Question

The direction cosines of the normal to the plane $x+2y-3z+4=0$ are

• A. $\frac{-1}{\sqrt{14}},\frac{-2}{\sqrt{14}},\frac{3}{\sqrt{14}}$
• B. $\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}$
• C. $\frac{-1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}$
• D. $\frac{1}{\sqrt{14}},\frac{-2}{\sqrt{14}},\frac{-3}{\sqrt{14}}$

Answer 1

The correct option is A $\frac{-1}{\sqrt{14}},\frac{-2}{\sqrt{14}},\frac{3}{\sqrt{14}}$

Clearly, the normal to the plane has DR's \equiv $(1,2,-3)$

$\therefore$ DC's are $\equiv \pm (\frac{1}{\sqrt{1^2+2^2+(-3{)}^2}},\frac{2}{\sqrt{14}},\frac{-3}{\sqrt{14}})$

$=\pm (\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{-3}{\sqrt{14}})$

$\therefore A$