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If $\vec{a}$ is a position vector of a point $(1,-3)$ and $A$ is another point $(-1,5)$, then what are the coordinates of the point $B$ such that $\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{a}}$ ?
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The correct answer is:
$(0,2)$
Let the coordinates of $\mathrm{B}$ be $(\mathrm{x}, \mathrm{y})$.
$\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}-3 \hat{\mathrm{j}}$
P.V. of $\mathrm{A}$ is $(-1,5) \mathrm{so}, \overrightarrow{\mathrm{OA}}=\hat{\mathrm{i}}+5 \hat{\mathrm{j}}, \overrightarrow{\mathrm{OB}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}$
$\therefore \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{a}}$
$\Rightarrow(\mathrm{x}+1) \hat{\mathrm{i}}+(\mathrm{y}-5) \overline{\mathrm{j}}=\overline{\mathrm{i}}-3 \overline{\mathrm{j}}$
$\Rightarrow \mathrm{x}+1=1$ and $\mathrm{y}-5=-3$
$\Rightarrow \mathrm{x}=0$ and $\mathrm{y}=2$
$\therefore \quad$ Coordinates of $\mathrm{B}$ are $(0,2)$.
$\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}-3 \hat{\mathrm{j}}$
P.V. of $\mathrm{A}$ is $(-1,5) \mathrm{so}, \overrightarrow{\mathrm{OA}}=\hat{\mathrm{i}}+5 \hat{\mathrm{j}}, \overrightarrow{\mathrm{OB}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}$
$\therefore \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{a}}$
$\Rightarrow(\mathrm{x}+1) \hat{\mathrm{i}}+(\mathrm{y}-5) \overline{\mathrm{j}}=\overline{\mathrm{i}}-3 \overline{\mathrm{j}}$
$\Rightarrow \mathrm{x}+1=1$ and $\mathrm{y}-5=-3$
$\Rightarrow \mathrm{x}=0$ and $\mathrm{y}=2$
$\therefore \quad$ Coordinates of $\mathrm{B}$ are $(0,2)$.
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