Prove the following
Question:

Let $a=\hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$ be two vectors. If $\vec{c}$ is a vector such that $\vec{b} \times \vec{c}=\vec{b} \times \vec{a}$ and $\vec{c} \cdot \vec{a}=0$, then $\vec{c} \cdot \vec{b}$ is equal to:

  1. $-\frac{3}{2}$

  2. $\frac{1}{2}$

  3. $-\frac{1}{2}$

  4. $-1$


Correct Option: , 3

Solution:

$\vec{a} \times(\vec{b} \times \vec{c})=\vec{a} \times(\vec{b} \times \vec{a})$

$\Rightarrow-(\vec{a} \cdot \vec{b}) \vec{c}=(\vec{a} \cdot \vec{a}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{a}$

$\Rightarrow-4 \vec{c}=6(\hat{i}-\hat{j}+\hat{k})-4(\hat{i}-2 \hat{j}-\hat{k})$

$\Rightarrow-4 \vec{c}=2 \hat{i}-2 \hat{j}+2 \hat{k}$

$\Rightarrow \vec{c}=\frac{1}{2}(\hat{i}+\hat{j}+\hat{k})$

$\Rightarrow \vec{b} \cdot \vec{c}=-\frac{1}{2}$

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