Vector Algebra Class 12 Maths NCERT Solutions

Step-by-step solution

1. 1Choose a scale, e.g. 10 km per cm, and draw the north line (y-axis) as the reference.
2. 2The displacement is an arrow of length 4 cm making 30° with the north direction, swung towards the west.

Step-by-step solution

1. 1A scalar has only magnitude; a vector has magnitude and direction.
2. 2(i) 10 kg — magnitude only → scalar. (ii) 2 metres north-west — has direction → vector.
3. 3(iii) 40°, (iv) 40 watt, (v) 10⁻¹⁹ coulomb carry no direction → scalars.
4. 4(vi) 20 m/s² (an acceleration value of the vector kind) → vector.

Step-by-step solution

1. 1Coinitial vectors share the same initial point — here a and d start from the same point.
2. 2Equal vectors have the same magnitude and direction — here b and d are equal.
3. 3Collinear but not equal — a and c lie on the same line but face opposite directions, so they are not equal.

Step-by-step solution

1. 1(i) a and −a lie on the same line (opposite directions), so they are collinear → True.
2. 2(ii) Collinearity says nothing about lengths — the vectors need not be equal in magnitude → False.
3. 3(iii) Equal magnitude does not force a common line → False.
4. 4(iv) Collinear vectors of equal magnitude may point in opposite directions (a and −a) → False.

Step-by-step solution

1. 1|a| = √(1² + 1² + 1²) = √3.
2. 2|b| = √(4 + 49 + 9) = √62.
3. 3|c| = √(1/3 + 1/3 + 1/3) = √1 = 1.

Step-by-step solution

1. 1Take a = î + ĵ + k̂ and b = î + ĵ − k̂.
2. 2Both have magnitude √(1 + 1 + 1) = √3, but none of their components match exactly, so the vectors are different.

Step-by-step solution

1. 1Multiply any vector by a positive scalar: let a = î + 2ĵ + 3k̂ and b = 2î + 4ĵ + 6k̂ = 2a.
2. 2A positive scalar multiple has the same direction (and never the reverse), so a and b are parallel.

Step-by-step solution

1. 1Vectors are equal exactly when all corresponding components are equal.
2. 2Coefficient of î: x = 2; coefficient of ĵ: y = 3.

Step-by-step solution

1. 1A vector joining two points is terminal − initial.
2. 2Vector = (−5 − 2)î + (7 − 1)ĵ = −7î + 6ĵ.

Step-by-step solution

1. 1Add coefficients along each direction.
2. 2î-component: 1 − 2 + 1 = 0; ĵ-component: −2 + 4 − 6 = −4; k̂-component: 1 + 5 − 7 = −1.

Step-by-step solution

1. 1|a| = √(1 + 1 + 4) = √6.
2. 2The unit vector is a/|a|.

Step-by-step solution

1. 1PQ = (4 − 1)î + (5 − 2)ĵ + (6 − 3)k̂ = 3î + 3ĵ + 3k̂.
2. 2|PQ| = √(9 + 9 + 9) = 3√3.
3. 3Unit vector = (3î + 3ĵ + 3k̂)/(3√3).

Step-by-step solution

1. 1a + b = (2 − 1)î + (−1 + 1)ĵ + (2 − 1)k̂ = î + k̂.
2. 2|a + b| = √(1 + 1) = √2.
3. 3Unit vector = (î + k̂)/√2.

Step-by-step solution

1. 1|5î − ĵ + 2k̂| = √(25 + 1 + 4) = √30.
2. 2The unit vector in that direction is (5î − ĵ + 2k̂)/√30.
3. 3Multiply by 8 to get magnitude 8.

Step-by-step solution

1. 1−4î + 6ĵ − 8k̂ = −2(2î − 3ĵ + 4k̂).
2. 2One vector is a scalar multiple of the other, so they lie on the same line — they are collinear.

Step-by-step solution

1. 1|a| = √(1 + 4 + 9) = √14.
2. 2Direction cosines are the components divided by the magnitude.

