Question

If the line of intersection of the planes ax + by = 3 and ax + by + cz = 0, a> 0 makes an angle 30° with the plane y – z + 2 = 0, then the direction cosines of the line are :

• A. $\frac{1}{\sqrt2}, \frac{1}{\sqrt2}, 0$
• B. $\frac{1}{\sqrt2}, -\frac{1}{\sqrt2}, 0$
• C. $\frac{1}{\sqrt5}, -\frac{2}{\sqrt5}, 0$
• D. $\frac{1}{2}, -\frac{\sqrt3}{2}, 0$

Answer 1

Answer: (B)

$\frac{1}{\sqrt2}, -\frac{1}{\sqrt2}, 0$

Answer 2

The correct answer is (B) : $\frac{1}{\sqrt2}, -\frac{1}{\sqrt2}, 0$
$P_1 : ax+by+0z = 3$, normal vector : $\vec{n}_1 = (a,b,0)$
$P_2 : ax+by+cz = 0$, normal vector $: \vec{n}_2 = (a,b,c)$
Vector parallel to the line of intersection $= \stackrel{→}{n_1}×\stackrel{→}{n_2}$
$\stackrel{→}{n_1}×\stackrel{→}{n_2}$ $= (bc, -ac, 0)$
Vector normal to $0.x+y-z+2 = 0$ is $\stackrel{→}{n_3} = (0,1,-1)$
Angle between line and plane is 30°
$⇒ |\frac{0-ac+0}{\sqrt{b^2c^2+c^2a^2}\sqrt2}| = \frac{1}{2}$
$⇒ a^2 = b^2$
Hence, $\stackrel{→}{n_1}×\stackrel{→}{n_2}$ $= (ac, -ac, 0)$
Direction ratios $= (\frac{1}{\sqrt2}, -\frac{1}{\sqrt2}, 0)$