<p>The direction ratios of normal to the plane through the points (0, –1, 0) and (0, 0, 1) and making an angle \(\pi/4\) with the plane \(y - z + 5 = 0\) are:</p>
<p>2, –1, 1</p>
<p>\(2, \sqrt{2}, -\sqrt{2}\)</p>
<p>\(2\sqrt{3}, 1, -1\)</p>
<p>\(\sqrt{2}, 1, -1\)</p>
Step-by-Step Solution
Key Concept: The normal to the required plane must be perpendicular to the vector joining the two given points, and the angle between two planes equals the angle between their normals, using cos(π/4) = 1/√2.
Step 1: Let the normal to the required plane be n = (a, b, c). The plane passes through (0, –1, 0) and (0, 0, 1), so the vector joining them is v = (0, 1, 1). Step 2: Since the plane contains both points, its normal must be perpendicular to v : 0·a + 1·b + 1·c = 0 ⟹ b + c = 0 ⟹ c = –b Step 3: The normal to plane y – z + 5 = 0 is n_1 = (0, 1, –1). The angle between planes is π/4: cos(π/4) = | n · n_1 |/(| n || n_1 |) 1/√2 = |0·a + b·1 + c·(–1)|/(√(a^2 + b^2 + c^2)·√2) 1/√2 = |b – c|/(√(a^2 + b^2 + c^2)·√2) Step 4: Substitute c = –b: 1/√2 = |b – (–b)|/(√(a^2 + b^2 + b^2)·√2) = 2|b|/(√(a^2 + 2b^2)·√2) 1 = 2|b|/√(a^2 + 2b^2) a^2 + 2b^2 = 4b^2 ⟹ a^2 = 2b^2 ⟹ a = ±√2·b Step 5: Taking b = 1, c = –1, a = √2 (or a = –√2), the direction ratios are (√2, 1, –1) or proportionally (√2, 1, –1) . ∴ Answer: B
Correct Answer: B