Exercise 10.2 answers: Vector Algebra

Open NCERT p. 354Matches NCERT’s answer

1. For a⃗ = î+ĵ+k̂: magnitude = √(1²+1²+1²) = √3.
2. For b⃗ = 2î−7ĵ−3k̂: magnitude = √(2²+(−7)²+(−3)²) = √(4+49+9) = √62.
3. For c⃗ = (1/√3)î+(1/√3)ĵ−(1/√3)k̂: magnitude = √(1/3+1/3+1/3) = √1 = 1.

Answer|a⃗| = √3, |b⃗| = √62, |c⃗| = 1

Watch this explained “Its length, already known”, 0:53 into Magnitude, the three angles with the axes, and the cosines and ratios they give

Open NCERT p. 354Checked by computerAnswers can differ: one example

1. Many pairs work: any two vectors of the same length that point in different directions. Here is one pair.
2. Take a⃗ = î, that is (1, 0, 0). Its magnitude is √(1² + 0² + 0²) = 1.
3. Take b⃗ = ĵ, that is (0, 1, 0). Its magnitude is √(0² + 1² + 0²) = 1.
4. Both have magnitude 1, but a⃗ points along the x-axis and b⃗ along the y-axis, so a⃗ ≠ b⃗.

AnswerOne example: a⃗ = î = (1, 0, 0) and b⃗ = ĵ = (0, 1, 0), both of magnitude 1 (many other pairs are also correct).

Watch this explained “Same length is not the same vector”, 12:03 into Splitting a vector along the axes so the algebra becomes coordinate arithmetic

Open NCERT p. 354Checked by computerAnswers can differ: one example

1. Many pairs work: any vector and a positive multiple of it. Here is one pair.
2. Take a⃗ = î + ĵ + k̂, that is (1, 1, 1).
3. Take b⃗ = 2a⃗ = 2î + 2ĵ + 2k̂, that is (2, 2, 2).
4. Multiplying by a positive number changes only the length, not the direction, so a⃗ and b⃗ point the same way; their magnitudes are √3 and 2√3, so they are different vectors.

AnswerOne example: a⃗ = î + ĵ + k̂ = (1, 1, 1) and b⃗ = 2î + 2ĵ + 2k̂ = (2, 2, 2) (many other pairs are also correct).

Watch this explained “Five multiples of one arrow”, 4:52 into Stretching by a scalar, and dividing a vector by its own length

Open NCERT p. 354Matches NCERT’s answer

1. Given: 2î+3ĵ and xî+yĵ.
2. Two vectors are equal only when their matching components are equal.
3. Matching the î-parts: x = 2.
4. Matching the ĵ-parts: y = 3.

Answerx = 2, y = 3

Watch this explained “One equality is worth three equations”, 11:21 into Splitting a vector along the axes so the algebra becomes coordinate arithmetic

Open NCERT p. 354Matches NCERT’s answer

1. Vector = terminal point − initial point = (−5 − 2, 7 − 1) = (−7, 6).
2. The scalar components are the plain numbers: −7 and 6.
3. The vector components are these numbers attached to the axis directions: −7î and 6ĵ.

AnswerScalar components: −7, 6. Vector components: −7î and 6ĵ.

Watch this explained “Two lists, one word apart”, 4:13 into Splitting a vector along the axes so the algebra becomes coordinate arithmetic

Open NCERT p. 354Matches NCERT’s answer

1. î-parts: 1 + (−2) + 1 = 0.
2. ĵ-parts: −2 + 4 + (−6) = −4.
3. k̂-parts: 1 + 5 + (−7) = −1.

Answera⃗+b⃗+c⃗ = 0î − 4ĵ − k̂

Watch this explained “Four rules, no argument”, 7:23 into Splitting a vector along the axes so the algebra becomes coordinate arithmetic

Open NCERT p. 354Matches NCERT’s answer

1. Magnitude of a⃗ = √(1²+1²+2²) = √6.
2. Unit vector = a⃗ ÷ √6 = (1/√6)î + (1/√6)ĵ + (2/√6)k̂.
3. Written with a rational denominator: (√6/6)î + (√6/6)ĵ + (√6/3)k̂.

Answerâ = (√6/6)î + (√6/6)ĵ + (√6/3)k̂

Watch this explained “Divide it by its own length”, 15:54 into Stretching by a scalar, and dividing a vector by its own length

Open NCERT p. 354Matches NCERT’s answer

1. PQ = Q − P = (4−1, 5−2, 6−3) = (3, 3, 3).
2. Magnitude of PQ = √(3²+3²+3²) = √27 = 3√3.
3. Unit vector = PQ ÷ 3√3 = (√3/3)î + (√3/3)ĵ + (√3/3)k̂.

