}-3\hat{j}-10\hat{k}

The magnitude is given by,

=> |\vec{d}| = \sqrt{17^2+(-3)^2+(-10)^2}

=> |\vec{d}| = \sqrt{289+9+100}

=> |\vec{d}| = \sqrt{398}

Question 6: If \vec{PQ} = 3\hat{i}+2\hat{j}-\hat{k}   and the coordinates of P are (1,-1,2), find the coordinates of Q.

Solution:

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

And, \vec{P} = \hat{i}-\hat{j}+2\hat{k}

=> \vec{PQ} = \vec{Q}-\vec{P}

=> \vec{Q} = \vec{PQ}+ \vec{P}

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

=> \vec{Q} = 4\hat{i}+\hat{j}+\hat{k}

=> Thus the coordinates of Q are (4,1,1).

Question 7: Prove that the points \hat{i}-\hat{j}4\hat{i}-3\hat{j}+\hat{k} and 2\hat{i}-4\hat{j}+5\hat{k} are the vertices of a right-angled triangle.

Solution:

Let,

=> \vec{A} = \hat{i}-\hat{j}

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

=> \vec{C} = 2\hat{i}-4\hat{j}+5\hat{k}

Thus, the 3 sides of the triangle are,

=> \vec{AB} = \vec{B} - \vec{A}

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

=> \vec{AB} = 3\hat{i}-2\hat{j}+\hat{k}

=> \vec{BC} = \vec{C} - \vec{B}

=> \vec{BC} = (2\hat{i}-4\hat{j}+5\hat{k})-(4\hat{i}-3\hat{j}+\hat{k})

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

=> \vec{CA} = \vec{A} -\vec{C}

=> \vec{CA} = ( \hat{i}-\hat{j})-(2\hat{i}-4\hat{j}+5\hat{k})

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

The lengths of every side are given by their magnitude,

=> |\vec{AB}| = \sqrt{3^2+(-2)^2+1^2} = \sqrt{14}

=> |\vec{BC}| = \sqrt{(-2)^2+(-1)^2+4^2} = \sqrt{21}

=> |\vec{CA}| = \sqrt{(-1)^2+3^2+(-5)^2} = \sqrt{35}

As we can see,

=> |\vec{CA}|^2 = |\vec{AB}|^2+|\vec{BC}|^2

=> These 3 points form a right-angled triangle.

Question 8: If the vertices A, B and C of a triangle ABC are the points with position vectors a_1\hat{i}+a_2\hat{j}+a_3\hat{k}  b_1\hat{i}+b_2\hat{j}+b_3\hat{k}  c_1\hat{i}+c_2\hat{j}+c_3\hat{k}   respectively, what are the vectors determined by its sides? Find the length of these vectors.

Solution:

Let,

=> \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}

=> \vec{c} = c_1\hat{i}+c_2\hat{j}+c_3\hat{k}

The sides of the triangle are given as,

=> \vec{AB} = \vec{b} - \vec{a}

=> \vec{AB} = ( b_1\hat{i}+b_2\hat{j}+b_3\hat{k})-(a_1\hat{i}+a_2\hat{j}+a_3\hat{k})

=> \vec{AB} = (b_1-a_1)\hat{i}+(b_2-a_2)\hat{j}+(b_3-a_3)\hat{k}

=> \vec{BC} = \vec{c}-\vec{b}

=> \vec{BC} = ( c_1\hat{i}+c_2\hat{j}+c_3\hat{k})-(b_1\hat{i}+b_2\hat{j}+b_3\hat{k})

=> \vec{BC} = (c_1-b_1)\hat{i}+(c_2-b_2)\hat{j}+(c_3-b_3)\hat{k}

=> \vec{CA} = \vec{a}-\vec{c}

=> \vec{CA} = ( a_1\hat{i}+a_2\hat{j}+a_3\hat{k})-(c_1\hat{i}+c_2\hat{j}+c_3\hat{k})

=> \vec{CA} = (a_1-c_1)\hat{i}+(a_2-c_2)\hat{j}+(a_3-c_3)\hat{k}

The lengths of the sides are,

=> |\vec{AB}| = \sqrt{(b_1-a_1)^2+(b_2-a_2)^2+(b_3-a_3)^2}

=> |\vec{BC}| = \sqrt{(c_1-b_1)^2+(c_2-b_2)^2+(c_3-b_3)^2}

=> |\vec{CA}| = \sqrt{(a_1-c_1)^2+(a_2-c_2)^2+(a_3-c_3)^2}

Question 9: Find the vector from the origin O to the centroid of the triangle whose vertices are (1,-1,2), (2,1,3), and (-1,2,-1).

