Question

$a,b$ and $c$ are three vectors such that $|a|=1,|b|=2,|c|=3$ and $b,c$ are perpendicular. If projection of $b$ on $a$ is the same as the projection of $c$ on $a$, then $|a-b+c|$ is equal to

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

Answer 1

Answer: (C)

Given that,
$|a|=1,|b|=2,|c|=3$
$\because b$ and $c$ are perpendicular.
$\therefore b\cdot c=0...(i)$
And projection of $b$ on $a=\frac{a\cdot b}{|a|}$
Projection of $c$ on $a=\frac{a\cdot c}{|a|}$
$\because$ Both projections are same.
$\therefore \frac{a\cdot b}{|a|}=\frac{a\cdot c}{|a|}$
$\Rightarrow a\cdot b=a\cdot c$
Then $|a-b+c{|}^2=|a{|}^2+|b{|}^2+|c{|}^2-2a\cdot b-2b\cdot c+2c\cdot a$
$=(1{)}^2+(2{)}^2+(3{)}^2-2a\cdot b-0+2a\cdot c$
$=1+4+9-2a\cdot b+2a\cdot b=14$
from Eqs. (i) and (ii)]
$\therefore |a-b+c|=\sqrt{14}$