Let $\vec{a}=2 \hat{i}+\hat{j}-2 \hat{k}, \vec{b}=\hat{i}+\hat{j}$ and $\vec{c}$ be a vector such that…
Let $\vec{a}=2 \hat{i}+\hat{j}-2 \hat{k}, \vec{b}=\hat{i}+\hat{j}$ and $\vec{c}$ be a vector such that $|\vec{c}-\vec{a}|=3$. If $\overrightarrow{\mathrm{p}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$, then the angle between $\overrightarrow{\mathrm{p}}$ and $\overrightarrow{\mathrm{c}}$ is $\frac{\pi}{6}$ and $|\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{c}}|=3$. Thus $\vec{a} . \vec{c}$ is equal to-
$\frac{1}{8}$
$1$
$2$
$4$
Solution
Given $\vec{a}=2 \hat{i}+\hat{j}-2 \widehat{k}, \vec{b}=\hat{i}+\hat{j},|\vec{c}-\vec{a}|=3$
$\vec{p}=\vec{a} \times \vec{b},|\vec{p} \times \vec{c}|=3$ and angle between $\vec{p}$ and $\vec{c}$ is $\frac{\pi}{6}$
Here, $\vec{p}=\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \widehat{k} \\ 2 & 1 & -2 \\ 1 & 1 & 0\end{array}\right|=2 \hat{i}-2 \hat{j}+\widehat{k} \Rightarrow|\vec{p}|=3$
also $|\vec{a}|=3$ and $|\vec{b}|=\sqrt{2}$
$\begin{aligned} & \text { Now } \because|\vec{p} \times \vec{c}|=3 \Rightarrow|\vec{p}||\vec{c}| \sin \frac{\pi}{6}=3 \\ & \Rightarrow 3|\vec{c}| \times \frac{1}{2}=3 \\ & \Rightarrow|\vec{c}|=2 \\ & \text { and }|\vec{c}-\vec{a}|=3 \Rightarrow|\vec{c}-\vec{a}|^2=9 \\ & \Rightarrow|\vec{c}|^2+|\vec{a}|^2-2 \vec{c} \cdot \vec{a}=9 \\ & \Rightarrow 4+9-2 \vec{a} \cdot \vec{c}=9\end{aligned}$
$\Rightarrow \vec{a} \cdot \vec{c}=2$