Step-by-step solution

1. 1AB = (−1 − 1)î + (−2 − 2)ĵ + (1 + 3)k̂ = −2î − 4ĵ + 4k̂.
2. 2|AB| = √(4 + 16 + 16) = √36 = 6.
3. 3Divide each component by 6.

Step-by-step solution

1. 1|î + ĵ + k̂| = √3, so the direction cosines are 1/√3, 1/√3, 1/√3.
2. 2cos α = cos β = cos γ = 1/√3, hence α = β = γ — equal inclinations to the three axes.

Step-by-step solution

1. 1Internal division: R = (2Q + 1P)/(2 + 1).
2. 22Q + P = 2(−î + ĵ + k̂) + (î + 2ĵ − k̂) = (−2î + 2ĵ + 2k̂) + (î + 2ĵ − k̂) = −î + 4ĵ + k̂.
3. 3So R = (−î + 4ĵ + k̂)/3.
4. 4External division: R = (2Q − 1P)/(2 − 1) = 2(−î + ĵ + k̂) − (î + 2ĵ − k̂) = −3î + 0ĵ + 3k̂.

Step-by-step solution

1. 1Mid point = (P + Q)/2.
2. 2= ((2 + 4)/2, (3 + 1)/2, (4 − 2)/2) = (3, 2, 1).

Step-by-step solution

1. 1AB = b − a = −î + 3ĵ + 5k̂, so |AB|² = 1 + 9 + 25 = 35.
2. 2BC = c − b = −î − 2ĵ − 6k̂, so |BC|² = 1 + 4 + 36 = 41.
3. 3CA = a − c = 2î − ĵ + k̂, so |CA|² = 4 + 1 + 1 = 6.
4. 435 + 6 = 41, i.e. |AB|² + |CA|² = |BC|² — Pythagoras holds, so the angle at A is right.

Step-by-step solution

1. 1(A) The sum of the side vectors around a triangle is the zero vector → true.
2. 2(B) AC = AB + BC, so AB + BC − AC = 0 → true.
3. 3(C) AB + BC − CA = AC − CA = 2AC ≠ 0 → not true.
4. 4(D) AB − CB + CA = AB + BC + CA = 0 → true.

Step-by-step solution

1. 1(A) Collinear vectors are scalar multiples: b = λa → correct.
2. 2(C) Scalar multiples have proportional components → correct.
3. 3(B) a = −b is only true when the magnitudes are equal → incorrect in general.
4. 4(D) Collinear vectors may perfectly well point in opposite directions → incorrect.

Step-by-step solution

1. 1cos θ = (a·b)/(|a||b|).
2. 2cos θ = √6/(√3 × 2) = √6/(2√3) = 1/√2.
3. 3θ = π/4.

Step-by-step solution

1. 1a·b = 1·3 + (−2)(−2) + 3·1 = 3 + 4 + 3 = 10.
2. 2|a| = √(1 + 4 + 9) = √14 and |b| = √(9 + 4 + 1) = √14.
3. 3cos θ = 10/(√14·√14) = 10/14 = 5/7.

Step-by-step solution

1. 1Projection of a on b = (a·b)/|b|.
2. 2a·b = (1)(1) + (−1)(1) + (0)(0) = 0.
3. 3Projection = 0/√2 = 0.

Step-by-step solution

1. 1Projection of a on b = (a·b)/|b|.
2. 2a·b = 7 − 3 + 56 = 60.
3. 3|b| = √(49 + 1 + 64) = √114.

Step-by-step solution

1. 1|2î + 3ĵ + 6k̂|² = 4 + 9 + 36 = 49, so the first vector has magnitude √49/7 = 1; likewise the other two squares of norms are 49.
2. 2(2î + 3ĵ + 6k̂)·(3î − 6ĵ + 2k̂) = 6 − 18 + 12 = 0.
3. 3(3î − 6ĵ + 2k̂)·(6î + 2ĵ − 3k̂) = 18 − 12 − 6 = 0.
4. 4(6î + 2ĵ − 3k̂)·(2î + 3ĵ + 6k̂) = 12 + 6 − 18 = 0.

