Question

The direction ratios of normal to the plane through the points \((0, -1, 0)\) and \((0, 0, 1)\) and making an angle \(\frac{\pi}{4}\) with the plane \(y - z + 5 = 0\) are :

• A. \(2, -1, 1\)
• B. \(2\sqrt{3}, 1, -1\)
• C. \(\sqrt{2}, 1, -1\)
• D. \(\sqrt{2}, -\sqrt{2}\)

Hint:

Let the normal be \((a, b, c)\). It is perpendicular to the direction of the chord \((0,1,1)\) joining the two points, giving \(b + c = 0\); then use the angle formula with the normal \((0, 1, -1)\).

Answer 1

The Correct Option is C

Let the normal of the plane be \((a, b, c)\).
The plane contains both points, so it contains the direction \((0, 0, 1) - (0, -1, 0) = (0, 1, 1)\). Hence \(0\cdot a + b + c = 0 \Rightarrow c = -b\).
The angle between two planes equals the angle between their normals; the other normal is \((0, 1, -1)\):
\(\cos\frac{\pi}{4} = \frac{|b - c|}{\sqrt{2}\sqrt{a^2 + b^2 + c^2}}\Rightarrow \frac{1}{\sqrt{2}} = \frac{|2b|}{\sqrt{2}\sqrt{a^2 + 2b^2}}\).
So \(a^2 + 2b^2 = 4b^2 \Rightarrow a^2 = 2b^2 \Rightarrow a = \pm\sqrt{2}\,b\).
Taking \(b = 1\): \((a, b, c) = (\sqrt{2}, 1, -1)\).
Answer: (3) \(\sqrt{2}, 1, -1\)