$\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}+\hat{j}+\hat{k}$ are two vectors and $\vec{c}$ is a unit vector lying in the plane of $\vec{a}$ and $\vec{b}$. If $\vec{c}$ is perpendicular $\vec{b}$. then $\vec{c} \cdot(\hat{i}+\hat{j}+2 \hat{k})=$

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Aicharya · Accepted Answer

Vector lying in the plane of $\vec{a} \& \vec{b}$ is $\begin{aligned} & \vec{c}=\lambda(\hat{i}-\hat{j}+\hat{k})+\mu(2 \hat{i}+\hat{j}+\hat{k}) \ & \vec{c}=(\lambda+2 \mu) \hat{i}+(\mu-\lambda) \hat{j}+(\mu+\lambda) \hat{k} \ & \vec{c} . \vec{b}=2(\lambda+2 \mu)+\mu-\lambda+\mu+\lambda=0 \ & =6 \mu+2 \lambda=0 \end{aligned}$ $\begin{aligned} & \lambda=-3 \mu \ & \vec{c}=-\mu \hat{i}+4 \mu \hat{j}-2 \mu \hat{k} ;|\vec{c}|=1=\sqrt{\mu^2+4 \mu^2+16 \mu^2} \ & 1=\sqrt{21} \mu \Rightarrow \mu=\frac{1}{\sqrt{21}} ; \vec{c}=\frac{-1}{\sqrt{21}} \hat{i}+\frac{4}{\sqrt{21}} \hat{j}-\frac{2}{\sqrt{21}} \hat{k} \ & \vec{c} .(\hat{i}+\hat{j}+2 \hat{k})=\frac{1}{\sqrt{21}}\end{aligned}$

$\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}+\hat{j}+\hat{k}$ are two vectors and $\vec{c}$ is a unit…

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