Question

If $\overline{a},\overline{b},\overline{c}$ are mutually perpendicular vectors having magnitudes $1,2,3$ respectively, then $[\overline{a}+\overline{b}+\overline{c}\overline{b}-\overline{a}-\overline{c}]=$

• A. $0$
• B. $6$
• C. $12$
• D. $18$

Answer 1

Answer: (C)

$12$

$|\overline{a}|=1,|\overline{b}|=2,|\overline{c}|=3\overline{a}.\overline{b}=\overline{b}.\overline{c}=\overline{a}.\overline{c}=0$
Now,
$[\overline{a}+\overline{b}+\overline{c}$ $\overline{b}-\overline{a}-\overline{c}]$

$=(\overline{a}+\overline{b}+\overline{c})\left[\right(\overline{b}-\overline{a})$ $\times \overline{c}]$
$=(\overline{a}+\overline{b}+\overline{c})(\overline{b}$ $\times \overline{c}-\overline{a}$ $\times \overline{c})=2[\overline{a}\overline{b}\overline{c}]=2\overline{a}.(\overline{b}$ $\times \overline{c})$
$=2|\overline{a}|.|\overline{b}$ $\times \overline{c}|=2|\overline{a}||\overline{b}||\overline{c}|sin{90}^o=2\times 1\times 2\times 3=12$
Therefore Correct Answer is $C$