Question

Vectors $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ are of the same length and when taken pairwise, they form equal angles. If $\overrightarrow{a}=\hat{i}+\hat{j}$ and $\overrightarrow{b}=\hat{j}+\hat{k}$, then vector $\overrightarrow{c}$ can be

• A. $\hat{i}+\hat{k}$
• B. $\frac{1}{3}\hat{i}-\frac{4}{3}\hat{j}-\frac{1}{3}\hat{k}$
• C. $-\frac{1}{3}\hat{i}+\frac{4}{3}\hat{j}-\frac{1}{3}\hat{k}$
• D. $\hat{i}+2\hat{j}+\hat{k}$

Answer 1

Answer: (A), (C)

$-\frac{1}{3}\hat{i}+\frac{4}{3}\hat{j}-\frac{1}{3}\hat{k}$

Let $\overrightarrow{c}=x\hat{i}+y\hat{j}+z\hat{k}$
Then $|\overrightarrow{a}|=|\overrightarrow{b}|=|\overrightarrow{c}|\Rightarrow x^2+y^2+z^2=2$
It is given that the angles between the vectors taken in pairs are equal, say $\theta$.
Therefore, $\cos \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}=\frac{0+1+0}{\sqrt{2}\sqrt{2}}=\frac{1}{2}$
$\Rightarrow \frac{\overrightarrow{a}.\overrightarrow{c}}{|\overrightarrow{a}||\overrightarrow{c}|}=\frac{1}{2}$ and $\frac{\overrightarrow{b}.\overrightarrow{c}}{|\overrightarrow{b}||\overrightarrow{c}|}=\frac{1}{2}$
$\Rightarrow \frac{x+y}{2}=\frac{1}{2}$ and $\frac{y+z}{2}=\frac{1}{2}$
$\Rightarrow x+y=1$ and $y+z=1$
$\Rightarrow y=1-x$ and $z=1-y=1-(1-x)=x$

$x^2+y^2+z^2=2\Rightarrow x^2+(1-x{)}^2+x^2=2$
$\Rightarrow (3x+1)(x-1)=0$
$\Rightarrow x=1,-\frac{1}{3}$
Now, $y=1-x$
$\Rightarrow y=0$ for $x=1$ and $y=\frac{4}{3}$ for $x=-\frac{1}{3}$
Hence, $\overrightarrow{c}=\hat{i}+\hat{k}$ or $\overrightarrow{c}=-\frac{1}{3}\hat{i}+\frac{4}{3}\hat{j}-\frac{1}{3}\hat{k}$