Let L be the line of intersection of the planes 2x+3y+z=1 and x+3y+2z=2.

Question

Let $L$ be the line of intersection of the planes $2x+3y+z=1$ and $x+3y+2z=2$. If $L$ makes an angle $\alpha$ with the positive $x$-axis, then $\cos\alpha$ equals

Accepted Answer

Let the direction cosines of line $L$ be $l,m,n,$

then $2l+3m+n=0$ $$\,\,\,\,\,\,\,....\left( i \right)$$

and $$l + 3m + 2n = 0\,\,\,\,\,\,\,\,\,\,....\left( {ii} \right)$$

on solving equation $(i)$ and $(ii),$  we get

$${l \over {6 - 3}} = {m \over {1 - 4}} = {n \over {6 - 3}}$$

$$\,\,\,\,\,\,\,\,\,\, \Rightarrow {l \over 3} = {m \over { - 3}} = {n \over 3}$$

Now $$ \Rightarrow {l \over 3} = {m \over { - 3}} = {n \over 3} = {{\sqrt {{l^2} + {m^2} + {n^2}} } \over {\sqrt {{3^2} + {{\left( { - 3} \right)}^2} + {3^2}} }}$$

As  ${l^2} + {m^2} + {n^2} = 1$

$$\therefore$$ $${l \over 3} = {m \over { - 3}} = {n \over 3} = {1 \over {\sqrt {27} }}$$

$$ \Rightarrow l = {3 \over {\sqrt {27} }} = {1 \over {\sqrt 3 }},\,\,m =  - {1 \over {\sqrt 3 }},n = {1 \over {\sqrt 3 }}$$

Line $L,$  makes an angle $$\alpha $$ with $+ve$ $x$-axis

$$\therefore$$ $$l = \cos \,\alpha \,\,\,\, \Rightarrow \,\,\,\cos \alpha \,\, = {1 \over {\sqrt 3 }}$$

Solutions · 1