If the line of intersection of the planes a x+b y=3 and a x+b y+c z=0, a 0 makes an angle…

Question

If the line of intersection of the planes $a x+b y=3$ and $a x+b y+c z=0$, a $>0$ makes an angle $$30^{\circ}$$ with the plane $y-z+2=0$, then the direction cosines of the line are :

Accepted Answer

${P_1}:ax + by + 0z = 3$, normal vector : $${\overrightarrow n _1} = (a,b,0)$$

${P_2}:ax + by + cz = 0$, normal vector : $${\overrightarrow n _2} = (a,b,c)$$

Vector parallel to the line of intersection $$ = {\overrightarrow n _1} \times {\overrightarrow n _2}$$

$${\overrightarrow n _1} \times {\overrightarrow n _2} = (bc, - ac,0)$$

Vector normal to $$0\,.\,x + y - z + 2 = 0$$ is $${\overrightarrow n _3} = (0,1, - 1)$$

Angle between line and plane is 30$$^\circ$$

$$ \Rightarrow \left| {{{0 - ac + 0} \over {\sqrt {{b^2}{c^2} + {c^2}{a^2}} \sqrt 2 }}} \right| = {1 \over 2}$$

$$ \Rightarrow {a^2} = {b^2}$$

Hence, $${\overrightarrow n _1} \times {\overrightarrow n _2} = (ac, - ac,0)$$

Direction ratios $$ = \left( {{1 \over {\sqrt 2 }}, - {1 \over {\sqrt 2 }},0} \right)$$

Solutions · 1