AnswerUnit vector along PQ = (√3/3)î + (√3/3)ĵ + (√3/3)k̂

Watch this explained “Divide it by its own length”, 15:54 into Stretching by a scalar, and dividing a vector by its own length

Open NCERT p. 354Matches NCERT’s answer

1. a⃗+b⃗ = (2−1, −1+1, 2−1) = (1, 0, 1).
2. Magnitude of a⃗+b⃗ = √(1²+0²+1²) = √2.
3. Unit vector = (1, 0, 1) ÷ √2 = (√2/2)î + 0ĵ + (√2/2)k̂.

AnswerUnit vector along a⃗+b⃗ = (√2/2)î + (√2/2)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. 354Matches NCERT’s answer

1. Magnitude of 5î−ĵ+2k̂ = √(5²+(−1)²+2²) = √30.
2. Unit vector along it = (5î−ĵ+2k̂)/√30.
3. Multiply the unit vector by 8: 8×(5/√30)î + 8×(−1/√30)ĵ + 8×(2/√30)k̂.
4. With rational denominators: (4√30/3)î − (4√30/15)ĵ + (8√30/15)k̂.

Answer(4√30/3)î − (4√30/15)ĵ + (8√30/15)k̂

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. 354One way to think about it

1. Given: 2î − 3ĵ + 4k̂ and −4î + 6ĵ − 8k̂.
2. Compare −4î+6ĵ−8k̂ with 2î−3ĵ+4k̂, part by part.
3. −4 = −2×2, 6 = −2×(−3), −8 = −2×4 — every part matches the same multiplier, −2.
4. So −4î+6ĵ−8k̂ = −2×(2î−3ĵ+4k̂), one vector is a scalar multiple of the other.
5. A vector that is a scalar multiple of another always lies along the same or opposite direction, so the two vectors are collinear.

In short−4î+6ĵ−8k̂ = −2(2î−3ĵ+4k̂), so the two vectors are collinear.

Watch this explained “Collinear, as one equation”, 8:50 into Stretching by a scalar, and dividing a vector by its own length

Open NCERT p. 354Matches NCERT’s answer

1. Given: î + 2ĵ + 3k̂.
2. Magnitude of î+2ĵ+3k̂ = √(1²+2²+3²) = √14.
3. Divide each part by √14: 1/√14, 2/√14, 3/√14.
4. With rational denominators: √14/14, √14/7, 3√14/14.

AnswerDirection cosines: √14/14, √14/7, 3√14/14

Watch this explained “One, one, minus two”, 9:39 into Magnitude, the three angles with the axes, and the cosines and ratios they give

Open NCERT p. 354Matches NCERT’s answer

1. AB = B − A = (−1−1, −2−2, 1−(−3)) = (−2, −4, 4).
2. Magnitude of AB = √((−2)²+(−4)²+4²) = √36 = 6.
3. Divide each part by 6: −2/6, −4/6, 4/6 = −1/3, −2/3, 2/3.

AnswerDirection cosines: −1/3, −2/3, 2/3

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. 354One way to think about it

1. Given: î + ĵ + k̂.
2. Magnitude of î+ĵ+k̂ = √(1²+1²+1²) = √3.
3. Direction cosines = 1/√3, 1/√3, 1/√3 — the cosines of the angles made with OX, OY, OZ.
4. All three direction cosines are equal, so all three angles are equal.
5. Equal angles with all three axes means the vector is equally inclined to OX, OY and OZ.

In shortThe direction cosines of î+ĵ+k̂ are all 1/√3, so it makes equal angles with OX, OY and OZ — it is equally inclined to all three axes.

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. 354Matches NCERT’s answer

(i) internally

1. P is at (1, 2, −1) and Q is at (−1, 1, 1).
2. For a point cutting PQ in the ratio m : n internally, its position vector is (m×Q + n×P) ÷ (m + n).
3. Here m : n = 2 : 1, so R = (2×Q + 1×P) ÷ 3.
4. 2×Q = (−2, 2, 2) and 1×P = (1, 2, −1); adding gives (−1, 4, 1).
5. Divide by 3: R = (−1/3, 4/3, 1/3).

