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° with the plane y-z+2=0, then the direction cosines of the line are :

Solution

Maths sir · 4 steps

Subtract the planes: z is constant, so the direction is (l, m, 0), and the 30° fixes only |m|.

1. Line stays flat. Subtract ax + by = 3 from ax + by + cz = 0 to get cz = -3. So z is constant along the line, and its direction cosines are (l, m, 0) with l² + m² = 1.
2. Line and plane. The angle between a line and a plane is the complement of the angle between the line and the plane's normal, so it uses sine. The normal of y - z + 2 = 0 is (0, 1, -1), with length √2.

   sin 30° = (|l · 0 + m · 1 + 0 · (-1)|)/(1 × √2) = |m|/√2

3. Find l and m. 1/2 = |m|/√2 gives |m| = 1/√2, and then l² = 1 - 1/2, so l = 1/√2 after choosing the positive sense. The line is also perpendicular to (a, b, 0), so al + bm = 0, which gives b = ∓ a. The condition a > 0 does not fix the sign of b, so both signs of m survive.
4. So. The direction cosines are 1/√2, ±1/√2, 0, which is option (b).

The idea to keep. Line with plane means sine, measured against the plane's normal: sin θ = (|d⃗ · n⃗|)/|d⃗||n⃗|. If the data fixes only |m|, keep both signs.

Answer from the NTA final answer key. Solution by PrepWiser.