Let $a, b$ and $c$ be unit vectors such that $a$ is perpendicular to the plane containing $\mathbf{b}$ and…
Let $a, b$ and $c$ be unit vectors such that $a$ is perpendicular to the plane containing $\mathbf{b}$ and $\mathbf{c}$ and angle between $\mathbf{b}$ and $\mathbf{c}$ is $\frac{\pi}{3}$. Then,$|\mathbf{a}+\mathbf{b}+\mathbf{c}|=$
$3$
$1$
$2$
$4$
Solution
As given $\mathbf{a}$ is perpendicular to $\mathbf{b}$ and $\mathbf{c}$
$\Rightarrow \mathbf{a} \cdot \mathbf{b}=0$ and $\mathbf{a} \cdot \mathbf{c}=0$ and angle between $\mathbf{b}$ and
$c=\frac{\pi}{3}$
$\therefore \mathbf{b} \cdot \mathbf{c}=|\mathbf{b}||\mathbf{c}| \cos \frac{\pi}{3}=1 \cdot 1 \cdot \frac{1}{2}$
$=\frac{1}{2} \quad[\because b$ and $c$ are unit vectors $]$
Now,
$\begin{aligned}|\mathbf{a}+\mathbf{b}+\mathbf{c}|^2=|\mathbf{a}|^2+|\mathbf{b}|^2 & +|\mathbf{c}|^2 \\ & +2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a})\end{aligned}$
$\begin{aligned} & =1+1+1+2\left(0+\frac{1}{2}+0\right)=1+1+1+1=4 \\ & |a+b+c|=\sqrt{4}=2 \text { units }\end{aligned}$