Exercise 10.3 answers: Vector Algebra

Open NCERT p. 361Matches NCERT’s answer

1. Given: a⃗ and b⃗ with magnitudes √3 and 2, respectively having a⃗ · b⃗ = √6.
2. The angle θ between two vectors satisfies cos θ = (a⃗·b⃗) ÷ (|a⃗||b⃗|).
3. Here a⃗·b⃗ = √6, |a⃗| = √3, |b⃗| = 2, so cos θ = √6 ÷ (√3 × 2).
4. √6 ÷ √3 = √2, so cos θ = √2 ÷ 2 = 1/√2.
5. cos θ = 1/√2 gives θ = 45° = π/4.

Answerθ = π/4 (45°)

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. 361Matches NCERT’s answer

1. Given: î − 2ĵ + 3k̂ and 3î − 2ĵ + k̂.
2. Write the vectors as (1, −2, 3) and (3, −2, 1).
3. Dot product = 1×3 + (−2)×(−2) + 3×1 = 3 + 4 + 3 = 10.
4. |first vector| = √(1+4+9) = √14; |second vector| = √(9+4+1) = √14.
5. cos θ = 10 ÷ (√14 × √14) = 10 ÷ 14 = 5/7.
6. θ = cos⁻¹(5/7).

Answerθ = cos⁻¹(5/7)

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. 361Matches NCERT’s answer

1. Given: î − ĵ on the vector î + ĵ.
2. The projection of u⃗ on v⃗ is (u⃗·v⃗) ÷ |v⃗|.
3. u⃗ = (1, −1, 0), v⃗ = (1, 1, 0); dot product = 1×1 + (−1)×1 + 0 = 1 − 1 = 0.
4. Since the dot product is 0, the projection is 0 ÷ |v⃗| = 0.

AnswerProjection = 0 — the two vectors are perpendicular, so there is no shadow along v⃗.

Watch this explained “Look before you compute”, 10:17 into Projection: how much of one vector points along another

Open NCERT p. 361Matches NCERT’s answer

1. Given: î + 3ĵ + 7k̂ on the vector 7î − ĵ + 8k̂.
2. The projection of u⃗ on v⃗ is (u⃗·v⃗) ÷ |v⃗|.
3. u⃗ = (1, 3, 7), v⃗ = (7, −1, 8); dot product = 1×7 + 3×(−1) + 7×8 = 7 − 3 + 56 = 60.
4. |v⃗| = √(49+1+64) = √114.
5. Projection = 60 ÷ √114, which tidies to (10√114) ÷ 19 after rationalising.

AnswerProjection = 60/√114, i.e. (10√114)/19

Watch this explained “Look before you compute”, 10:17 into Projection: how much of one vector points along another

Open NCERT p. 361One way to think about it

1. For a unit vector, u⃗·u⃗ should equal 1.
2. First vector: (1/7)²×(2²+3²+6²) = (1/49)×(4+9+36) = 49/49 = 1 — a unit vector.
3. Second vector: (1/7)²×(3²+(−6)²+2²) = (1/49)×(9+36+4) = 49/49 = 1 — a unit vector.
4. Third vector: (1/7)²×(6²+2²+(−3)²) = (1/49)×(36+4+9) = 49/49 = 1 — a unit vector.
5. For mutual perpendicularity, each pair's dot product should be 0.
6. First · Second = (1/49)×(2×3 + 3×(−6) + 6×2) = (1/49)×(6−18+12) = 0.
7. Second · Third = (1/49)×(3×6 + (−6)×2 + 2×(−3)) = (1/49)×(18−12−6) = 0.
8. Third · First = (1/49)×(6×2 + 2×3 + (−3)×6) = (1/49)×(12+6−18) = 0.
9. All three dot products are 0, so the vectors are mutually perpendicular.

In shortAll three vectors have magnitude 1, and every pair's dot product is 0 — so they are unit vectors and mutually perpendicular.

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. 362Matches NCERT’s answer

1. Given: |a⃗| and |b⃗|, if (a⃗+b⃗)·(a⃗−b⃗)=8 and |a⃗|=8|b⃗|.
2. (a⃗+b⃗)·(a⃗−b⃗) = a⃗·a⃗ − a⃗·b⃗ + b⃗·a⃗ − b⃗·b⃗ = |a⃗|² − |b⃗|², since the middle terms cancel (a⃗·b⃗=b⃗·a⃗).
3. So |a⃗|² − |b⃗|² = 8.
4. Given |a⃗| = 8|b⃗|, substitute: (8|b⃗|)² − |b⃗|² = 8, i.e. 64|b⃗|² − |b⃗|² = 8, so 63|b⃗|² = 8.
5. |b⃗|² = 8/63, so |b⃗| = √(8/63) = 2√14/21.
6. |a⃗| = 8|b⃗| = 16√14/21.

Answer|b⃗| = 2√14/21 and |a⃗| = 16√14/21.

