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

The Correct Option is B

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 \(= (\frac{1}{\sqrt2}, -\frac{1}{\sqrt2}, 0)\)