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

Answer: 14

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