Exercise 10.1
Represent graphically a displacement of 40 km, 30° east of north.
Answer
Draw a directed line segment (arrow) making an angle of 30° measured from North towards East. The length of the arrow represents 40 km on a chosen scale. The starting point is the tail (initial position) and the arrowhead is the head (final position). This directed line segment represents the displacement vector.
Classify as scalar or vector: (i) Time period (ii) Distance (iii) Force (iv) Velocity (v) Work done
Answer
(i) Time period — Scalar (has only magnitude)
(ii) Distance — Scalar (has only magnitude)
(iii) Force — Vector (has magnitude and direction)
(iv) Velocity — Vector (has magnitude and direction)
(v) Work done — Scalar (dot product of force and displacement)
Classify as scalar or vector: (i) 10 kg (ii) 2 metres north-west (iii) 40° (iv) 40 watt (v) 10⁻¹⁹ coulomb (vi) 20 m/s²
Answer
(i) 10 kg — Scalar
(ii) 2 metres north-west — Vector
(iii) 40° — Scalar
(iv) 40 watt — Scalar
(v) 10⁻¹⁹ coulomb — Scalar
(vi) 20 m/s² — Scalar (if magnitude only) or Vector (if acceleration with direction); in context it is a scalar magnitude.
In the figure, identify the following vectors: (i) Coinitial (ii) Equal (iii) Collinear but not equal
Answer
(i) Coinitial vectors: a⃗ and d⃗ (same initial point)
(ii) Equal vectors: b⃗ and d⃗ (same magnitude and direction)
(iii) Collinear but not equal: a⃗ and c⃗ (parallel but different magnitude or opposite direction)
Q5. Answer True or False
(i) a⃗ and −a⃗ are collinear. → True (they lie on parallel/same lines)
(ii) Two collinear vectors are always equal in magnitude. → False
(iii) Two vectors having the same magnitude are collinear. → False
(iv) Two collinear vectors having the same magnitude are equal. → False (they could be opposite in direction)
Exercise 10.2
Compute the magnitude of: (i) a⃗ = î + ĵ + k̂ ... actually: a⃗ + b⃗ where a⃗ = î − ĵ + k̂, b⃗ = 2î + ĵ − k̂...
Answer
Compute the magnitude of the following vectors
(i) a⃗ = î + ĵ + k̂: |a⃗| = √(1²+1²+1²) = √3
(ii) a⃗ = 2î − 7ĵ − 3k̂: |a⃗| = √(4+49+9) = √62 ≈ √62
(iii) a⃗ = (1/√3)î + (1/√3)ĵ − (1/√3)k̂: |a⃗| = √(1/3+1/3+1/3) = 1
Write two different vectors having the same magnitude.
Answer
a⃗ = î + 2ĵ + 3k̂ and b⃗ = 2î + ĵ + 3k̂. Both have magnitude |a⃗| = |b⃗| = √(1+4+9) = √14. But they are different vectors (different directions).
Write two different vectors having the same direction.
Answer
a⃗ = î + ĵ + k̂ and b⃗ = 2î + 2ĵ + 2k̂. Both have direction cosines (1/√3, 1/√3, 1/√3). They are parallel (same direction) but different vectors.
Find values of x, y, z so that a⃗ = b⃗: a⃗ = 2î + yĵ + zk̂, b⃗ = xî + ĵ − 2k̂
Answer
For a⃗ = b⃗: corresponding components must be equal.
⇒ x = 2, y = 1, z = −2.
Find a⃗ + b⃗ for a⃗ = î + 2ĵ, b⃗ = −2î + ĵ (Point not specified but the question asks to find).
Answer
a⃗ + b⃗ = (1−2)î + (2+1)ĵ = −î + 3ĵ
Find a vector with initial point A(2,3,0) and terminal point B(3,−1,4). Find |AB⃗|.
Answer
AB⃗ = (3−2)î + (−1−3)ĵ + (4−0)k̂ = î − 4ĵ + 4k̂
|AB⃗| = √(1+16+16) = √33
Find a⃗ + b⃗, a⃗ − b⃗, and |a⃗ + b⃗| for a⃗ = 2î + 3ĵ − k̂, b⃗ = î − 2ĵ + k̂.
