Question

Let $\vec{b}=\hat{i}+\hat{j}+\lambda \hat{k}, \lambda \in R$. If $\vec{a}$ is a vector such that $\vec{ a } \times \vec{ b }=13 \hat{ i }-\hat{ j }-4 \hat{ k } $ and $ \vec{ a } \cdot \vec{ 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 1

$(\vec{a} \times \vec{b}) \cdot \vec{b}=0$
$\Rightarrow 13-1-4 \lambda=0$
$\Rightarrow \lambda=3$
$\Rightarrow \vec{b}=\hat{i}+\hat{j}+3 \hat{k}$
$\Rightarrow \vec{a} \times \vec{b}=13 \hat{i}-\hat{j}-4 \hat{k}$
$\Rightarrow(\vec{a} \times \vec{b}) \times \vec{b}=(13 \hat{i}-\hat{j}-4 \hat{k}) \times(\hat{i}+\hat{j}+3 \hat{k})$
$\Rightarrow-21 \vec{b}-11 \vec{a}=\hat{i}-43 \hat{ j }+14 \hat{k}$
$\Rightarrow \vec{a}=-2 \hat{i}+2 \hat{j}-7 \hat{k}$
Now $(\vec{ b }-\vec{ a }) \cdot(\hat{ k }-\hat{ j })+(\vec{ b }+\vec{ a }) \cdot(\hat{ i }-\hat{ k })=14$