Question

The vectors $\overline{a},\overline{b},\overline{c}$ are equal in length and taken pairwise they make equal angles. If $\overline{a}=\overline{i}+\overline{j}$, $\overline{b}=\overline{j}+\overline{k}$ and $\overline{c}$ makes an obtuse angle with $x$-axis, then $\overline{c}=$

• A. $\overline{i}+\overline{k}$
• B. $-\overline{i}+4\overline{j}-\overline{k}$
• C. $\frac{1}{3}(-\overline{i}+4\overline{j}-\overline{k})$
• D. $\frac{1}{3}(\overline{i}-4\overline{j}+\overline{k})$

Answer 1

The correct option is C $\frac{1}{3}(-\overline{i}+4\overline{j}-\overline{k})$

$\left|\overrightarrow{c}\right|=\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=\sqrt{2}$
$\overrightarrow{a}-\overrightarrow{b}=\overrightarrow{b}-\overrightarrow{c}=\overrightarrow{c}-\overrightarrow{a}$
$\overrightarrow{a}-\overrightarrow{b}=\left|a\right|\left|\overrightarrow{b}\right|\cos \theta$
$\frac{1}{2}=cos\theta$
$\theta ={60}^0$
$(\overrightarrow{c}-\overrightarrow{b})=1$ $\overrightarrow{c}.\overrightarrow{a}=1$
Let $\overrightarrow{c}=d\hat{i}+m\hat{j}+x\hat{k}$
$\overrightarrow{c}-\overrightarrow{b}=m+x=1$.......1
$\overrightarrow{c}-\overrightarrow{a}=d+m=1$.......2
$d^2+m^2+2^2=2$.......3
Solve $1,2,3$
$d=\frac{-1}{3}=x$
$m=\frac{4}{3}$
$\overrightarrow{c}=\frac{1}{3}(-\hat{i}+4\hat{j}-\hat{k})$