AnswerR = −(1/3)î + (4/3)ĵ + (1/3)k̂

(ii) externally

1. For external division in the ratio m : n, the position vector is (m×Q − n×P) ÷ (m − n).
2. Here m : n = 2 : 1, so R = (2×Q − 1×P) ÷ (2 − 1) = 2×Q − P.
3. 2×Q = (−2, 2, 2) and P = (1, 2, −1); subtracting gives (−3, 0, 3).

AnswerR = −3î + 3k̂

Watch this explained “One segment, one ratio, two answers”, 13:02 into Position vectors, the vector joining two points, and dividing a segment in a ratio

Open NCERT p. 355Matches NCERT’s answer

1. The midpoint of P and Q is the average of their position vectors: (P + Q) ÷ 2.
2. P + Q = (2+4, 3+1, 4−2) = (6, 4, 2).
3. Divide by 2: (3, 2, 1).

AnswerMidpoint = 3î + 2ĵ + k̂

Watch this explained “The midpoint is not a separate result”, 12:03 into Position vectors, the vector joining two points, and dividing a segment in a ratio

Open NCERT p. 355One way to think about it

1. Position vectors: A = (3, −4, −4), B = (2, −1, 1), C = (1, −3, −5).
2. Find the sides as vectors: AB = B − A, BC = C − B, CA = A − C.
3. AB = (−1, 3, 5), BC = (−1, −2, −6), CA = (2, −1, 1).
4. Find the squared length of each side: |AB|² = 1+9+25 = 35, |BC|² = 1+4+36 = 41, |CA|² = 4+1+1 = 6.
5. Check Pythagoras: 35 + 6 = 41, so |AB|² + |CA|² = |BC|².
6. This matches the converse of Pythagoras' theorem, so the triangle has a right angle — at the vertex common to sides AB and CA, which is A.

In short|AB|² + |CA|² = |BC|², so triangle ABC is right-angled at A.

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. 355Checked by computerReads two ways: both answers shown

1. By the triangle law, AB⃗ + BC⃗ = AC⃗.
2. (A) AB⃗ + BC⃗ + CA⃗ = AC⃗ + CA⃗ = 0⃗. True.
3. (B) AB⃗ + BC⃗ − AC⃗ = AC⃗ − AC⃗ = 0⃗. True.
4. (C) is printed exactly the same as (B), so as printed it is also true.
5. (D) −CB⃗ = BC⃗, so AB⃗ − CB⃗ + CA⃗ = AB⃗ + BC⃗ + CA⃗ = 0⃗. True.
6. So, as printed, all four options are true and none is 'not true'. Two options are never meant to be the same, so (C) is a misprint.
7. NCERT's answer key gives (C), which can only be the answer if (C) was meant to differ from (B). The likely intended (C) is AB⃗ + BC⃗ − CA⃗ = 0⃗ (CA⃗ in place of AC⃗).
8. Then AB⃗ + BC⃗ − CA⃗ = AC⃗ + AC⃗ = 2AC⃗, which is not 0⃗ because A and C are different points. So the corrected (C) is not true.

AnswerRead with (C) as AB⃗ + BC⃗ − CA⃗ = 0⃗, the likely intended option AB⃗ + BC⃗ − CA⃗ = 0⃗ (NCERT's answer key): (C). As printed, (B) and (C) are identical and all four options are true, so none is 'not true'.

Watch this explained “Which of these is not true?”, 14:13 into Two laws for adding, why they agree, and what addition obeys

Open NCERT p. 355Matches NCERT’s answer

1. Given: a⃗ and b⃗.
2. Collinear vectors means one is a scalar multiple of the other: b⃗ = λa⃗ for SOME real number λ — λ can be any value, positive, negative, or ±1.
3. (A) says exactly this, so it is correct — NOT one of the incorrect options.
4. (B) says a⃗ = ±b⃗, i.e. it needs λ = ±1 only. Take a⃗ = î and b⃗ = −2î (λ = −2): a⃗ ≠ b⃗ and a⃗ ≠ −b⃗. So (B) is incorrect.
5. (C) says the components are NOT proportional — but b⃗ = λa⃗ means every component of b⃗ is λ times the matching component of a⃗, so they ARE proportional. (C)'s claim is false, so (C) is incorrect.
6. (D) needs same direction (λ > 0) and different magnitude (λ ≠ 1). With a⃗ = î, b⃗ = −2î, the vectors point in OPPOSITE directions, so (D) is also incorrect.
7. Only (A) holds for every pair of collinear vectors; (B), (C) and (D) each fail for this one example, so they are the incorrect statements.

Answer(B), (C) and (D) are incorrect.

Watch this explained “Which of these are incorrect?”, 10:23 into Stretching by a scalar, and dividing a vector by its own length