Question
Let $\vec{b}=\hat{i}+\hat{j}+\lambda \hat{k}, \lambda \in R$. If $\vec{a}$ is a vector such that $\overrightarrow{ a } \times \overrightarrow{ b }=13 \hat{ i }-\hat{ j }-4 \hat{ k } \quad$ and $\quad \overrightarrow{ a } \cdot \overrightarrow{ b }+21=0$, then $(\vec{b}-\vec{a}) \cdot(\hat{k}-\hat{j})+(\vec{b}+\vec{a}) \cdot(\hat{i}-\hat{k})$ is equal to

Answer

c
$(\overrightarrow{ a } \times \overrightarrow{ b }) \cdot \overrightarrow{ b }=0$

$\Rightarrow 13-1-4 \lambda=0 \Rightarrow \lambda=3$

$\Rightarrow \overrightarrow{ b }=\hat{ i }+\hat{ j }+3 \hat{ k } \Rightarrow \overrightarrow{ a } \times \overrightarrow{ b }=13 \hat{ i }-\hat{ j }-4 \hat{ k }$

$\Rightarrow(\overrightarrow{ a } \times \overrightarrow{ b }) \times \overrightarrow{ b }=(13 \hat{ i }-\hat{ j }-4 \hat{ k }) \times(\hat{ i }+\hat{ j }+3 \hat{ k })$

$\Rightarrow-21 \overrightarrow{ b }-11 \overrightarrow{ a }=\hat{ i }-43 \hat{ j }+14 \hat{ k }$

$\Rightarrow \overrightarrow{ a }=-2 \hat{ i }+2 \hat{ j }-7 \hat{ k }$

Now $(\overrightarrow{ b }-\overrightarrow{ a }) \cdot(\hat{ k }-\hat{ j })+(\overrightarrow{ b }+\overrightarrow{ a }) \cdot(\hat{ i }-\hat{ k })=14$

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