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Class 12 RD Sharma Solutions – Chapter 23 Algebra of Vectors – Exercise 23.6 | Set 1

  • Last Updated : 28 Mar, 2021

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}

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=> |\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}




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