Question

If $A$ has coordinates $(-1,5)$ and $\overrightarrow{a}$ is a position vector whose tip is $(1,-3)$. Then the coordinates of the point B such that $\overrightarrow{AB}=\overrightarrow{a}$ is

• A. $(1,2)$
• B. $(0,2)$
• C. $(0,-2)$
• D. $(2,0)$

Answer 1

The correct option is B $(0,2)$

Let $O$ be the origin and let $P(1,-3)$ be the tip of the position vector $\overrightarrow{a}$.
Then, $\overrightarrow{a}=\overrightarrow{OP}=\hat{i}-3\hat{j}$
Let the coordinates of $B$ be $(x,y)$ and $A$ has coordinates $(-1,5)$.
$\therefore \overrightarrow{AB}=\text{Position vector of}B-\text{Position vector of}A\Rightarrow \overrightarrow{AB}=(x\hat{i}+y\hat{j})-(-\hat{i}+5\hat{j})\Rightarrow \overrightarrow{AB}=(x+1)\hat{i}+(y-5)\hat{j}$
Now, $\overrightarrow{AB}=\overrightarrow{a}$ $\Rightarrow (x+1)\hat{i}+(y-5)\hat{j}=\hat{i}-3\hat{j}$ $\Rightarrow x+1=1$ and $y-5=-3$
$\Rightarrow x=0$ and $y=2$
Hence, the coordinates of $B$ are $(0,2)$.