Question

A person goes $2km$ east, then $3km$ north, then $4km$ west and then $1km$ 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 $y=\frac{20}{5}=4$
$\therefore \vec{B}=\left(3,4\right)$