Question

The component of vector $\overrightarrow{A}=a_x\hat{i}+a_y\hat{j}+a_z\hat{k}$ along the direction of $\hat{i}-\hat{j}$ is:

• A. $a_x-a_y+a_z$
• B. $a_x-a_y$
• C. $\frac{a_x-a_y}{\sqrt{2}}$
• D. $(a_x+a_y+a_z)$

Answer 1

The correct option is C $\frac{a_x-a_y}{\sqrt{2}}$

Given, $\overrightarrow{A}=a_x\hat{i}+a_y\hat{j}+a_z\hat{k}$

and let $\overrightarrow{B}=\hat{i}-\hat{j}$

The component of vector $\overrightarrow{A}$ along $\overrightarrow{B}$ is $co{m}_A=Acos\theta$

where $\theta$ angle between vector A and B.

So, $\cos \theta =\frac{\overrightarrow{A}.\overrightarrow{B}}{AB}$

Thus, $co{m}_A=\frac{\overrightarrow{A}.\overrightarrow{B}}{B}=\frac{(a_x\hat{i}+a_y\hat{j}+a_z\hat{k}).(\hat{i}-\hat{j})}{\sqrt{1^2+(-1{)}^2}}=\frac{a_x-a_y}{\sqrt{2}}$