Question

The angle $\theta$ between the vector $\overline{p}=\hat{i}+\hat{j}+\hat{k}$ and unit vector along x-axis is

• A. ${\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$
• B. ${\cos }^{-1}\left(\frac{1}{\sqrt{2}}\right)$
• C. ${\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right)$
• D. ${\cos }^{-1}\left(\frac{1}{2}\right)$

Answer 1

The correct option is A ${\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$

The angle between two vectors is given by:

$\cos \theta =\frac{\overrightarrow{A}.\overrightarrow{B}}{\parallel A\parallel \parallel B\parallel }$.

Now, the given vectors are:

$\overrightarrow{A}=\hat{i}+\hat{j}+\hat{k}\Rightarrow \Vert \overrightarrow{A}\Vert =\sqrt{1^2+1^2+1^2}=\sqrt{3}\overrightarrow{B}=\hat{i}\Rightarrow \Vert \overrightarrow{B}\Vert =\sqrt{1^2}=1$.

$\Rightarrow \overrightarrow{A}.\overrightarrow{B}=1$

Thus, we have: $\theta ={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$