RowQ
The Vault
RowQ
The Vault
CBSE Class 12 Maths · 11 questions · 26 marks
Vectors capture quantities that have both magnitude and direction, and this chapter builds the algebra needed to manipulate them — addition, scalar multiplication, and two very different products. The dot product measures how aligned two vectors are and gives angles and projections, while the cross product produces a new vector perpendicular to both, whose magnitude equals the area it sweeps out. Together they set up the geometry used throughout three-dimensional coordinate geometry in the next chapter.
The magnitude of the vector a = 3î - 4ĵ + 12k̂ is:
Answer
The magnitude is 13. |a| = √(3² + (-4)² + 12²) = √(9+16+144) = √169 = 13.
If a·b = 0 for two nonzero vectors a and b, then a and b are:
Answer
a and b are perpendicular. Since a·b = |a||b|cosθ and both |a| and |b| are nonzero, a·b = 0 forces cosθ = 0, so θ = 90°, meaning a and b are perpendicular.
The area of the parallelogram with adjacent sides a = î + ĵ and b = î - ĵ is:
Answer
The area is 2 sq units. a×b = determinant with rows î ĵ k̂, (1,1,0), (1,-1,0) = î(1×0-0×(-1)) - ĵ(1×0-0×1) + k̂(1×(-1)-1×1) = 0î - 0ĵ - 2k̂. |a×b| = √(0²+0²+(-2)²) = 2, which is the area of the parallelogram.
The projection of a = 2î + 3ĵ + k̂ on b = î + 2ĵ + 2k̂ is:
Answer
The projection is 10/3. a·b = 2(1) + 3(2) + 1(2) = 2+6+2 = 10. |b| = √(1²+2²+2²) = √9 = 3. Projection of a on b = (a·b)/|b| = 10/3.
Assertion (A): The vectors a = 2î + 3ĵ - k̂ and b = -6î - 9ĵ + 3k̂ are parallel. Reason (R): Two vectors are parallel if and only if one is a scalar multiple of the other.
Answer
Both A and R are true and R is the correct explanation of A. Reason R correctly states the general condition for parallel vectors. For Assertion A, notice b = -3a, since -3(2î+3ĵ-k̂) = -6î-9ĵ+3k̂, exactly matching b. Since b is a scalar multiple of a, they are parallel — confirming A precisely via the criterion in R.
Find the unit vector in the direction of a = 4î - 3ĵ + 12k̂.
Answer
First find the magnitude: |a| = √(4² + (-3)² + 12²) = √(16+9+144) = √169 = 13. The unit vector is â = a/|a| = (4/13)î - (3/13)ĵ + (12/13)k̂.
Find λ if the vectors a = 2î + λĵ + k̂ and b = 4î - 2ĵ - 2k̂ are perpendicular.
Answer
Two vectors are perpendicular when their dot product is 0. a·b = 2(4) + λ(-2) + 1(-2) = 8 - 2λ - 2 = 6 - 2λ. Setting this to 0: 6 - 2λ = 0, so λ = 3.
Find the angle between the vectors a = î + ĵ - k̂ and b = î - ĵ + k̂.
Answer
a·b = 1(1) + 1(-1) + (-1)(1) = 1 - 1 - 1 = -1. |a| = √(1+1+1) = √3, and |b| = √(1+1+1) = √3. cosθ = (a·b)/(|a||b|) = -1/(√3×√3) = -1/3. So the angle between a and b is θ = cos⁻¹(-1/3).
Find a×b if a = î + 2ĵ + 3k̂ and b = 3î - ĵ + 2k̂, and hence find a unit vector perpendicular to both a and b.
Answer
a×b is the determinant with rows î ĵ k̂, (1,2,3), (3,-1,2). î component: (2×2 - 3×(-1)) = 4+3 = 7. ĵ component: -(1×2 - 3×3) = -(2-9) = 7. k̂ component: (1×(-1) - 2×3) = -1-6 = -7. So a×b = 7î + 7ĵ - 7k̂. |a×b| = √(7²+7²+(-7)²) = √147 = 7√3. The unit vector perpendicular to both a and b is (a×b)/|a×b| = (7î+7ĵ-7k̂)/(7√3) = (î+ĵ-k̂)/√3.
Show that the vectors a = î - 2ĵ + 3k̂, b = -2î + 3ĵ - 4k̂ and c = î - 3ĵ + 5k̂ are coplanar.
Answer
Three vectors are coplanar exactly when their scalar triple product a·(b×c) equals 0. First find b×c, the determinant with rows î ĵ k̂, (-2,3,-4), (1,-3,5). î component: (3×5 - (-4)×(-3)) = 15-12 = 3. ĵ component: -((-2)×5 - (-4)×1) = -(-10+4) = 6. k̂ component: ((-2)×(-3) - 3×1) = 6-3 = 3. So b×c = 3î + 6ĵ + 3k̂. Now a·(b×c) = 1(3) + (-2)(6) + 3(3) = 3 - 12 + 9 = 0. Since the scalar triple product is 0, the vectors a, b and c are coplanar.
Read the following and answer the questions that follow: A drone's displacement from its launch point is represented by the vector d = 3î + 4ĵ + 12k̂ (in metres), and a gust of wind exerts a force vector f = î - 2ĵ + 2k̂ (in newtons) on it. (a) Find |d|, the magnitude of the displacement. (b) Find the unit vector in the direction of d. (c) Find d·f. (d) Find the scalar projection of f on d.
Answer
(a) |d| = √(3² + 4² + 12²) = √(9+16+144) = √169 = 13 metres. (b) The unit vector is d/|d| = (3/13)î + (4/13)ĵ + (12/13)k̂. (c) d·f = 3(1) + 4(-2) + 12(2) = 3 - 8 + 24 = 19. (d) The scalar projection of f on d = (d·f)/|d| = 19/13.
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