Step-by-step solution

1. 1(a + b)·(a − b) = |a|² − |b|² = 8.
2. 2Put |a| = 8|b|: 64|b|² − |b|² = 63|b|² = 8.
3. 3|b|² = 8/63, so |b| = √(8/63) and |a| = 8√(8/63).

Step-by-step solution

1. 1Expand: 6(a·a) + 21(a·b) − 10(b·a) − 35(b·b).
2. 2Use a·a = |a|² and a·b = b·a.

Step-by-step solution

1. 1a·b = |a||b| cos 60° = |a|² × ½ (magnitudes equal).
2. 2½|a|² = ½ ⇒ |a|² = 1.

Step-by-step solution

1. 1(x − a)·(x + a) = |x|² − |a|² = |x|² − 1 (|a| = 1).
2. 2|x|² − 1 = 12 ⇒ |x|² = 13.

Step-by-step solution

1. 1a + λb = (2 − λ)î + (2 + 2λ)ĵ + (3 + λ)k̂.
2. 2Perpendicularity: (a + λb)·c = 0.
3. 3(2 − λ)(3) + (2 + 2λ)(1) = 6 − 3λ + 2 + 2λ = 8 − λ = 0.

Step-by-step solution

1. 1Dot the two vectors: (|a| b + |b| a)·(|a| b − |b| a).
2. 2= |a|²(b·b) − |a||b|(b·a) + |a||b|(a·b) − |b|²(a·a).
3. 3= |a|²|b|² − |a||b|(a·b) + |a||b|(a·b) − |b|²|a|² = 0.

Step-by-step solution

1. 1a·a = |a|² = 0 forces a = 0 (the zero vector).
2. 2Then a·b = 0·b = 0 holds for every vector b — no restriction.

Step-by-step solution

1. 1|a + b + c|² = 0.
2. 20 = |a|² + |b|² + |c|² + 2(a·b + b·c + c·a) = 3 + 2S.

Step-by-step solution

1. 1Take a = î and b = ĵ, both non-zero vectors.
2. 2a·b = 1·1 + 0·0 + 0·0 = 0, yet neither vector is the zero vector.

Step-by-step solution

1. 1∠ABC is the angle between BA and BC.
2. 2BA = A − B = 2î + 2ĵ + 3k̂ and BC = C − B = î + ĵ + 2k̂.
3. 3BA·BC = 2 + 2 + 6 = 10; |BA| = √17, |BC| = √6.
4. 4cos(∠ABC) = 10/(√17√6) = 10/√102.

Step-by-step solution

1. 1AB = (2 − 1)î + (6 − 2)ĵ + (3 − 7)k̂ = î + 4ĵ − 4k̂.
2. 2BC = (3 − 2)î + (10 − 6)ĵ + (−1 − 3)k̂ = î + 4ĵ − 4k̂ = AB.
3. 3The shared point B and the equal directions place A, B, C on one line.

Step-by-step solution

1. 1Treat them as position vectors of points A, B, C.
2. 2AB = b − a = −î − 2ĵ − 6k̂, |AB|² = 41.
3. 3BC = c − b = 2î − ĵ + k̂, |BC|² = 6.
4. 4CA = a − c = −î + 3ĵ + 5k̂, |CA|² = 35.
5. 541 = 6 + 35 ⇒ |AB|² = |BC|² + |CA|², right angle at C.

Step-by-step solution

1. 1|λa| = |λ| |a| = |λ| a.
2. 2For a unit vector: |λ| a = 1 ⇒ a = 1/|λ|.

Step-by-step solution

1. 1a × b = ((−7)(2) − (7)(−2))î + ((7)(3) − (1)(2))ĵ + ((1)(−2) − (−7)(3))k̂.
2. 2= (−14 + 14)î + (21 − 2)ĵ + (−2 + 21)k̂ = 19ĵ + 19k̂.
3. 3|a × b| = √(0 + 361 + 361) = √722 = 19√2.

