Question

Let $\vec{x},\vec{y}$ and $\vec{z}$ be three vectors each of magnitude $\sqrt{2}$ and the angle between each pair of them is $\frac{\pi }{3}$. If $\vec{a}$ is a non-zero vector perpendicular to $\vec{x}$ and $\vec{y}\times z$ and $\vec{b}$ is a non-zero vector perpendicular to $\vec{y}$ and $\vec{z}\times \vec{x}$, then

• A. $\vec{b}=(\vec{b}\cdot \vec{z})(\vec{z}-\vec{x})$
• B. $\vec{a}=(\vec{a}\cdot \vec{y})(\vec{y}-\vec{z})$
• C. $\vec{a}\cdot \vec{b}=-(\vec{a}\cdot \vec{y})(\vec{b}\cdot \vec{z})$
• D. $\vec{a}=(\vec{a}\cdot \vec{y})(\vec{z}-\vec{y})$

Answer 1

Answer: (A), (B), (C)

$\vec{a}$ is in direction of $\vec{x}\times (\vec{y}\times \vec{z})$
i.e. $(\bar{x}\cdot \bar{z})\vec{y}-(\bar{x}\cdot \vec{y})\bar{z}$
$\Rightarrow \vec{a}={\lambda }_1\left[2\times \frac{1}{2}(\bar{y}-\bar{z})\right]$
$\vec{a}={\lambda }_1(\vec{y}-\vec{z})\dots$ (1)
Now $\vec{a}\cdot \vec{y}={\lambda }_1(\vec{y}\cdot \vec{y}-\vec{y}\cdot \vec{z})$
$={\lambda }_1(2-1)\Rightarrow {\lambda }_1=\vec{a}\cdot \vec{y}\dots$(2)
From (1) and $(2),\vec{a}=\vec{a}\cdot \vec{y}(\vec{y}-\vec{z})$
Similarly, $\vec{b}=(\vec{b}\cdot \vec{z})(\vec{z}-\vec{x})$
Now, $\vec{a}\cdot \vec{b}=(\vec{a}\cdot \vec{y})(\vec{b}\cdot \vec{z})[(\vec{y}-\vec{z})\cdot (\vec{z}-\vec{x})]$
$=(\vec{a}\cdot \vec{y})(\vec{b}\cdot \bar{z})[1-1-2+1]$
$=-(\vec{a}\cdot \vec{y})(\vec{b}\cdot \vec{z})$