Question

The vector $\overrightarrow{a}$,$\overrightarrow{b}$ and $\overrightarrow{c}$ are equal in length and taken pairwise, they make equal angles. If$\overrightarrow{a}=\hat{i}+\hat{j}$,$\overrightarrow{b}=\hat{j}+\hat{k}$, and $\overrightarrow{c}$ makes an obtuse angle with the base vector $\hat{i}$ and $\overrightarrow{c}$ is equal to?

• A. $\hat{i}+2\hat{j}+\hat{k}$
• B. $-\frac{1}{3}\hat{i}+\frac{4}{3}\hat{j}-\frac{1}{3}\hat{k}$
• C. $\hat{i}-\hat{j}+\hat{k}$
• D. None of these

Answer 1

The correct option is B

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

Explanation for correct option

Given, vectors $\overrightarrow{a}=\hat{i}+\hat{j}$, $\overrightarrow{b}=\hat{j}+\hat{k}$, and $\overrightarrow{c}$

$$\begin{gathered} \left|a\right|=\sqrt{2} \\ \left|b\right|=\sqrt{2} \\ \left|c\right|=\sqrt{2} \end{gathered}$$

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

Since, c makes an obtuse angle with $\hat{i}$, then $\overrightarrow{c}.\overrightarrow{i}=x<0$

Also, the angles between the vector are equal.

Therefore,

${\cos }^{-1}\left(\frac{\overrightarrow{a}·\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}\right)={\cos }^{-1}\left(\frac{\overrightarrow{a}·\overrightarrow{c}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{c}\right|}\right)={\cos }^{-1}\left(\frac{\overrightarrow{b}·\overrightarrow{c}}{\left|\overrightarrow{b}\right|\left|\overrightarrow{c}\right|}\right)$
Now,

$$\begin{gathered} \overrightarrow{a}.\overrightarrow{b}=1 \\ \overrightarrow{a}.\overrightarrow{c}=x+y \\ \overrightarrow{b}.\overrightarrow{c}=z+y \end{gathered}$$

${\cos }^{-1}\left(\frac{\overrightarrow{a}·\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}\right)={\cos }^{-1}\left(\frac{1}{2}\right)={\cos }^{-1}\left(\frac{c_1+c_2}{2}\right)={\cos }^{-1}\left(\frac{c_3+c_2}{2}\right)$
Thus,

$x+y=1$ and $z+y=1$, thus,

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

Now by substituting the values we get;

$$\begin{gathered} \Rightarrow 2=x^2+y^2+z^2 \\ \Rightarrow 2=x^2+{(1-x)}^2+x^2 \end{gathered}$$

So, $x=1,-\frac{1}{3}$

Since, $x$ should be less than $0$, therefore, $x=-\frac{1}{3}$

$\therefore z=-\frac{1}{3}$

$y=1-x=1+\frac{1}{3}=\frac{4}{3}$

Therefore, the vector is: $\overrightarrow{c}=-\frac{1}{3}\hat{i}+\frac{4}{3}\hat{j}-\frac{1}{3}\hat{k}$

Hence, the option (B) is correct