Exercise 10.4 answers: Vector Algebra

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1. Given: |a × b|, if a = î − 7ĵ + 7k̂ and b = 3î − 2ĵ + 2k̂.
2. Write a⃗ = (1, −7, 7) and b⃗ = (3, −2, 2).
3. Find the first part of a⃗×b⃗: (−7)(2) − (7)(−2) = −14+14 = 0.
4. Find the second part: (7)(3) − (1)(2) = 21−2 = 19.
5. Find the third part: (1)(−2) − (−7)(3) = −2+21 = 19.
6. So a⃗ × b⃗ = (0, 19, 19).
7. |a⃗ × b⃗| = √(0²+19²+19²) = √722 = 19√2.

Answer|a⃗ × b⃗| = 19√2

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Open NCERT p. 368Checked by computerAnswers can differ: one example

1. Find a⃗ + b⃗ = (3+1)î + (2+2)ĵ + (2−2)k̂ = 4î + 4ĵ + 0k̂.
2. Find a⃗ − b⃗ = (3−1)î + (2−2)ĵ + (2−(−2))k̂ = 2î + 0ĵ + 4k̂.
3. A vector perpendicular to both is their cross product: (a⃗+b⃗) × (a⃗−b⃗) = î(4×4 − 0×0) − ĵ(4×4 − 0×2) + k̂(4×0 − 4×2) = 16î − 16ĵ − 8k̂.
4. Its length is √(16² + 16² + 8²) = √576 = 24.
5. Divide by 24: (16/24)î − (16/24)ĵ − (8/24)k̂ = (2/3)î − (2/3)ĵ − (1/3)k̂.
6. The opposite vector, −(2/3)î + (2/3)ĵ + (1/3)k̂, is also a unit vector perpendicular to both (it comes from crossing in the other order), so either one is a correct answer.

Answer(2/3)î − (2/3)ĵ − (1/3)k̂; its negative, −(2/3)î + (2/3)ĵ + (1/3)k̂, is equally correct.

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Open NCERT p. 368Matches NCERT’s answer

1. For a unit vector, the squares of its three direction cosines add to 1: cos²l + cos²m + cos²n = 1.
2. Here cos²(π/3) + cos²(π/4) + cos²θ = 1, i.e. (1/2)² + (√2/2)² + cos²θ = 1.
3. 1/4 + 1/2 + cos²θ = 1, so cos²θ = 1/4, giving cosθ = ±1/2.
4. Since θ is acute, cosθ = 1/2, so θ = π/3 (60°).
5. For a unit vector, its components are its direction cosines: (cosπ/3, cosπ/4, cosπ/3) = (1/2, √2/2, 1/2).

Answerθ = π/3 (60°), and a⃗ = (1/2)î + (√2/2)ĵ + (1/2)k̂.

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Open NCERT p. 368One way to think about it

1. Given: (a−b) × (a+b) = 2(a × b).
2. Expand the left side using the distributive law of the cross product over addition: (a⃗−b⃗)×(a⃗+b⃗) = (a⃗−b⃗)×a⃗ + (a⃗−b⃗)×b⃗.
3. Expand each piece: (a⃗−b⃗)×a⃗ = a⃗×a⃗ − b⃗×a⃗, and (a⃗−b⃗)×b⃗ = a⃗×b⃗ − b⃗×b⃗.
4. Any vector crossed with itself is the zero vector, so a⃗×a⃗ = 0⃗ and b⃗×b⃗ = 0⃗.
5. This leaves −b⃗×a⃗ + a⃗×b⃗.
6. Swapping the two vectors in a cross product negates the answer, so −b⃗×a⃗ = a⃗×b⃗.
7. So the total is a⃗×b⃗ + a⃗×b⃗ = 2(a⃗×b⃗), which is the right side.

In short(a⃗ − b⃗) × (a⃗ + b⃗) = 2(a⃗ × b⃗), proved by expanding with the distributive law and using a⃗×a⃗ = b⃗×b⃗ = 0⃗.

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Open NCERT p. 368Matches NCERT’s answer

1. Given: λ and μ if (2î + 6ĵ + 27k̂) × (î + λĵ + μk̂) = 0.
2. A cross product of 0⃗ means the two vectors are parallel (one is a scalar multiple of the other).
3. So their components must be in the same ratio: 2/1 = 6/λ = 27/μ.
4. From 2/1 = 6/λ, λ = 6/2 = 3.
5. From 2/1 = 27/μ, μ = 27/2.

Answerλ = 3 and μ = 27/2.

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Open NCERT p. 368One way to think about it

1. Given: a·b = 0 and a × b = 0. What can you conclude about the vectors a and b.
2. a⃗·b⃗ = 0 means a⃗ and b⃗ are perpendicular to each other, OR one of them is the zero vector.
3. a⃗ × b⃗ = 0⃗ means a⃗ and b⃗ are parallel to each other, OR one of them is the zero vector.
4. Two nonzero vectors cannot be both perpendicular and parallel to each other at the same time.
5. So the only way both conditions can hold together is if at least one of a⃗ and b⃗ is the zero vector.