Solution:

The position of the centroid is given by,

=> (x, y, z) = (\dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3},\dfrac{z_1+z_2+z_3}{3})

=> (x, y, z) = (\dfrac{1+2+(-1)}{3},\dfrac{(-1)+1+2}{3},\dfrac{2+3+(-1)}{3})

=> (x, y, z) = (\dfrac{2}{3},\dfrac{2}{3},\dfrac{4}{3})

The vector to the centroid from O is,

=> \vec{c} = \dfrac{2}{3}\hat{i}+\dfrac{2}{3}\hat{j}+\dfrac{4}{3}\hat{k}

Question 10: Find the position vector of a point R which divides the line segment joining points p(\hat{i}+2\hat{j}+\hat{k}  ) and q(-\hat{i}+\hat{j}+\hat{k}) in the ratio 2:1.

(i) Internally

Solution:

The position vectors of a point that divides a line segment internally are given by,

=> \vec{OR} = \dfrac{m\vec{Q}+n\vec{P}}{m+n}  , where \dfrac{m}{n}=\dfrac{2}{1}

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

=> \vec{OR} = \dfrac{-\hat{i}+4\hat{j}+3\hat{k}}{3}

(ii) Externally

Solution:

The position vectors of a point that divides a line segment externally are given by,

=> \vec{OR} = \dfrac{m\vec{Q}-n\vec{P}}{m-n}  , where \dfrac{m}{n}=\dfrac{2}{1}

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

=> \vec{OR} = 3\hat{i}-\hat{k}

Question 11: Find the position vector of the mid-point of the vector joining the points P(2\hat{i}-3\hat{j}+4\hat{k} ) and Q(4\hat{i}+\hat{j}-2\hat{k} ).
Solution:

The mid-point of the line segment joining 2 vectors is given by:
=> \vec{R} = \dfrac{\vec{P}+\vec{Q}}{2}
=> \vec{R} = \dfrac{(2\hat{i}-3\hat{j}+4\hat{k})+(4\hat{i}+\hat{j}-2\hat{k})}{2}
=> \vec{R} = \dfrac{6\hat{i}-2\hat{j}+2\hat{k}}{2}
=> \vec{R} = 3\hat{i}-\hat{j}+\hat{k}
Question 12: Find the unit vector in the direction of the vector \vec{PQ}, where P and Q are the points (1,2,3) and (4,5,6).
Solution:
Let,
=> \vec{p} = \hat{i}+2\hat{j}+3\hat{k}
=> \vec{q} = 4\hat{i} + 5\hat{j}+6\hat{k}
=> \vec{PQ} = \vec{q}-\vec{p}
=> \vec{PQ} = (4\hat{i} + 5\hat{j}+6\hat{k})-(\hat{i}+2\hat{j}+3\hat{k})
=> \vec{PQ} = 3\hat{i}+3\hat{j}+3\hat{k}
Unit vector is,
=> \hat{PQ} = \dfrac{\vec{PQ}}{|\vec{PQ}|}
=> \hat{PQ} = \dfrac{1}{\sqrt{3^2+3^2+3^2}}(3\hat{i}+3\hat{j}+3\hat{k})
=> \hat{PQ} = \dfrac{1}{3\sqrt{3}}3(\hat{i}+\hat{j}+\hat{k})
=> \hat{PQ} = \dfrac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})
Question 13: Show that the points A(2\hat{i}-\hat{j}+\hat{k}), B(\hat{i}-3\hat{j}-5\hat{k}), C(3\hat{i}-4\hat{j}-4\hat{k}) are the vertices of a right-angled triangle.
Solution:
Let,
=> \vec{a} = 2\hat{i}-\hat{j}+\hat{k}
=> \vec{b} = \hat{i}-3\hat{j}-5\hat{k}
=> \vec{c} = 3\hat{i}-4\hat{j}-4\hat{k}
The line segments are,
=> \vec{AB} = \vec{b}-\vec{a}
=> \vec{AB} = (\hat{i}-3\hat{j}-5\hat{k})-(2\hat{i}-\hat{j}+\hat{k})
=> \vec{AB} = -\hat{i}-2\hat{j}-6\hat{k}
=> \vec{BC} = \vec{c}-\vec{b}
=> \vec{BC} = (3\hat{i}-4\hat{j}-4\hat{k})-(\hat{i}-3\hat{j}-5\hat{k})
=> \vec{BC} = 2\hat{i}-\hat{j}+\hat{k}
=> \vec{CA} = \vec{a}-\vec{c}
=> \vec{CA} = (2\hat{i}-\hat{j}+\hat{k})-(3\hat{i}-4\hat{j}-4\hat{k})
=> \vec{CA} = -\hat{i}+3\hat{j}+5\hat{k}
The magnitudes of the sides are,
=> |\vec{AB}| = \sqrt{(-1)^2+(-2)^2+(-6)^2} = \sqrt{41}
=> |\vec{BC}| = \sqrt{2^2+(-1)^2+1^2} = \sqrt{6}
=> |\vec{CA}| = \sqrt{(-1)^2+3^2+5^2} = \sqrt{35}
As we can see that |\vec{AB}|^2 = |\vec {BC}|^2+|\vec{CA}|^2
=> Thus, ABC is a right-angled triangle.
Question 14: Find the position vector of the mid-point of the vector joining the points P(2, 3, 4) and Q(4, 1, -2).
Solution:
Let,
=> \vec{p} = 2\hat{i}+3\hat{j}+4\hat{k}
=> \vec{q} = 4\hat{i}+\hat{j}-2\hat{k}
 The mid-point of the line segment joining 2 vectors is given by:
=> \vec{r} = \dfrac{\vec{p}+\vec{q}}{2}
=> \vec{r} = \dfrac{(2\hat{i}+3\hat{j}+4\hat{k})+( 4\hat{i}+\hat{j}-2\hat{k})}{2}
=> \vec{r} = \dfrac{6\hat{i}+4\hat{j}+2\hat{k}}{2}
=> \vec{r} = 3\hat{i}+2\hat{j}+\hat{k}
Question 15: Find the value of x for which x(\hat{i}+\hat{j}+\hat{k}) is a unit vector.
Solution:
The magnitude of the given vector is,
=> |x(\hat{i}+\hat{j}+\hat{k})| = \sqrt{x^2+x^2+x^2}
=> |x(\hat{i}+\hat{j}+\hat{k})| = \sqrt{3x^2}
=> |x(\hat{i}+\hat{j}+\hat{k})| = \pm x\sqrt{3}
For it to be a unit vector,
=> |x(\hat{i}+\hat{j}+\hat{k})| = 1
=> x\sqrt{3} = \pm 1
=> x = \pm \dfrac{1}{\sqrt{3}}
Question 16: If \vec{a}=\hat{i}+\hat{j}+\hat{k}\vec{b} = 2\hat{i}-\hat{j}+3\hat{k} and \vec{c}=\hat{i}-2\hat{j}+\hat{k}, find a unit vector parallel to 2\vec{a}-\vec{b}+3\vec{c} .
