Question

The direction cosines of a line parallel to the planes $3x+4y+z=0$ and $x-2y-3z=5$ are

• A. $(-1,1,-1)$
• B. $(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$
• C. $(-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}})$
• D. no line possible

Answer 1

The correct option is C $(-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}})$

Given equations of the planes are $3x+4y+z=0$ and $x-2y-3z=5$
required line is parallel to the given palnes,i.e perpendicular to the normals to the planes whose direction ratios are
$(3,4,1)$ and $(1,-2,-3)$ respectively
let $(a,b,c)$ be direction ratios of the line.
$\Rightarrow 3a+4b+c=0anda-2b-3c=0$
$\Rightarrow a=-b=c$
$\therefore$ direction cosines of the line are $(-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}})$ or $(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$