CBSE Important Questions Class 12 Maths Chapter 10 – Vector Algebra

Class 12 Maths Chapter 10 Vector Algebra is part of the CBSE Class 12 Maths syllabus. A vector is a quantity that has both magnitude and direction. Students learn the types of vectors, how to add and subtract vectors, the position vector, direction cosines and direction ratios, the scalar (dot) product, the vector (cross) product, and the section formula for vectors.

These CBSE important questions for Class 12 Maths Chapter 10 cover finding the magnitude and unit vector, direction cosines, the angle between two vectors using the dot product, the projection of one vector on another, the area of a triangle and a parallelogram using the cross product, and the condition for two vectors to be perpendicular or parallel. Every question has a clear step-by-step solution in simple English, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. Define a vector. Name the different types of vectors.

Answer: A vector is a quantity that has both magnitude and direction, such as velocity, force or displacement. A quantity with only magnitude, such as mass or temperature, is a scalar.
Types of vectors:
• Zero vector: magnitude zero, with no definite direction
• Unit vector: magnitude exactly 1
• Equal vectors: same magnitude and same direction
• Collinear vectors: parallel to the same line
• Coplanar vectors: lying in the same plane
• Negative of a vector: same magnitude but opposite direction

Q2. Find the magnitude and the unit vector in the direction of a⃗ = 2î + 3ĵ + 6k̂.

Answer: Magnitude:
|a⃗| = √(2² + 3² + 6²)
= √(4 + 9 + 36)
= √49 = 7
Unit vector: â = a⃗ / |a⃗|
= (2î + 3ĵ + 6k̂)/7, that is (2/7)î + (3/7)ĵ + (6/7)k̂

Q3. Define direction cosines and direction ratios. Find the direction cosines of the vector î + 2ĵ + 3k̂.

Answer: If a vector makes angles α, β and γ with the positive x, y and z axes, then cosα, cosβ and cosγ are its direction cosines, written l, m and n.
The components themselves (a, b, c) are the direction ratios.
They are related by l = a/|r⃗|, m = b/|r⃗|, n = c/|r⃗|, and l² + m² + n² = 1.
For î + 2ĵ + 3k̂:
|r⃗| = √(1 + 4 + 9) = √14
Direction cosines = 1/√14, 2/√14, 3/√14

Q4. Write the formula for the scalar (dot) product and find the angle between a⃗ = î + ĵ + k̂ and b⃗ = î − ĵ + k̂.

Answer: Dot product: a⃗ · b⃗ = |a⃗||b⃗| cosθ
In component form, it is the sum of the products of corresponding components.
Calculation:
a⃗ · b⃗ = (1)(1) + (1)(−1) + (1)(1) = 1
|a⃗| = √3 and |b⃗| = √3
cosθ = 1 / (√3 × √3) = 1/3
θ = cos⁻¹(1/3), which is about 70.5°.

Q5. Find the projection of the vector a⃗ = 2î + 3ĵ + 2k̂ on b⃗ = î + 2ĵ + k̂.

Answer: The projection of a⃗ on b⃗ is given by
Projection = (a⃗ · b⃗) / |b⃗|
Dot product: (2)(1) + (3)(2) + (2)(1) = 2 + 6 + 2 = 10
Magnitude of b⃗: √(1 + 4 + 1) = √6
Projection = 10/√6, which can be rationalised as (10√6)/6 = (5√6)/3.

Q6. Write the condition for two vectors to be perpendicular and parallel.

Answer: Perpendicular: Two non-zero vectors a⃗ and b⃗ are perpendicular when the angle between them is 90°, so cosθ = 0.
This means a⃗ · b⃗ = 0.
Parallel: Two vectors are parallel when the angle between them is 0° or 180°, so sinθ = 0.
This means a⃗ × b⃗ = 0⃗, or equivalently a⃗ = λb⃗ for some scalar λ, which means their components are proportional.

Q7. Find the value of λ for which the vectors 2î + λĵ + k̂ and î − 2ĵ + 3k̂ are perpendicular.

Answer: For perpendicular vectors, the dot product is zero.
(2)(1) + (λ)(−2) + (1)(3) = 0
2 − 2λ + 3 = 0
5 = 2λ
λ = 5/2

Q8. Write the formula for the vector (cross) product. Find a⃗ × b⃗ where a⃗ = î + ĵ + k̂ and b⃗ = 2î + ĵ + 3k̂.

