The direction cosines of the normal to the plane $x-y+z=4$ are : |
$-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$ $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$ $\frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$ $\frac{-1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$ |
$\frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$ |
for plane : Ax + By + Cz = D → its normal vector is $A\hat{i} + B\hat{j} + C\hat{k}$ so for x - y + z = 4 ⇒ $\vec{n} = \hat{i} - \hat{j} + \hat{k}$ as $|\vec{n}|$ = $\sqrt{1^2 + (-1)^2 + 1^2}$ = $\sqrt{3}$ $\hat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{\hat{i} - \hat{j} + \hat{k}}{\sqrt{3}}$ So direction cosins are $\frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$ |