Miscellaneous Exercise answers: Vector Algebra

Open NCERT p. 372Matches NCERT’s answer

1. A unit vector lying flat in the XY-plane, making angle θ with the positive x-axis, is (cosθ, sinθ, 0).
2. Here θ = 30°, so cos30° = √3/2 and sin30° = 1/2.
3. The unit vector is (√3/2, 1/2, 0).

Answer(√3/2) î + (1/2) ĵ

Watch this explained “The one place components come from an angle”, 16:36 into Splitting a vector along the axes so the algebra becomes coordinate arithmetic

Open NCERT p. 372Matches NCERT’s answer

1. The vector joining two points is terminal point minus initial point: PQ = Q − P.
2. In components, PQ = (x₂−x₁, y₂−y₁, z₂−z₁). These are the scalar components.
3. The magnitude of PQ is the square root of the sum of the squares of these components: √[(x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²].

AnswerScalar components: (x₂−x₁, y₂−y₁, z₂−z₁). Magnitude: √[(x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²].

Watch this explained “Between two points, in that order”, 10:30 into Magnitude, the three angles with the axes, and the cosines and ratios they give

Open NCERT p. 372Checked by computer

1. Take î pointing east and ĵ pointing north, starting from the girl's starting point O.
2. First walk, 4 km west: −4î.
3. Second walk, 3 km in a direction 30° east of north (turned 30° from north towards east): 3 sin 30° î + 3 cos 30° ĵ = (3/2)î + (3√3/2)ĵ.
4. Displacement = sum of the two walks = (−4 + 3/2)î + (3√3/2)ĵ = −(5/2)î + (3√3/2)ĵ.
5. Magnitude = √[(5/2)² + (3√3/2)²] = √(25/4 + 27/4) = √(52/4) = √13 km.
6. Direction: the displacement goes 5/2 km west and 3√3/2 km north, so its angle with the west direction, turned towards north, is tan⁻¹[(3√3/2) ÷ (5/2)] = tan⁻¹(3√3/5), about 46.1°.

AnswerDisplacement = −(5/2)î + (3√3/2)ĵ km: magnitude √13 km (about 3.61 km), direction tan⁻¹(3√3/5) ≈ 46.1° north of west.

Watch this explained “The same walk, with numbers”, 15:34 into Two laws for adding, why they agree, and what addition obeys

Open NCERT p. 372Matches NCERT’s answer

1. Given: a⃗ = b⃗ + c⃗, then is it true that |a⃗| = |b⃗| + |c⃗|.
2. a⃗ = b⃗ + c⃗ means a⃗ is the vector sum of b⃗ and c⃗.
3. The triangle inequality says |b⃗ + c⃗| ≤ |b⃗| + |c⃗|, with equality only when b⃗ and c⃗ point the same way.
4. Take b⃗ = î, c⃗ = ĵ — these do not point the same way.
5. Then a⃗ = î + ĵ, so |a⃗| = √2, but |b⃗| + |c⃗| = 1 + 1 = 2.
6. √2 ≠ 2, so the two sides are not always equal.

AnswerNo. |a⃗| ≤ |b⃗| + |c⃗| always, with equality only when b⃗ and c⃗ point the same way — e.g. b⃗=î, c⃗=ĵ gives |a⃗|=√2 ≠ 2.

Watch this explained “The length of a sum is not the sum of lengths”, 16:23 into Two laws for adding, why they agree, and what addition obeys

Open NCERT p. 372Checked by computer

1. x(î+ĵ+k̂) has components (x, x, x).
2. Its magnitude is √(x² + x² + x²) = |x|√3.
3. A unit vector has magnitude 1, so |x|√3 = 1.
4. |x| = 1/√3, so x = 1/√3 or x = −1/√3.

Answerx = ±1/√3 (that is, x = 1/√3 or x = −1/√3).

Watch this explained “The item with two answers”, 20:39 into Stretching by a scalar, and dividing a vector by its own length

Open NCERT p. 372Matches NCERT’s answer

1. Resultant = a⃗ + b⃗ = (2+1)î + (3−2)ĵ + (−1+1)k̂ = 3î + ĵ.
2. Its magnitude = √(3² + 1²) = √10.
3. Unit vector along the resultant = (3î + ĵ)/√10.
4. Multiply by 5 to get a vector of magnitude 5: 5(3, 1, 0)/√10.
5. Simplify: (15/√10, 5/√10, 0) = (3√10/2, √10/2, 0).