Answer: Cross product: a⃗ × b⃗ = |a⃗||b⃗| sinθ n̂, where n̂ is a unit vector perpendicular to both.
In component form it is evaluated as a 3 × 3 determinant with î, ĵ, k̂ in the first row.
Calculation:
î component: (1)(3) − (1)(1) = 2
ĵ component: −[(1)(3) − (1)(2)] = −1
k̂ component: (1)(1) − (1)(2) = −1
So a⃗ × b⃗ = 2î − ĵ − k̂

Q9. Find the area of the triangle whose two adjacent sides are a⃗ = 3î + ĵ + 4k̂ and b⃗ = î − ĵ + k̂.

Answer: The area of a triangle with adjacent sides a⃗ and b⃗ is ½|a⃗ × b⃗|.
Cross product:
î: (1)(1) − (4)(−1) = 5
ĵ: −[(3)(1) − (4)(1)] = 1
k̂: (3)(−1) − (1)(1) = −4
a⃗ × b⃗ = 5î + ĵ − 4k̂
|a⃗ × b⃗| = √(25 + 1 + 16) = √42
Area = (√42)/2 square units

Q10. Find the area of the parallelogram whose adjacent sides are a⃗ = î + 2ĵ + 3k̂ and b⃗ = 3î − 2ĵ + k̂.

Answer: The area of a parallelogram with adjacent sides a⃗ and b⃗ is |a⃗ × b⃗|.
Cross product:
î: (2)(1) − (3)(−2) = 8
ĵ: −[(1)(1) − (3)(3)] = 8
k̂: (1)(−2) − (2)(3) = −8
a⃗ × b⃗ = 8î + 8ĵ − 8k̂
|a⃗ × b⃗| = √(64 + 64 + 64) = 8√3
Area = 8√3 square units

Q11. Find the position vector of the point which divides the line joining A(2î + 3ĵ) and B(−î + ĵ) in the ratio 2 : 1 internally.

Answer: The section formula for internal division in the ratio m : n is
r⃗ = (m b⃗ + n a⃗) / (m + n)
Here m = 2, n = 1, a⃗ = 2î + 3ĵ and b⃗ = −î + ĵ.
r⃗ = [2(−î + ĵ) + 1(2î + 3ĵ)] / 3
= [−2î + 2ĵ + 2î + 3ĵ] / 3
= 5ĵ/3
So the position vector is (5/3)ĵ.

Q12. Show that the vectors 2î − 3ĵ + 4k̂ and −4î + 6ĵ − 8k̂ are collinear.

Answer: Two vectors are collinear (parallel) if one is a scalar multiple of the other.
Look at the second vector:
−4î + 6ĵ − 8k̂ = −2(2î − 3ĵ + 4k̂)
So b⃗ = −2a⃗, which means λ = −2.
Since one vector is a scalar multiple of the other, the two vectors are collinear.
(The negative value of λ tells us they point in opposite directions.)

Related Study Material on Extramarks

Q1.

If  a ^  is a unit vector and ( x )( x + a ^ )=15, then the value of  | x | is ____. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXafv3ySLgzGmvETj2BSbqefm0B1jxALjhiov2Daebbfv3ySLgzGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@7F17@

Opt.

Since a^ is a unit vector, so |a^|=1.                  (xa^)(x+a^)=15x.x+x.<m

Ans.

Since a^ is a unit vector, so |a^|=1.                 (xa)(x+a^)=15x .x+x.a^a^.x ˆa^.a^=15                  |x|21=15[µx .a^=a^.x  and a^.a^=1]                              |x|2=16                               |x|=4 If a^ is a unit vector and (x ’a^)(x +a^)=15, then the value of |x| is 4¯.

Q2.

Leta=3i^+j^+2k^andb=i^+2j^+5k^.Findavectorwhichisperpendiculartobothaandbandhavingamagnitudeof780.

Opt.

Thevectorwhichisperpendiculartoaandbisa×b=ij

Ans.

Thevectorwhichisperpendiculartoaandb isa—b =ijk312125               =i5ˆ4ˆj15’2+k6ˆ1               =i13j+5kTheunitvectorinthedirectionofa—b=a—ba—b                                                                          =iˆ13j+5k1+169+25                                                                          =i13j+5^195Therequiredvector=780—i13j+5k195                                 =2195—i13j+5k195                                =2i’13j+5k                                 =2i26j+10k

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