Question

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

• A. $(1,0,1)$
• B. $(0,1,1)$
• C. $(\frac{1}{3},-\frac{4}{3},\frac{1}{3})$
• D. $(-\frac{1}{3},\frac{4}{3},-\frac{1}{3})$

Answer 1

Answer: (A), (D)

$(-\frac{1}{3},\frac{4}{3},-\frac{1}{3})$

Let $\overrightarrow{c}=x\hat{i}+y\hat{j}+z\hat{k}$
So, angle between them : $\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}=\frac{\overrightarrow{b}\cdot \overrightarrow{c}}{|\overrightarrow{b}||\overrightarrow{c}|}=\frac{\overrightarrow{c}\cdot \overrightarrow{a}}{|\overrightarrow{c}||\overrightarrow{a}|}\Rightarrow 1=y+z=x+y\cdots \left(A\right)\because |\overrightarrow{a}|=|\overrightarrow{b}|=|\overrightarrow{c}|\Rightarrow x=z\cdots \left(i\right)$
(from $A$)

and $x^2+y^2+z^2=2$
$\Rightarrow x^2+(1-x{)}^2+x^2=2$
solving for $x,$ we get $x=-\frac{1}{3},1$
So, components are $(1,0,1)$ or $(-\frac{1}{3},\frac{4}{3},-\frac{1}{3})$