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. 362Matches NCERT’s answer

1. Expand using distributivity: (3a⃗−5b⃗)·(2a⃗+7b⃗) = 3a⃗·2a⃗ + 3a⃗·7b⃗ − 5b⃗·2a⃗ − 5b⃗·7b⃗.
2. = 6(a⃗·a⃗) + 21(a⃗·b⃗) − 10(a⃗·b⃗) − 35(b⃗·b⃗), using a⃗·b⃗ = b⃗·a⃗.
3. = 6|a⃗|² + 11(a⃗·b⃗) − 35|b⃗|², collecting the two middle terms (21−10 = 11).

Answer6|a⃗|² + 11(a⃗·b⃗) − 35|b⃗|²

Watch this explained “Two properties”, 10:11 into A product that returns a number, and the angle you can extract from it

Open NCERT p. 362Matches NCERT’s answer

1. Given: a⃗ and b⃗, having the same magnitude and such that the angle between them is 60°.
2. Scalar product formula: a⃗·b⃗ = |a⃗||b⃗|cos θ.
3. Since |a⃗| = |b⃗|, call this common magnitude m, so a⃗·b⃗ = m² cos 60° = m² × (1/2).
4. Given a⃗·b⃗ = 1/2, so m² × (1/2) = 1/2, giving m² = 1.
5. So m = 1, since a magnitude cannot be negative.

AnswerEach vector has magnitude 1.

Watch this explained “A length, a length, and a cosine”, 1:03 into A product that returns a number, and the angle you can extract from it

Open NCERT p. 362Matches NCERT’s answer

1. Given: |x⃗|, if for a unit vector a⃗, (x⃗−a⃗)·(x⃗+a⃗)=12.
2. (x⃗−a⃗)·(x⃗+a⃗) = x⃗·x⃗ − a⃗·a⃗ = |x⃗|² − |a⃗|², since the cross terms cancel.
3. Since a⃗ is a unit vector, |a⃗|² = 1.
4. So |x⃗|² − 1 = 12, giving |x⃗|² = 13.
5. |x⃗| = √13.

Answer|x⃗| = √13

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. 362Matches NCERT’s answer

1. Given: a⃗=2î+2ĵ+3k̂, b⃗=−î+2ĵ+k̂ and c⃗=3î+ĵ are such that a⃗+λb⃗ is perpendicular to c⃗.
2. a⃗ = (2,2,3), b⃗ = (−1,2,1), c⃗ = (3,1,0).
3. a⃗+λb⃗ perpendicular to c⃗ means (a⃗+λb⃗)·c⃗ = 0.
4. a⃗·c⃗ = 2×3+2×1+3×0 = 6+2+0 = 8.
5. b⃗·c⃗ = −1×3+2×1+1×0 = −3+2+0 = −1.
6. So 8 + λ×(−1) = 0, giving λ = 8.

Answerλ = 8

Watch this explained “The formula at work”, 13:07 into A product that returns a number, and the angle you can extract from it

Open NCERT p. 362One way to think about it

1. Given: |a⃗|b⃗+|b⃗|a⃗ is perpendicular to |a⃗|b⃗−|b⃗|a⃗, for any two nonzero vectors a⃗ and b⃗.
2. Two vectors are perpendicular exactly when their dot product is 0.
3. (|a⃗|b⃗+|b⃗|a⃗)·(|a⃗|b⃗−|b⃗|a⃗) = |a⃗|²(b⃗·b⃗) − |a⃗||b⃗|(b⃗·a⃗) + |b⃗||a⃗|(a⃗·b⃗) − |b⃗|²(a⃗·a⃗).
4. b⃗·a⃗ = a⃗·b⃗, so the middle two terms cancel: −|a⃗||b⃗|(a⃗·b⃗) + |a⃗||b⃗|(a⃗·b⃗) = 0.
5. What remains is |a⃗|²(b⃗·b⃗) − |b⃗|²(a⃗·a⃗) = |a⃗|²|b⃗|² − |b⃗|²|a⃗|² = 0.
6. The dot product is 0 for any a⃗ and b⃗, so the two vectors are always perpendicular.

In shortThe dot product works out to 0 for any nonzero a⃗ and b⃗, so the two vectors are always perpendicular.

Watch this explained “The formula at work”, 13:07 into A product that returns a number, and the angle you can extract from it

Open NCERT p. 362One way to think about it

1. Given: a⃗·a⃗=0 and a⃗·b⃗=0, then what can be concluded about the vector b⃗.
2. a⃗·a⃗ = |a⃗|², so a⃗·a⃗=0 means |a⃗|² = 0, i.e. |a⃗| = 0.
3. A vector with magnitude 0 is the zero vector, so a⃗ = 0⃗.
4. Now look at the second condition, a⃗·b⃗ = 0. Since a⃗ is already the zero vector, 0⃗·b⃗ = 0 is true no matter what b⃗ is.
5. So the second condition gives no extra information about b⃗ — it would hold for ANY vector b⃗.

In shortNothing can be concluded about b⃗ — it can be any vector at all (including the zero vector). The first condition alone already forces a⃗ to be the zero vector, and once that is true, a⃗·b⃗=0 holds automatically, whatever b⃗ is.

