Let the vectors a=-i+j+3k and b=i+3j+k.

Question

Let the vectors $\vec{a}=-\hat{i}+\hat{j}+3\hat{k}$ and $\vec{b}=\hat{i}+3\hat{j}+\hat{k}$. For some $\lambda ,\mu \in R$, let $\vec{c}=\lambda \vec{a}+\mu \vec{b}$.
If $\vec{c}\cdot (3\hat{i}-6\hat{j}+2\hat{k})=10$ and $\vec{c}\cdot (\hat{i}+\hat{j}+\hat{k})=-2$, then $|\vec{c}{|}^{2}$ is equal to :

Accepted Answer

$\vec{c}=\lambda \vec{a}+\mu \vec{b}$

$\vec{c}=(\mu -\lambda )\hat{i}+(\lambda +3\mu )\hat{j}+(3\lambda +\mu )\hat{k}$

$\vec{c}\cdot (3\hat{i}-6\hat{j}+2\hat{k})=10$

$3(\mu -\lambda )-6(\lambda +3\mu )+2(3\lambda +\mu )=10$

$-3\lambda -13\mu =10\, \ldots (1)$

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

$5\mu +3\lambda =-2\, \ldots (2)$

Solving (1) \& (2)

$\lambda =1,\mu =-1$

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

$|\vec{c}{|}^{2}=12$

Solutions · 1