In shortAt least one of a⃗ and b⃗ must be the zero vector (a⃗ = 0⃗ or b⃗ = 0⃗).

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Open NCERT p. 368One way to think about it

1. Given: a, b, c be given as a₁î+a₂ĵ+a₃k̂, b₁î+b₂ĵ+b₃k̂, c₁î+c₂ĵ+c₃k̂.
2. b⃗ + c⃗ = (b₁+c₁)î + (b₂+c₂)ĵ + (b₃+c₃)k̂.
3. By the determinant rule, a⃗ × (b⃗+c⃗) = î[a₂(b₃+c₃) − a₃(b₂+c₂)] − ĵ[a₁(b₃+c₃) − a₃(b₁+c₁)] + k̂[a₁(b₂+c₂) − a₂(b₁+c₁)].
4. Open the brackets and group the b-terms and the c-terms: = î[(a₂b₃ − a₃b₂) + (a₂c₃ − a₃c₂)] − ĵ[(a₁b₃ − a₃b₁) + (a₁c₃ − a₃c₁)] + k̂[(a₁b₂ − a₂b₁) + (a₁c₂ − a₂c₁)].
5. The b-terms alone are î(a₂b₃ − a₃b₂) − ĵ(a₁b₃ − a₃b₁) + k̂(a₁b₂ − a₂b₁), which is a⃗ × b⃗.
6. The c-terms alone are î(a₂c₃ − a₃c₂) − ĵ(a₁c₃ − a₃c₁) + k̂(a₁c₂ − a₂c₁), which is a⃗ × c⃗.
7. So a⃗ × (b⃗ + c⃗) = a⃗ × b⃗ + a⃗ × c⃗.

In shorta⃗ × (b⃗ + c⃗) = a⃗ × b⃗ + a⃗ × c⃗: each component of the left side splits into the matching component of a⃗ × b⃗ plus that of a⃗ × c⃗.

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Open NCERT p. 368Checked by computerAnswers can differ: one example

1. Given: a = 0 or b = 0, then a × b = 0.
2. The converse would say: if a⃗ × b⃗ = 0⃗, then a⃗ = 0⃗ or b⃗ = 0⃗.
3. This is false — two nonzero vectors that are parallel to each other also give a cross product of 0⃗.
4. Example: a⃗ = î = (1, 0, 0) and b⃗ = 2î = (2, 0, 0). Neither is the zero vector.
5. a⃗ × b⃗ = (1,0,0) × (2,0,0) = (0,0,0) = 0⃗, because parallel vectors always give a zero cross product.
6. So a⃗ × b⃗ = 0⃗ even though neither a⃗ nor b⃗ is 0⃗, which disproves the converse.

AnswerNo, the converse is false. For example, a⃗ = î and b⃗ = 2î are both nonzero, but a⃗ × b⃗ = 0⃗ because they are parallel.

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Open NCERT p. 368Matches NCERT’s answer

1. Find AB = B − A = (1, 2, 3) and AC = C − A = (0, 4, 3).
2. The area of triangle ABC is half the length of AB × AC.
3. Find AB × AC: first part = 2×3−3×4 = −6; second part = −(1×3−3×0) = −3; third part = 1×4−2×0 = 4.
4. So AB × AC = (−6, −3, 4), and its length is √(36+9+16) = √61.
5. Area = (1/2)√61.

AnswerArea = √61/2 square units.

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Open NCERT p. 369Matches NCERT’s answer

1. The area of a parallelogram with adjacent sides a⃗ and b⃗ is |a⃗ × b⃗|.
2. Find a⃗ × b⃗: first part = (−1)(1)−(3)(−7) = −1+21 = 20; second part = −[(1)(1)−(3)(2)] = 5; third part = (1)(−7)−(−1)(2) = −5.
3. So a⃗ × b⃗ = (20, 5, −5).
4. |a⃗ × b⃗| = √(400+25+25) = √450 = 15√2.

AnswerArea = 15√2 square units.

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Open NCERT p. 369Matches NCERT’s answer

1. |a⃗ × b⃗| = |a⃗||b⃗| sinθ, where θ is the angle between a⃗ and b⃗.
2. For a⃗×b⃗ to be a unit vector, this must equal 1: 3 × (√2/3) × sinθ = 1.
3. This simplifies to √2 sinθ = 1, so sinθ = 1/√2.
4. sinθ = 1/√2 gives θ = π/4.

Answer(B) π/4

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Open NCERT p. 369Matches NCERT’s answer

1. Read off the position vectors: A(−1, 1/2, 4), B(1, 1/2, 4), C(1, −1/2, 4), D(−1, −1/2, 4).
2. Find two adjacent sides of the rectangle: AB = B − A = (2, 0, 0), and BC = C − B = (0, −1, 0).
3. The area of the rectangle is |AB × BC|.
4. AB × BC = (2,0,0) × (0,−1,0) = (0, 0, −2), so |AB × BC| = 2.

Answer(C) 2

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