Question

If $\overrightarrow{n}$ is a vector of magnitude $\sqrt{3}$ and is equally inclined with an acute angle with the coordinate axes. Find the vector and cartesian forms of equation of a plane which passes through $(2,1,-1)$ and is normal to $\overrightarrow{n}$.

Answer 1

Let $\alpha ,\beta ,\gamma$ angles of $\overline{n}$

Given angels are equal So: $\alpha =\beta =\gamma$

ie: $\cos \alpha =\cos \beta =\cos \gamma$

So direction cosines $\Rightarrow l=m=n\Rightarrow l=l=l$

As we know $l^2+m^2+n^2=n\Rightarrow 8l^2=1$

$l=1/\sqrt{3}$

Given magnitude $=\sqrt{3}$

$\overline{n}=\sqrt{3}(\frac{1}{\sqrt{3}}\hat{i}+\frac{1}{\sqrt{3}}\hat{j}+\frac{1}{\sqrt{3}}\hat{k})=\hat{i}+\hat{j}+\hat{k}$

Given $(2,1,-1)\Rightarrow \overline{a}=2\hat{i}+\hat{j}+\hat{k}$ & $\overline{n}=\hat{i}+\hat{j}+\hat{k}$

get plane equation $\overline{r}.n=\overline{a}.\overline{n}$

$\overline{r}(\hat{i}+\hat{j}+\hat{k})=(2\hat{i}+\hat{j}-\hat{k})(\hat{i}+\hat{j}+\hat{k})$

$=2+1-1$

$\overline{r}(\hat{i}+\hat{j}+\hat{k})=2$

$(x\hat{i}+y\hat{j}+z\hat{k})(\hat{i}+\hat{j}+\hat{k})=2$ $[\overline{r}=x\hat{i}+y\hat{j}+z\hat{k}]$

$x+y+z=2$ As we want Question form