Question

Arrange the value of $|a+b+c|$ in ascending order

$A$ .If $\hat{a},\hat{b},\hat{c}$ are mutually perpendicular unit vectors
$B$. If $\hat{a},\hat{b},\hat{c}$ are vectors of lengths 2,3,4 respectively and $\hat{a},\hat{b},\hat{c}$ are perpendicular to $\hat{b}+\hat{c},\hat{c}+\hat{a},\hat{a}+\hat{b}$ respectively.
$C$. If $\hat{a},\hat{b},\hat{c}$are vectors of lengths 4,4,5 respectively and $\hat{a},\hat{b},\hat{c}$ are perpendicular to $\hat{b}+\hat{c}$,$\hat{c}+\hat{a}$,$\hat{a}+\hat{b}$ respectively.

• A. $A,B,C$
• B. $C,B,A$
• C. $B,C,A$
• D. $B,A,C$

Answer 1

The correct option is A $A,B,C$

In $A$ :
$\hat{a}.\hat{b}=\hat{b}.\hat{c}=\hat{a}.\hat{c}=0$ Hence,$|\hat{a}+\hat{b}+\hat{c}{|}^2=|\hat{a}{|}^2+|\hat{b}{|}^2+|\hat{c}{|}^2$
Hence,$|\hat{a}+\hat{b}+\hat{c}{|}^2$=3
In Option $B$ :
$\hat{a}.\hat{b}+\hat{b}.\hat{c}=\hat{a}.\hat{c}+\hat{b}.\hat{c}=\hat{a}.\hat{b}+\hat{a}.\hat{c}=0$
Hence,$|\hat{a}+\hat{b}+\hat{c}{|}^2=|\hat{a}{|}^2+|\hat{b}{|}^2+|\hat{c}{|}^2=29$
Similarly in Option $C$ :
$|\hat{a}+\hat{b}+\hat{c}{|}^2=|\hat{a}{|}^2+|\hat{b}{|}^2+|\hat{c}{|}^2=57$