Answer(3√10/2) î + (√10/2) ĵ, i.e. the vector (3√10/2, √10/2, 0).

Watch this explained “From length one to any length”, 18:27 into Stretching by a scalar, and dividing a vector by its own length

Open NCERT p. 372Matches NCERT’s answer

1. 2a⃗ = 2î+2ĵ+2k̂; −b⃗ = −2î+ĵ−3k̂; 3c⃗ = 3î−6ĵ+3k̂.
2. Add them: 2a⃗−b⃗+3c⃗ = (2−2+3)î + (2+1−6)ĵ + (2−3+3)k̂ = 3î−3ĵ+2k̂.
3. Magnitude = √(9+9+4) = √22.
4. Unit vector = (3, −3, 2)/√22.

Answer(3/√22) î − (3/√22) ĵ + (2/√22) k̂.

Watch this explained “A unit vector along a sum”, 19:21 into Stretching by a scalar, and dividing a vector by its own length

Open NCERT p. 372Matches NCERT’s answer

1. AB = B − A = (5−1, 0−(−2), −2−(−8)) = (4, 2, 6).
2. BC = C − B = (11−5, 3−0, 7−(−2)) = (6, 3, 9).
3. BC = (3/2)·AB — same direction, so A, B, C lie on one line: collinear.
4. |AB| : |BC| = 1 : 3/2 = 2 : 3, so B divides AC in the ratio 2 : 3.

AnswerA, B, C are collinear; B divides AC in the ratio 2 : 3.

Watch this explained “Three points on a line, and the ratio”, 17:30 into Position vectors, the vector joining two points, and dividing a segment in a ratio

Open NCERT p. 372Matches NCERT’s answer

1. Position vector of P is 2a⃗+b⃗, of Q is a⃗−3b⃗.
2. For external division in the ratio 1 : 2, R = (1·Q − 2·P)/(1−2).
3. R = [(a⃗−3b⃗) − 2(2a⃗+b⃗)] / (−1) = (−3a⃗−5b⃗)/(−1) = 3a⃗+5b⃗.
4. Midpoint of R and Q = (R+Q)/2 = (3a⃗+5b⃗+a⃗−3b⃗)/2 = (4a⃗+2b⃗)/2 = 2a⃗+b⃗.
5. This is exactly the position vector of P, so P is the midpoint of RQ.

AnswerPosition vector of R is 3a⃗ + 5b⃗; and (R+Q)/2 = 2a⃗+b⃗ = P, so P is the midpoint of RQ.

Watch this explained “The one place both formulas are checked together”, 18:25 into Position vectors, the vector joining two points, and dividing a segment in a ratio

Open NCERT p. 372Matches NCERT’s answer

1. A parallelogram has two diagonals. The one through the corner where the two given sides start is their sum: (2+1)î + (−4−2)ĵ + (5−3)k̂ = 3î − 6ĵ + 2k̂. This is the diagonal the question means.
2. Its length is √(3² + (−6)² + 2²) = √(9 + 36 + 4) = √49 = 7.
3. Unit vector along this diagonal = (3î − 6ĵ + 2k̂)/7 = (3/7)î − (6/7)ĵ + (2/7)k̂.
4. (The other diagonal is the difference of the sides, ±(î − 2ĵ + 8k̂), of length √69; its unit vector would be ±(î − 2ĵ + 8k̂)/√69.)
5. Area of the parallelogram = |(side 1) × (side 2)|, the cross product of the two SIDES, not the diagonals.
6. (2î − 4ĵ + 5k̂) × (î − 2ĵ − 3k̂) = ((−4)(−3) − (5)(−2))î − ((2)(−3) − (5)(1))ĵ + ((2)(−2) − (−4)(1))k̂ = 22î + 11ĵ + 0k̂.
7. Its length = √(22² + 11²) = √(484 + 121) = √605 = 11√5.

AnswerUnit vector along the diagonal = (3/7)î − (6/7)ĵ + (2/7)k̂; area = 11√5 square units.

