Question

$\overrightarrow{u},\overrightarrow{v},\overrightarrow{w}$ be such that $\left|\overrightarrow{u}\right|=1,\left|\overrightarrow{v}\right|=2,\left|\overrightarrow{w}\right|=3.$ If the projection of $\overrightarrow{v}$ along $\overrightarrow{u}$ is equal to that of $\overrightarrow{w}$ along $\overrightarrow{u}$ and vectors $\overrightarrow{v},\overrightarrow{w}$ are perpendicular to each other then $|\overrightarrow{u}-\overrightarrow{v}+\overrightarrow{w}|$ equals

• A. $2$
• B. $\sqrt{7}$
• C. $\sqrt{14}$
• D. $14$

Answer 1

The correct option is C $\sqrt{14}$

$|\overrightarrow{u}|=1,|\overrightarrow{v}|=2,|\overrightarrow{w}|=3$

According to the information given in the question, we can infer that $|\overrightarrow{w}|\times$ the angle between $\overrightarrow{u}$ and $\overrightarrow{w}$ and $|\overrightarrow{v}|\times$ the angle between $\overrightarrow{u}$ and $\overrightarrow{v}$ are the same.

Now, let us take $\overrightarrow{w}-\overrightarrow{v}$ as $\overrightarrow{b}$

We need to find

$|\overrightarrow{u}+\overrightarrow{b}|$

$=\sqrt{u^2+b^2+2(\overrightarrow{u}.\overrightarrow{b})}$

$=\sqrt{1+b^2+0}$

$=\sqrt{1+|\overrightarrow{v}{|}^2+|\overrightarrow{w}{|}^2+0}$

$=\sqrt{1+4+9}$

$=\sqrt{1}4$