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  • Provide solution for RD sharma maths class 12 chapter 22 Algebra of Vectors exercise 22.5 question 8

Provide solution for RD sharma maths class 12 chapter 22 Algebra of Vectors exercise 22.5 question 8

Answers (1)

(9, -4)

Hint: position vector \vec{AB} is given by \vec{AB} =position vector of B- position vector of A

Given: coordinates of A(4, -1) , a \vec{a} is the position vector whose tip i is (5, -3)

Solution:

Let O be the origin of point (0, 0)

Position P(5, -3) be the tip of the position vector \vec{a}

Then the \vec{a} ,\vec{OP} =Position vector of p- position vector of o.

As we know position vector of point (x, y) is given by \left ( x\hat{i} + y\hat{j}\right ) where \hat{i} and \hat{j} are the unit vectors.
Then

\vec{P}=5\hat{i}+-3\hat{j}=5\hat{i}-3\hat{j}\\ \vec{O}=0\hat{i}+0\hat{j} \\ \vec{OP}=5\hat{i}-3\hat{j}-0\hat{i}+0\hat{j}\\ \vec{OP}=5\hat{i}-3\hat{j}-0\hat{i}-0\hat{j}\\ =5-0\hat{i}+-3-0\hat{j}\\ \vec{OP}=5\hat{i}-3\hat{j}

Let the coordinate of B be (x,y) and A has coordinate (4,-1)

Therefore

\vec{A}=4\hat{i}+-1\hat{j}=4\hat{i}-\hat{j}\\ \vec{B}=x\hat{i}+y\hat{j}

AB=\vec{AB} Position vectors of B – position vector of A

\vec{AB}=x\hat{i}+y\hat{j}-\left (4\hat{i}-\hat{j} \right )\\ =x\hat{i}+y\hat{j}-4\hat{i}+\hat{j}\\ =\left (x-4 \right )\hat{i}+\hat{j}\left (y+1 \right )

Now,

\vec{AB}=\vec{a}\\ \left ( x-4 \right )\hat{i}+\left ( y+1 \right )\hat{j}=5\hat{i}-3\hat{j}

Comparing coefficient of \hat{i}  &  \hat{j}
x-4=5\\ x=5+4\\ x=9\\ x+1=-3\\ y=-3-1\\ y=-4

Hence, the coordinate of B are (9,-4)

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