Question

A person goes $ 2\, km $ east, then $ 3 \,km $ north, then $ 4 \,km $ west and then $ 1\, km $ north, starting from the origin. This point is taken as vector $ \vec{A} $ . The vector $ \vec{B} $ such that $ 3 \vec{A} + 5 \vec{B} = (9, 32) $ , is

• A. $ (4 , 3) $
• B. $ (-3, 4) $
• C. $ (-4, 3) $
• D. $ (3, 4) $

Answer 1

Answer: (D)

$\vec{A}=2\hat{i}+4\hat{j}$
Let $\vec{B}=x\hat{i}+y \hat{j} $
Now, $3 \vec{A}+5\vec{B}=9\backslash, \hat{i}+32\,\hat{j}$
$\Rightarrow -6 \hat{i}+12\hat{j}+5x \hat{i}+5y \hat{j}$
$=9 \,\hat{i}+32\,\hat{j}$
Comparing coefficients, we get
$5x-6=9$ and $5y+12=32$
$\Rightarrow x=\frac{15}{5}=3$ and $y=\frac{20}{5}=4 $
$\therefore \vec{B}=\left(3,4\right)$