Watch this explained “A product of nought, and what it does not say”, 23:28 into A product that returns a number, and the angle you can extract from it

Open NCERT p. 362Matches NCERT’s answer

1. Given: a⃗,b⃗,c⃗ are unit vectors such that a⃗+b⃗+c⃗=0⃗, find the value of a⃗·b⃗+b⃗·c⃗+c⃗·a⃗.
2. Dot both sides of a⃗+b⃗+c⃗=0⃗ with themselves: (a⃗+b⃗+c⃗)·(a⃗+b⃗+c⃗) = 0.
3. Expand the left side: a⃗·a⃗+b⃗·b⃗+c⃗·c⃗ + 2(a⃗·b⃗+b⃗·c⃗+c⃗·a⃗) = 0.
4. Since a⃗,b⃗,c⃗ are unit vectors, a⃗·a⃗=b⃗·b⃗=c⃗·c⃗=1, so 1+1+1+2(a⃗·b⃗+b⃗·c⃗+c⃗·a⃗) = 0.
5. 3 + 2(a⃗·b⃗+b⃗·c⃗+c⃗·a⃗) = 0, so a⃗·b⃗+b⃗·c⃗+c⃗·a⃗ = −3/2.

Answera⃗·b⃗+b⃗·c⃗+c⃗·a⃗ = −3/2

Watch this explained “One trick, forwards and backwards”, 21:38 into A product that returns a number, and the angle you can extract from it

Open NCERT p. 362Checked by computerAnswers can differ: one example

1. Given: a⃗=0⃗ or b⃗=0⃗, then a⃗·b⃗=0.
2. The converse would say: if a⃗·b⃗=0, then a⃗=0⃗ or b⃗=0⃗. This need not be true, because two nonzero perpendicular vectors also give a dot product of 0.
3. Take a⃗ = î = (1,0,0) and b⃗ = ĵ = (0,1,0). Both are nonzero.
4. a⃗·b⃗ = 1×0 + 0×1 + 0×0 = 0.
5. So a⃗·b⃗=0 even though neither a⃗ nor b⃗ is the zero vector — the converse fails.

AnswerFor example, a⃗ = î and b⃗ = ĵ: neither is the zero vector, but a⃗·b⃗ = 0. So a⃗·b⃗=0 does not force a⃗=0⃗ or b⃗=0⃗ — any pair of nonzero perpendicular vectors is a counterexample.

Watch this explained “A product of nought, and what it does not say”, 23:28 into A product that returns a number, and the angle you can extract from it

Open NCERT p. 362Matches NCERT’s answer

1. Find vector BA by subtracting B's coordinates from A's: BA = (1−(−1), 2−0, 3−0) = (2, 2, 3).
2. Find vector BC by subtracting B's coordinates from C's: BC = (0−(−1), 1−0, 2−0) = (1, 1, 2).
3. Find BA·BC = 2×1 + 2×1 + 3×2 = 10.
4. Find |BA| = √(2²+2²+3²) = √17 and |BC| = √(1²+1²+2²) = √6.
5. cos(∠ABC) = (BA·BC) ÷ (|BA|×|BC|) = 10/√102.

Answer∠ABC = cos⁻¹(10/√102), about 8.05°.

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. 362One way to think about it

1. Find vector AB = B − A = (2−1, 6−2, 3−7) = (1, 4, −4).
2. Find vector BC = C − B = (3−2, 10−6, −1−3) = (1, 4, −4).
3. AB and BC come out to be exactly the same vector.
4. Since AB and BC are the same vector and share the point B, A, B and C lie on one straight line.

In shortA, B and C are collinear (B lies between A and C, in fact B is the midpoint of AC).

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. 362One way to think about it

1. Read the three vectors as position vectors of points A(2,−1,1), B(1,−3,−5) and C(3,−4,−4).
2. Find AB = B−A = (−1,−2,−6), BC = C−B = (2,−1,1), CA = A−C = (−1,3,5).
3. Find the squared lengths: |AB|² = 1+4+36 = 41, |BC|² = 4+1+1 = 6, |CA|² = 1+9+25 = 35.
4. Check the two sides that meet at C: CB = −BC = (−2,1,−1), so CB·CA = (−2)(−1)+(1)(3)+(−1)(5) = 2+3−5 = 0.
5. A dot product of 0 means CB and CA are perpendicular, so the triangle has a right angle at C.

In shortThe three points form a right-angled triangle, with the right angle at C.

Watch this explained “A right angle without the tool for right angles”, 15:14 into Position vectors, the vector joining two points, and dividing a segment in a ratio

Open NCERT p. 362Matches NCERT’s answer

1. Given: is a nonzero vector of magnitude 'a' and λ a nonzero scalar, then λa.
2. The length of λa⃗ is |λ| times the length of a⃗, i.e. |λ|×a.
3. For λa⃗ to be a unit vector, this length must equal 1: |λ|×a = 1.
4. Solve for a: a = 1/|λ|.
5. This matches option (D).

Answer(D) a = 1/|λ|

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