Three Dimensional Geometry: JEE Advanced PYQ, 2007–2026

Answer: (D)

The body diagonal has direction b1 = (1,1,1) and passes through O. The face diagonal has direction b2 = (0,−1,1) and passes through (a, a, 0). Then b1 × b2 = (2, −1, −1) and |b1 × b2| = √6. With a2 − a1 = (a, a, 0), the dot product (a2 − a1)·(b1 × b2) = 2a − a = a. The distance is a/√6.

Answer: (A)

From l + m + n = 0, l = −(m + n), so l² = m² + 2mn + n². Setting l² = m² + n² gives mn = 0. If m = 0, then l = −n, and l² + n² = 1 gives (1/√2, 0, −1/√2). If n = 0, then l = −m, giving (1/√2, −1/√2, 0). Then cosθ = (1/√2)(1/√2) + 0 + 0 = 1/2, so θ = 60°.

Answer: (C)

The line of intersection of two planes is perpendicular to both normals, so its direction is n₁ × n₂ = (2, −1, 1) × (1, 1, −1) = (0, 3, 3) ∝ (0, 1, 1). Line M has direction (3 − 1, 1 − 0, 1 + 1) = (2, 1, 2). Then cos θ = |0 + 1 + 2| / (√2 · 3) = 1/√2, so θ = 45°.

Answer: (A)

Write both lines as direction vectors. m = OG = (2, 3, 6) with |m| = √(4 + 9 + 36) = 7. n goes from (2, 0, 0) to (0, 3, 0), so n = (−2, 3, 0) with |n| = √13. The dot product is m·n = −4 + 9 + 0 = 5. Then cos θ = |m·n| / (|m||n|) = 5/(7√13).

Answer: (D)

A general point on the line is (3 + 2t, −2 + 3t, 1 + 6t). Its distance from A is |t|√(2² + 3² + 6²) = 7|t|. Setting 7|t| = 14 gives t = ±2. Since z = 1 + 6t must exceed 1, t = 2, which gives P = (7, 4, 13).

Answer: (A)

Body diagonal: through O(0,0,0) with direction b₁ = (1,1,1). Face diagonal: through A(a,0,0) and C(0,a,0), with direction b₂ = (−1,1,0). The two lines are skew, because the body diagonal meets the plane z = 0 only at O and O is not on AC. b₁ × b₂ = (1·0 − 1·1, 1·(−1) − 1·0, 1·1 − 1·(−1)) = (−1, −1, 2), with magnitude √6. a₂ − a₁ = (a,0,0), so the triple product is −a. d = a/√6.

Answer: (A)

First convert direction ratios to direction cosines: √(4 + 9 + 36) = 7, so the cosines are ±(2/7, −3/7, 6/7). For an acute angle with the positive z-axis, n > 0, so take (2/7, −3/7, 6/7). Moving a distance 14 from A gives B = (1 + 14·2/7, −1 + 14·(−3/7), 2 + 14·6/7) = (1 + 4, −1 − 6, 2 + 12) = (5, −7, 14).

Answer: (A)

Take P = (1 + t, 2t, 3 − t) on L₁ and Q = (3 + 2s, 4 + s, 4 + s) on L₂. Then Q − P = (2 + 2s − t, 4 + s − 2t, 1 + s + t). It must be parallel to (1, 1, 1), so its components are equal. Equating the first two components gives s + t = 2. Equating the last two gives 3 = 3t, so t = 1 and s = 1. Then P = (2, 2, 2) and Q = (5, 5, 5), so Q − P = (3, 3, 3) and the length is 3√3. The shortest distance is only √3, so the answer must exceed it.

Answer: (A)

Three points are collinear when AC is a scalar multiple of AB. AB = (3−1, 0+2, −1−3) = (2, 2, −4) and AC = (4, λ+2, μ−3). The x-components give the factor 2, so λ + 2 = 4 and μ − 3 = −8. Hence λ = 2 and μ = −5, and λ + μ = −3.

Answer: (A)

First rewrite the line in standard form. (−1 − y)/1 = (y + 1)/(−1) and (6 − 2z)/(−4) = (z − 3)/2. So the line is (x − 1)/2 = (y + 1)/(−1) = (z − 3)/2, with direction d = (2, −1, 2) through A(1, −1, 3). A general point is F(1 + 2t, −1 − t, 3 + 2t). Then PF = (2t − 1, 1 − t, 2t − 3). Setting PF · d = 0 gives 2(2t − 1) − (1 − t) + 2(2t − 3) = 9t − 9 = 0, so t = 1 and F = (3, −2, 5). Check: PF = (1, 0, −1) and PF · d = 2 − 2 = 0. F is the midpoint of PP', so P' = 2F − P = (6 − 2, −4 + 2, 10 − 6) = (4, −2, 4).