Step-by-step solution

1. 1a + b = 4î + 4ĵ and a − b = 2î + 4k̂.
2. 2(a + b) × (a − b) = (4·4 − 0·0)î + (0·2 − 4·4)ĵ + (4·4 − 4·2)k̂ = 16î − 16ĵ + 8k̂ — wait, recompute the third component: 4·4 − 4·2 = 16 − 8 = 8.
3. 3Actually: (a+b) × (a−b) = 16î − 16ĵ + 8k̂? Checking with determinants: a+b = (4,4,0), a−b = (2,0,4). First component: 4·4 − 0·0 = 16. Second: 0·2 − 4·4 = −16. Third: 4·0 − 4·2 = −8. So (16, −16, −8).
4. 4|(a+b) × (a−b)| = √(256 + 256 + 64) = √576 = 24.
5. 5Unit vector = (16î − 16ĵ − 8k̂)/24.

Step-by-step solution

1. 1For unit vectors, sum of squares of direction cosines = 1.
2. 2cos²(π/3) + cos²(π/4) + cos²θ = 1 ⇒ 1/4 + 1/2 + cos²θ = 1 ⇒ cos²θ = 1/4.
3. 3θ acute ⇒ cos θ = 1/2 ⇒ θ = π/3.
4. 4Components of a are the cosines of the three angles.

Step-by-step solution

1. 1(a − b) × (a + b) = a×a + a×b − b×a − b×b.
2. 2a×a = 0 and b×b = 0 (cross product of a vector with itself).
3. 3−b×a = a×b (anticommutativity).

Step-by-step solution

1. 1A zero cross product means the two vectors are parallel.
2. 2(2, 6, 27) = 2(1, λ, μ).
3. 3λ = 6/2 = 3 and μ = 27/2.

Step-by-step solution

1. 1a × b = 0 ⇒ a ∥ b (for non-zero vectors).
2. 2But parallel non-zero vectors have |a·b| = |a||b| ≠ 0, contradicting a·b = 0.
3. 3Hence at least one of a, b must be the zero vector.

Step-by-step solution

1. 1Write b + c = (b₁ + c₁)î + (b₂ + c₂)ĵ + (b₃ + c₃)k̂.
2. 2First component of a × (b + c) = a₂(b₃ + c₃) − a₃(b₂ + c₂) = (a₂b₃ − a₃b₂) + (a₂c₃ − a₃c₂).
3. 3The first bracket is the î-component of a × b and the second bracket the î-component of a × c.
4. 4The same splitting works for the ĵ and k̂ components, proving the identity.

Step-by-step solution

1. 1Converse claims a × b = 0 forces a = 0 or b = 0.
2. 2Take a = î and b = 2î — neither is the zero vector.
3. 3a × b = î × (2î) = 2(î × î) = 0.

Step-by-step solution

1. 1AB = î + 2ĵ + 3k̂ and AC = 4ĵ + 3k̂.
2. 2AB × AC = (2·3 − 3·4)î + (3·0 − 1·3)ĵ + (1·4 − 2·0)k̂ = −6î − 3ĵ + 4k̂.
3. 3|AB × AC| = √(36 + 9 + 16) = √61.
4. 4Area = ½|AB × AC|.

Step-by-step solution

1. 1a × b = ((−1)(1) − 3(−7))î + (3·2 − 1·1)ĵ + (1(−7) − (−1)(2))k̂.
2. 2= (−1 + 21)î + (6 − 1)ĵ + (−7 + 2)k̂ = 20î + 5ĵ − 5k̂.
3. 3Area = |a × b| = √(400 + 25 + 25) = √450.

Step-by-step solution

1. 1|a × b| = |a||b| sin θ = 3·(√2/3) sin θ = √2 sin θ.
2. 2Unit vector requires √2 sin θ = 1 ⇒ sin θ = 1/√2.
3. 3θ = π/4.

Step-by-step solution

1. 1AB = 2î and AD = −ĵ (the rectangle lies in the plane z = 4).
2. 2Area = |AB × AD| = |2î × (−ĵ)| = 2·|î × ĵ| = 2.