Question

A vector $\overrightarrow{n}$ is inclined to x-axis at ${45}^0$, to y-axis at ${64}^0$ and at an acute angle to z-axis. If $\overrightarrow{n}$ is a normal to a plane through the point $(\sqrt{2},-1,1),$ then the equation of the plane is:

Answer 1

$\alpha ={45}^0,\beta ={64}^0,\gamma =?\cos \alpha =\frac{1}{\sqrt{2}},\cos \beta =\cos {64}^0=0.438$

${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1\Rightarrow \gamma ={56}^0$

$\hat{n}=\cos \alpha \hat{i}+\cos \beta \hat{j}+\cos \gamma \hat{k}$

$\hat{n}=\frac{1}{\sqrt{2}}\hat{i}+0.438\hat{j}+0.56\hat{k}$

equation of plane is:$\overrightarrow{r}.\hat{n}=p$

$\overrightarrow{r}.\frac{1}{\sqrt{2}}\hat{i}+0.438\hat{j}+0.56\hat{k}=p(\sqrt{2},-1,1)$

$\therefore 1+0.56-0.438=p$

$p=1.122$

equation is $\frac{1}{\sqrt{2}}x+0.438y+0.56z=1.122$