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  • Provide solution for RD sharma maths class 12 chapter 22 Algebra of Vectors exercise 22.5 question 6

Provide solution for RD sharma maths class 12 chapter 22 Algebra of Vectors exercise 22.5 question 6

Answers (1)

the coordinates of D is (-4,-3)

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

Given;

Coordinates of A=(-2,-1)

Coordinates of B=(3,0)

Coordinates of C=(1,-2)

Let the coordinates of D is (x , y)

Since ABCD is a parallelogram,

AB=DC

We have

\vec{AB}=\vec{DC}

As we know position vector of point (x, y) is given by x\hat{i}=y\hat{j}where \hat{i} and \hat{j} are position vector of x, and y direction.

\vec{A}=-2\hat{i}-\hat{j}\\ \vec{B}=3\hat{i}+0\hat{j}\\ \vec{C}=\hat{i}-2\hat{j}\\ \vec{D}=x\hat{i}+y\hat{j}

Then the position vector \vec{AB} is given by
\vec{AB}= position vector of B - position vector of A

\left (3\hat{i}+0\hat{j} \right )-\left (-2\hat{i}-\hat{j} \right )\\ 3\hat{i}+0\hat{j} +2\hat{i}+\hat{j} \\ \vec{AB}=5\hat{i}+\hat{j}

Then we find \vec{DC}

\vec{DC}= position vector C- position vector of D

=\left (\hat{i}-2\hat{j} \right )-\left (x\hat{i}+y\hat{j} \right )\\ =\hat{i}-x\hat{i}-2\hat{j}+y\hat{j} \\ =\hat{i}\left ( 1-x \right )+\hat{j}\left ( -2-y \right )

We have \vec{AB}=\vec{DC}

Then:

5\hat{i}+\hat{j}=\left ( 1-x \right )\hat{i}+\left ( -2-y \right )\hat{j}

Then comparing coefficient of \hat{i} and \hat{j} 

               \begin{matrix} 5=1-x & 1=-2-y\\ x=1-5 &y=-2-1 \\ x=-4 & y=-3 \end{matrix}

Hence the coordinates of D is (-4, -3)

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