Class 12 Maths Chapter 10: Vector Algebra worksheet with answer key

Answer: (b) −6

→QP = →OP − →OQ = (3 + 2)î + (−1 − 5)ĵ + (4 − 1)k̂ = 5î − 6ĵ + 3k̂, so the ĵ-component is −6. 6 is the component of →PQ; 4 comes from adding the ĵ-coordinates.

Answer: (a) −67

l² + m² + n² = 1 gives n² = 1 − 449 − 949 = 3649, so n = ±67; with n < 0, n = −67. 67 ignores the condition n < 0.

Answer: (c) ĵ − î

k̂ × î = ĵ and ĵ × k̂ = î, so k̂ × î − ĵ × k̂ = ĵ − î. Reversing either product changes a sign.

Answer: (b) −98

|→a| = √(4 + 9 + 36) = 7. Opposite directions mean θ = π, so →a · →b = |→a||→b| cos π = −7 × 14 = −98. Forgetting cos π = −1 gives 98; 0 would need perpendicular vectors.

Answer: (a) 2

Perpendicular means the scalar product is 0: 8 − λ − 6 = 0, so λ = 2. Adding −6 as +6 gives 14.

|→a||→b| sin θ = 6√3 and |→a||→b| cos θ = 6, so tan θ = √3 and θ = π3 (cos θ > 0). Then 3|→b| cos π3 = 6 gives |→b| = 4.

• divides to get tan θ = √3, θ = π3 (1)
• |→b| = 4 (1)

In a parallelogram ABCD, →AD = →BC, so →OD = →OA + →OC − →OB = (1 + 4 − 3)î + (2 + 1 + 1)ĵ + (−1 + 5 − 2)k̂ = 2î + 4ĵ + 2k̂. Check: the mid-points of AC and BD are both 52î + 32ĵ + 2k̂.

• uses →AD = →BC (or diagonals bisecting): →OD = →OA + →OC − →OB (1)
• →OD = 2î + 4ĵ + 2k̂ (1)

Direction cosines must satisfy l² + m² + n² = 1. Here cos² 45° + cos² 60° + cos² 120° = 12 + 14 + 14 = 1, so yes; for example the unit vector 1√2î + 12ĵ − 12k̂ has these direction angles.

• states l² + m² + n² = 1 (1)
• computes the sum 1 and concludes yes (1)

|→b| = √(4 + 36 + 9) = 7. Projection of →a on →b = →a · →b|→b| = 2λ + 6 + 127 = 4, so 2λ + 18 = 28 and λ = 5. Then →a = 5î + ĵ + 4k̂, |→a| = √42, →a · →b = 28, and the projection of →b on →a is 28√42 = 2√423.

• |→b| = 7 and 2λ + 187 = 4 (1)
• λ = 5 (1)
• projection of →b on →a = 28√42 (1)

→AB = 2î − ĵ + 2k̂ and →AD = −î + 2ĵ + k̂. →OC = →OB + →AD = 2î + 3ĵ + 3k̂, so C(2, 3, 3). →AB × →AD = |î ĵ k̂; 2 −1 2; −1 2 1| = −5î − 4ĵ + 3k̂, and the area is √(25 + 16 + 9) = 5√2 square units.

• C(2, 3, 3) from →BC = →AD (1)
• →AB × →AD = −5î − 4ĵ + 3k̂ (1)
• area 5√2 square units (1)

(→a − →d) × (→b − →c) = →a × →b − →a × →c − →d × →b + →d × →c. Using →d × →b = −→b × →d and →d × →c = −→c × →d, this is →a × →b − →a × →c + →b × →d − →c × →d. Substituting →a × →b = →c × →d and →a × →c = →b × →d gives →c × →d − →b × →d + →b × →d − →c × →d = →0. Both vectors are non-zero, so their vector product being →0 means they are parallel.

• expands (→a − →d) × (→b − →c) by distributivity (1)
• uses anti-commutativity and the two given equalities to get →0 (1)
• concludes parallel since both are non-zero (1)