Watch this explained “A parallelogram has two diagonals”, 22:19 into A product that returns a vector, where it points, and the area it measures

Open NCERT p. 372One way to think about it

1. Let l, m, n be the direction cosines of the vector — the cosines of the angles it makes with OX, OY, OZ.
2. Equally inclined to the three axes means these three angles are equal, so l = m = n.
3. Every direction-cosine triple satisfies l² + m² + n² = 1.
4. Put l = m = n into this: 3l² = 1, so l² = 1/3, l = ±1/√3.
5. So l = m = n = ±1/√3.

In shortThe direction cosines are ±(1/√3, 1/√3, 1/√3).

Watch this explained “Equally inclined, and both signs”, 11:46 into Magnitude, the three angles with the axes, and the cosines and ratios they give

Open NCERT p. 372Checked by computer

1. A vector perpendicular to both a⃗ and b⃗ lies along a⃗ × b⃗.
2. a⃗ × b⃗ = (4×7−2×(−2), −(1×7−2×3), 1×(−2)−4×3) = (32, −1, −14).
3. Let d⃗ = t(32, −1, −14) for some scalar t.
4. c⃗ ⋅ d⃗ = t(2×32 + (−1)×(−1) + 4×(−14)) = t(64+1−56) = 9t.
5. Set 9t = 15, so t = 5/3.
6. d⃗ = (5/3)(32, −1, −14) = (160/3, −5/3, −70/3).
7. The answer key at the back of the book prints +70k̂, but that vector is not perpendicular to a⃗ or b⃗; the k̂ part must be −70/3, so d⃗ = (1/3)(160î − 5ĵ − 70k̂).

Answerd⃗ = (160/3) î − (5/3) ĵ − (70/3) k̂.

Watch this explained “Both of them, not one”, 18:52 into A product that returns a vector, where it points, and the area it measures

Open NCERT p. 372Matches NCERT’s answer

1. Sum of the two vectors: (2+λ, 6, −2).
2. Its magnitude = √((2+λ)² + 36 + 4) = √((2+λ)²+40).
3. Scalar product of î+ĵ+k̂ with the unit vector along this sum = [(2+λ)+6−2] / √((2+λ)²+40) = (λ+6)/√((2+λ)²+40).
4. Set this equal to 1: λ+6 = √((2+λ)²+40).
5. Square both sides: (λ+6)² = (2+λ)²+40, i.e. λ²+12λ+36 = λ²+4λ+44.
6. 8λ = 8, so λ = 1.

Answerλ = 1.

Watch this explained “One product, no angle”, 4:45 into Projection: how much of one vector points along another

Open NCERT p. 372One way to think about it

1. Note: the printed question says 'the vector c⃗⋅d⃗ = 15'. That is a misprint: c⃗⋅d⃗ is a number, not a vector, and no d⃗ is defined in this question. The vector meant, the only one the given facts are about, is a⃗ + b⃗ + c⃗. This is what we prove.
2. Let |a⃗| = |b⃗| = |c⃗| = k (equal magnitudes, k > 0) and a⃗⋅b⃗ = b⃗⋅c⃗ = c⃗⋅a⃗ = 0 (mutually perpendicular).
3. Let d⃗ = a⃗ + b⃗ + c⃗. Then d⃗⋅a⃗ = a⃗⋅a⃗ + b⃗⋅a⃗ + c⃗⋅a⃗ = k² + 0 + 0 = k². In the same way, d⃗⋅b⃗ = k² and d⃗⋅c⃗ = k².
4. |d⃗|² = d⃗⋅d⃗ = a⃗⋅a⃗ + b⃗⋅b⃗ + c⃗⋅c⃗ + 2(a⃗⋅b⃗ + b⃗⋅c⃗ + c⃗⋅a⃗) = k² + k² + k² + 0 = 3k², so |d⃗| = √3 k.
5. If α is the angle between d⃗ and a⃗, cos α = d⃗⋅a⃗ / (|d⃗||a⃗|) = k² / (√3 k × k) = 1/√3. In the same way, the cosines of the angles with b⃗ and with c⃗ are both 1/√3.
6. All three angles lie between 0 and π and have the same cosine, so they are equal: each is cos⁻¹(1/√3).

In shorta⃗ + b⃗ + c⃗ makes the same angle, cos⁻¹(1/√3), with each of a⃗, b⃗ and c⃗, so it is equally inclined to all three.

