A vector v⃗ in the first octant is inclined to the x-axis at 60°, to the y-axis at 45 and to the z-axis at an acute angle. If a plane passing through the points (√2,-1,1) and (a, b, c), is normal to v⃗, then :

Solution

Maths sir · 4 steps

Find the third direction cosine from the sum of squares, then the plane's equation follows at once.

1. Two known cosines. The direction cosines of v⃗ are l = cos 60° = 1/2 and m = cos 45° = 1/√2. The third one, n, comes from l² + m² + n² = 1.

   n² = 1 - 1/4 - 1/2 = 1/4 ⇒ n = 1/2

2. Pick the sign. The angle with the z-axis is acute, so n = +1/2. The normal is (1/2, 1/√2, 1/2). Multiply by 2 to get direction ratios (1, √2, 1).
3. Write the plane. The plane is x + √2 y + z = d. It passes through (√2, -1, 1), so d = √2 - √2 + 1 = 1, and the plane is x + √2 y + z = 1.
4. So. The point (a, b, c) lies on this plane, so a + √2 b + c = 1, which is option (d).

The idea to keep. Two angles of a line fix the third through l² + m² + n² = 1. Use the acute or obtuse condition to choose the sign. Before you pick an option, check that the given point satisfies the plane.

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