Vector Algebra: JEE Advanced PYQ, 2007–2026

Answer: (A)

A vector of magnitude m in the direction of a is m·a/|a|. Here |a| = √(9+16+144) = 13. For the opposite direction use −39/13 = −3, so the vector is −3a = −9i + 12j − 36k.

Answer: (A)

Adding unit vectors gives a rhombus whose diagonal bisects the angle. |a| = √(4+1+4) = 3 and |b| = √(4+9+36) = 7. So the bisector is along a/3 + b/7 = (7a + 3b)/21 = (20î + 2ĵ + 32k̂)/21, which is parallel to 10î + ĵ + 16k̂.

Answer: (A)

Perpendicular vectors have zero dot product. From (a + 3b)·(7a − 5b) = 0 we get 7|a|² + 16 a·b − 15|b|² = 0. From (a − 4b)·(7a − 2b) = 0 we get 7|a|² − 30 a·b + 8|b|² = 0. Subtracting the second from the first gives 46 a·b − 23|b|² = 0, so a·b = |b|²/2. Substituting into the first equation gives 7|a|² + 8|b|² − 15|b|² = 0, so |a| = |b|. Then cos θ = a·b/(|a||b|) = (|b|²/2)/|b|² = 1/2, so θ = 60°.

Answer: (A)

By the triangle inequality the maximum of |a+b+c| is 2 + 3 + 6 = 11, when all three point the same way. For the minimum, |a+b| ranges over [1, 5], and c has magnitude 6, so |a+b+c| ≥ 6 − 5 = 1, attained when a and b point along −c with total magnitude 5. The magnitude varies continuously, so every value in [1, 11] occurs. The integers 1, 2, …, 11 give 11 values.

Answer: (A)

Direction cosines of a vector are its components divided by its magnitude. PQ = (3−1)i + (−1−2)j + (5+1)k = 2i − 3j + 6k, and |PQ| = √49 = 7. So l = 2/7, m = −3/7, n = 6/7, and l² + m² + n² = 1 checks.

Answer: (A)

The z-component is 8 cos 60° = 4. The projection on the xy-plane has magnitude 8 sin 60° = 4√3. Splitting this projection by the 30° angle gives x = 4√3 cos 30° = 6 and y = 4√3 sin 30° = 2√3. The direction cosine with the y-axis is y/|r| = 2√3/8 = √3/4. As a check, 9/16 + 3/16 + 4/16 = 1.

Answer: (B)

d₁ × d₂ = (2·1 − 2·(−2))î + (2·2 − 1·1)ĵ + (1·(−2) − 2·2)k̂ = 6î + 3ĵ − 6k̂, of magnitude 9. The area is ½|d₁ × d₂| = 9/2.

Answer: (A)

A, B, C are collinear if AC = k·AB. AB = (4−2)a + (−1−1)b + (3+1)c = 2a − 2b + 4c. AC = (8−2)a + (λ−1)b + (11+1)c = 6a + (λ−1)b + 12c. The a-coefficients give k = 3, and the c-coefficients agree (12 = 3·4). Because a, b, c are non-coplanar, the b-coefficients must match: λ − 1 = 3(−2) = −6, so λ = −5.

Answer: (C)

By the triangle inequality, |a + b| ≤ |a| + |b| = 7. Then |a + b + c| ≥ |c| − |a + b| ≥ 10 − 7 = 3. Equality holds when a and b are in the same direction and c is in the opposite direction. The vectors cannot sum to zero because 10 > 3 + 4, so 0 is not attainable and the minimum is 3.

Answer: (A)

R lies beyond Q on line PQ with PR = 2RQ, so PQ = RQ and Q is the midpoint of PR. Hence b = (a + r)/2, so r = 2b − a = (−2 − 2, 8 − 1, 4 + 1) = (−4, 7, 5). Then |OR| = √(16 + 49 + 25) = √90 = 3√10. The same result comes from the external section formula with ratio 2:1.