Question

A vector $n$ of magnitude $8$ units is inclined to $x$-axis at ${45}^0$, $y$-axis at ${60}^0$ and at an acute angle with $z$-axis. If a plane passes through a point $(\sqrt{2},-1,1)$ and is normal to $\overrightarrow{n}$, find its equation in vector form

• A. $r.(\sqrt{2}i+j+k)=2$
• B. $r.(\sqrt{2}i+j+k)=4$
• C. $r.(\sqrt{2}i-j+k)=2$
• D. $r.(\sqrt{2}i-j+k)=4$

Answer 1

The correct option is A $r.(\sqrt{2}i+j+k)=2$

Let $\gamma$ be the angle made by $\overrightarrow{n}$ with z-axis, then direction cosines of $\overrightarrow{n},$ are

$l=\cos {45}^0=\frac{1}{\sqrt{2}},m=\cos {60}^0=\frac{1}{2}$ and $n=\cos \gamma$

$\therefore l^2+m^2+n^2=1\Rightarrow \frac{1}{2}+\frac{1}{4}+n^2=1$

$\Rightarrow n^2=\frac{1}{4};\Rightarrow n=\frac{1}{2}$ $($ neglecting $n=-\frac{1}{2}$ as $\gamma$ is acute, $\therefore n>0)$

We have $\left|\overrightarrow{n}\right|=8$

$\therefore \overrightarrow{n}=\left|\overrightarrow{n}\right|(l\hat{i}+m\hat{j}+n\overrightarrow{k})$

$\Rightarrow \overrightarrow{n}=8(\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{\sqrt{2}}\hat{k})$

$\Rightarrow 4\sqrt{2}\hat{i}+4\hat{l}+4\hat{k}$

The required plane passes through the point $(\sqrt{2},-1,1)$ having position vector $\overrightarrow{a}=\sqrt{2}\hat{i}-\hat{j}+\hat{k}$.

So, its vector equation is: $(\overrightarrow{r}-\overrightarrow{a}).\overrightarrow{n}=0$

$\Rightarrow \overrightarrow{r}.\overrightarrow{n}=\overrightarrow{a}.\overrightarrow{n}$

$\Rightarrow \overrightarrow{r}.(4\sqrt{2}\hat{i}+4\hat{j}+4\hat{k})=(\sqrt{2}\hat{i}-\hat{j}+\hat{k}).(4\sqrt{2}\hat{i}+4\hat{j}+4\hat{k})$

$\Rightarrow \overrightarrow{r}.(4\sqrt{2}\hat{i}+4\hat{j}+4\hat{k})$

$\Rightarrow \overrightarrow{r}.(\sqrt{2}\hat{i}+\hat{j}+\hat{k})=2$