Watch this explained “The angle, extracted”, 8:39 into A product that returns a number, and the angle you can extract from it

Open NCERT p. 373One way to think about it

1. Given: (a⃗ + b⃗) ⋅ (a⃗ + b⃗) = |a⃗|² + |b⃗|², if and only if a⃗, b⃗.
2. (a⃗+b⃗)⋅(a⃗+b⃗) = a⃗⋅a⃗ + 2a⃗⋅b⃗ + b⃗⋅b⃗ = |a⃗|² + 2a⃗⋅b⃗ + |b⃗|².
3. So the equation holds exactly when 2a⃗⋅b⃗ = 0, i.e. a⃗⋅b⃗ = 0.
4. Since a⃗ ≠ 0⃗ and b⃗ ≠ 0⃗, a⃗⋅b⃗=0 means a⃗ and b⃗ are perpendicular.
5. So the equation holds if and only if a⃗ and b⃗ are perpendicular.

In short(a⃗+b⃗)⋅(a⃗+b⃗) = |a⃗|²+|b⃗|² exactly when a⃗⋅b⃗=0, i.e. exactly when a⃗ ⊥ b⃗.

Watch this explained “Perpendicular, read both ways”, 5:20 into A product that returns a number, and the angle you can extract from it

Open NCERT p. 373Matches NCERT’s answer

1. Given: θ is the angle between two vectors a⃗ and b⃗, then a⃗⋅b⃗ ≥ 0.
2. a⃗⋅b⃗ = |a⃗||b⃗|cosθ, and lengths are never negative.
3. So the sign of the scalar product matches the sign of cosθ.
4. cosθ ≥ 0 exactly for 0 ≤ θ ≤ π/2, including both end values (cos 0 = 1, cos(π/2) = 0).
5. So the answer is (B).

Answer(B) 0 ≤ θ ≤ π/2

Watch this explained “The sign, and what it tells you”, 4:19 into A product that returns a number, and the angle you can extract from it

Open NCERT p. 373Matches NCERT’s answer

1. Given: a⃗ and b⃗ be two unit vectors and θ is the angle between them. Then a⃗ + b⃗.
2. |a⃗+b⃗|² = a⃗⋅a⃗ + 2a⃗⋅b⃗ + b⃗⋅b⃗ = |a⃗|² + 2|a⃗||b⃗|cosθ + |b⃗|².
3. a⃗, b⃗ are unit vectors, so |a⃗+b⃗|² = 1 + 2cosθ + 1 = 2 + 2cosθ.
4. For a⃗+b⃗ to be a unit vector, 2 + 2cosθ = 1, so cosθ = −1/2.
5. cosθ = −1/2 gives θ = 2π/3.
6. So the answer is (D).

Answer(D) θ = 2π/3

Watch this explained “Lengths out of products”, 14:19 into A product that returns a number, and the angle you can extract from it

Open NCERT p. 373Matches NCERT’s answer

1. For the standard basis, ĵ×k̂=î, î×k̂=−ĵ, î×ĵ=k̂.
2. î⋅(ĵ×k̂) = î⋅î = 1.
3. ĵ⋅(î×k̂) = ĵ⋅(−ĵ) = −1.
4. k̂⋅(î×ĵ) = k̂⋅k̂ = 1.
5. Sum = 1 − 1 + 1 = 1.
6. So the answer is (C).

Answer(C) 1

Watch this explained “The two products, side by side”, 23:50 into A product that returns a vector, where it points, and the area it measures

Open NCERT p. 373Matches NCERT’s answer

1. Given: θ is the angle between any two vectors a⃗ and b⃗, then |a⃗⋅b⃗| = |a⃗×b⃗| when θ.
2. |a⃗⋅b⃗| = |a⃗||b⃗||cosθ| and |a⃗×b⃗| = |a⃗||b⃗|sinθ, since sinθ ≥ 0 for 0 ≤ θ ≤ π.
3. Setting them equal: |cosθ| = sinθ.
4. At θ = π/4, |cos(π/4)| = 1/√2 = sin(π/4), so it holds.
5. It fails at θ = 0 (1 vs 0), θ = π/2 (0 vs 1) and θ = π (1 vs 0).
6. So the answer is (B).

Answer(B) π/4

Watch this explained “The two products, side by side”, 23:50 into A product that returns a vector, where it points, and the area it measures