Question

The vectors $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ are of the same length, taken pairwise and they form equal angles. If $\overrightarrow{a}=\hat{i}+\hat{j}$ and $\overrightarrow{b}=\hat{j}+\hat{k}$, then $\overrightarrow{c}$ is equal to

• A. $\hat{i}+\hat{k}$
• B. $\hat{i}+2\hat{j}+3\hat{k}$
• C. $-\hat{i}+\hat{j}+2\hat{k}$
• D. None of these

Answer 1

The correct option is A $\hat{i}+\hat{k}$

$\overrightarrow{c}=x\hat{i}+y\hat{j}+z\hat{k}$

Then, $\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=\left|\overrightarrow{c}\right|\Rightarrow x^2+y^2+z^2=2$ ...(1)

It is given that the angles between the vectors taken in pairs are equal, say $\theta$.

Therefore, $\cos \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}=\frac{0+1+0}{\sqrt{2}\sqrt{2}}=\frac{1}{2}$

Similarly $\frac{\overrightarrow{a}.\overrightarrow{c}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{c}\right|}=\frac{1}{2}$ and $\frac{\overrightarrow{b}.\overrightarrow{c}}{\left|\overrightarrow{b}\right|\left|\overrightarrow{c}\right|}=\frac{1}{2}$

$\Rightarrow \frac{x+y}{\sqrt{2}.\sqrt{2}}=\frac{1}{2}$ and $\frac{y+z}{\sqrt{2}.\sqrt{2}}=\frac{1}{2}$

$\Rightarrow x+y=1$ and $y+z=1$

$\Rightarrow y=1-x$ and $z=1-y=1-(1-x)=x$

$\therefore x^2+y^2+z^2=2$

$\Rightarrow 3x^2-2x-1=0\Rightarrow x=1,-\frac{1}{3}$

Now $y=1-x\Rightarrow y=0$ for $x=1$ and $y=\frac{1}{4}$ for $x=-\frac{1}{3}$

Hence, $\overrightarrow{c}=\hat{i}+0\hat{j}+\hat{k}$ and $\overrightarrow{c}=-\frac{1}{3}\hat{i}+\frac{4}{3}\hat{j}-\frac{1}{3}\hat{k}$