Solution:
Given, \vec{a}=\hat{i}+\hat{j}+\hat{k} \vec{b} = 2\hat{i}-\hat{j}+3\hat{k}  and \vec{c}=\hat{i}-2\hat{j}+\hat{k}
=> 2\vec{a}-\vec{b}+3\vec{c} = 2(\hat{i}+\hat{j}+\hat{k})-(2\hat{i}-\hat{j}+3\hat{k})+3(\hat{i}-2\hat{j}+\hat{k})
=> 2\vec{a}-\vec{b}+3\vec{c} = 3\hat{i}-3\hat{j}+2\hat{k}
Thus, the unit vector is,
=> \hat{p} = \dfrac{2\vec{a}-\vec{b}+3\vec{c}}{|2\vec{a}-\vec{b}+3\vec{c}|}
=> \hat{p} = \dfrac{1}{\sqrt{3^2+(-3)^2+2^2}}(3\hat{i}-3\hat{j}+2\hat{k})
=> \hat{p} = \dfrac{1}{\sqrt{22}}(3\hat{i}-3\hat{j}+2\hat{k})
Question 17: If \vec{a}=\hat{i}+\hat{j}+\hat{k} \vec{b}=4\hat{i}-2\hat{j}+3\hat{k} and \vec{c} = \vec{i}-2\hat{j}+\hat{k}, find a vector of magnitude 6 units which is parallel to the vector 2\vec{a}-\vec{b}+3\vec{c}.
Solution:
Given, \vec{a}=\hat{i}+\hat{j}+\hat{k} \vec{b}=4\hat{i}-2\hat{j}+3\hat{k} and \vec{c} = \vec{i}-2\hat{j}+\hat{k}
=> 2\vec{a}-\vec{b}+3\vec{c} = 2(\hat{i}+\hat{j}+\hat{k})-(4\hat{i}-2\hat{j}+3\hat{k})+3(\vec{i}-2\hat{j}+\hat{k})
=> 2\vec{a}-\vec{b}+3\vec{c} = \hat{i}-2\hat{j}+2\hat{k}
Unit vector in that direction is,
=> \hat{p} = \dfrac{2\vec{a}-\vec{b}+3\vec{c}}{|2\vec{a}-\vec{b}+3\vec{c}|}
=> \hat{p} = \dfrac{1}{\sqrt{1^2+(-2)^2+2^2}}(\hat{i}-2\hat{j}+2\hat{k})
=> \hat{p} = \dfrac{1}{3}(\hat{i}-2\hat{j}+2\hat{k})
Given that the vector has a magnitude of 6,
=> Required vectors are : \pm6\times\dfrac{1}{3}(\hat{i}-2\hat{j}+2\hat{k}) = \pm2(\hat{i}-2\hat{j}+2\hat{k})
Question 18: Find a vector of magnitude 5 units parallel to the resultant of the vector \vec{a}=2\hat{i}+3\hat{j}-\hat{k}and \vec{b} = \hat{i}-2\hat{j}+\hat{k}.
Solution:
Given, \vec{a}=2\hat{i}+3\hat{j}-\hat{k} and \vec{b} = \hat{i}-2\hat{j}+\hat{k}
The resultant vector will be given by,
=> \vec{r} = \vec{a}+\vec{b}
=> \vec{r} = (2\hat{i}+3\hat{j}-\hat{k})+(\hat{i}-2\hat{j}+\hat{k})
=> \vec{r} = 3\hat{i}+\hat{j}
Unit vector is,
=> \hat{p} = \dfrac{\vec{r}}{|\vec{r}|}
=> \hat{p} = \dfrac{1}{\sqrt{3^2+1^2}}(3\hat{i}+\hat{j})
=> \hat{p} = \dfrac{1}{\sqrt{10}}(3\hat{i}+\hat{j})
Given that the vector has a magnitude of 5,
=> Required vectors are: \pm5\times\dfrac{1}{\sqrt{10}}(3\hat{i}+\hat{j})= \pm\dfrac{5}{\sqrt{10}}(3\hat{i}+\hat{j})
Question 19: The two vectors \hat{j}+\hat{i} and 3\hat{i}+\hat{j}+4\hat{k} represent the sides \vec{AB} and \vec{AC} respectively of the triangle ABC. Find the length of the median through A.
Solution:
Let D be the point on BC, on which the median through A touches.
D is also the mid-point of BC.
The median \vec{AD} is thus given by:
=> \vec{AD} = \dfrac{\vec{B}+\vec{C}}{2}- \vec{A}
=> \vec{AD} = \dfrac{\vec{B}-\vec{A}+\vec{C}-\vec{A}}{2}
=> \vec{AD} = \dfrac{\vec{AB}+\vec{AC}}{2}
=> \vec{AD} = \dfrac{(\hat{j}+\hat{i})+(3\hat{i}+\hat{j}+4\hat{k})}{2}
=> \vec{AD} = \dfrac{4\hat{i}+2\hat{j}+4\hat{k}}{2}
=> \vec{AD} = 2\hat{i}+\hat{j}+2\hat{k}
Thus, the length of the median is,
=> |\vec{AD}| = \sqrt{2^2+1^2+2^2}
=> |\vec{AD}| = \sqrt{9}
=> |\vec{AD}| = 3  units

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