Question

$\overline{a},\overline{b},\overline{c}$ are three vectors such that $\left|\overline{a}\right|=1,\left|\overline{b}\right|=2,\left|\overline{c}\right|=3$ and $\overline{b},\overline{c}$ are perpendicular. IF projection of $\overline{b}$ on $\overline{a}$ is the same as the projection of $\overline{c}$ on $\overline{a}$, then $|\overline{a}-\overline{b}+\overline{c}|$

• A. $\sqrt{2}$
• B. $\sqrt{7}$
• C. $\sqrt{14}$
• D. $\sqrt{21}$

Answer 1

Answer: (C)

$\sqrt{14}$

It is given that projection of $\overrightarrow{b}$ on $\overrightarrow{a}$ is same as the projection of $\overrightarrow{c}$ on $\overrightarrow{a}$.

$⟹\overrightarrow{b}.\overrightarrow{a}=\overrightarrow{c}.\overrightarrow{a}$ ...[1]

Also, $\overrightarrow{b}$ and $\overrightarrow{c}$ are perpendicular

$⟹\overrightarrow{b}.\overrightarrow{c}=0$ ...[2]

Now,

$|\overrightarrow{a}-\overrightarrow{b}+\overrightarrow{c}{|}^2=|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2+|\overrightarrow{c}{|}^2-2\overrightarrow{b}.\overrightarrow{a}+2\overrightarrow{c}.\overrightarrow{a}-2\overrightarrow{b}.\overrightarrow{c}$

$|\overrightarrow{a}-\overrightarrow{b}+\overrightarrow{c}{|}^2=|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2+|\overrightarrow{c}{|}^2+0$ ...(Using [1] and [2])

$|\overrightarrow{a}-\overrightarrow{b}+\overrightarrow{c}{|}^2=1+4+9=14$

$⟹|\overrightarrow{a}-\overrightarrow{b}+\overrightarrow{c}|=\sqrt{14}$