Answer
a⃗ + b⃗ = 3î + ĵ + 0k̂ = 3î + ĵ, |a⃗ + b⃗| = √(9+1) = √10
a⃗ − b⃗ = î + 5ĵ − 2k̂
Find the unit vector in the direction of a⃗ = î + ĵ + 2k̂.
Answer
|a⃗| = √(1+1+4) = √6. Unit vector â = a⃗/|a⃗| = (1/√6)î + (1/√6)ĵ + (2/√6)k̂
Find a unit vector in the direction of PQ⃗ where P = (1,2,3) and Q = (4,5,6).
Answer
PQ⃗ = 3î + 3ĵ + 3k̂. |PQ⃗| = 3√3. Unit vector = (1/√3)î + (1/√3)ĵ + (1/√3)k̂
For a⃗ = 2î − 2ĵ + 3k̂, b⃗ = −î + 2ĵ + k̂, find |a⃗ + λb⃗| = 6 for what λ?... actually: For a⃗ = 2î − ĵ + 2k̂, find |a⃗|. Actually Q10 from NCERT: Given a⃗ = 2î − 2ĵ + 3k̂ and b⃗ = −î + 2ĵ + k̂, find the vector perpendicular to a⃗ + b⃗ and with magnitude...
Answer
NCERT Q10: Find a vector of magnitude 5 in the direction of a⃗ = î − 2ĵ + 2k̂.
Ans. |a⃗| = √(1+4+4) = 3. Unit vector â = (1/3)(î − 2ĵ + 2k̂). Required vector = 5â = (5/3)î − (10/3)ĵ + (10/3)k̂
Show that the vectors 2î − 3ĵ + 4k̂ and −4î + 6ĵ − 8k̂ are collinear.
Answer
Let a⃗ = 2î − 3ĵ + 4k̂ and b⃗ = −4î + 6ĵ − 8k̂. Then b⃗ = −2a⃗. Since b⃗ = λa⃗ with λ = −2, the vectors are collinear (parallel). ✓
Find the direction cosines of the vector î + 2ĵ + 3k̂.
Answer
Magnitude = √(1+4+9) = √14. Direction cosines: l = 1/√14, m = 2/√14, n = 3/√14
Find direction cosines of the vector joining A(1,2,−3) and B(−1,−2,1), directed from A to B.
Answer
AB⃗ = −2î − 4ĵ + 4k̂. |AB⃗| = √(4+16+16) = 6. DC: l = −1/3, m = −2/3, n = 2/3
Show that the points A(1,2,7), B(2,6,3), C(3,10,−1) are collinear.
Answer
AB⃗ = î + 4ĵ − 4k̂, BC⃗ = î + 4ĵ − 4k̂. Since AB⃗ = BC⃗, they are equal vectors, so A, B, C are collinear. ✓
Find the position vector of the midpoint M of A and B with position vectors a⃗ = 2î + 3ĵ and b⃗ = 4î + 5ĵ.
Answer
OM⃗ = (a⃗ + b⃗)/2 = (6î + 8ĵ)/2 = 3î + 4ĵ
Show the points A, B, C with position vectors a⃗ = 2î − ĵ + k̂, b⃗ = î − 3ĵ − 5k̂, c⃗ = 3î − 4ĵ − 4k̂ form a right-angled triangle.
Answer
AB⃗ = −î − 2ĵ − 6k̂, BC⃗ = 2î − ĵ + k̂, CA⃗ = −î + 3ĵ + 5k̂.
|AB⃗|² = 1+4+36 = 41, |BC⃗|² = 4+1+1 = 6, |CA⃗|² = 1+9+25 = 35.
|AB⃗|² = |BC⃗|² + |CA⃗|² since 41 = 6 + 35. So it is a right-angled triangle (right angle at C). ✓
Find the position vector of D if ABCD is a parallelogram and A, B, C have position vectors: a⃗ = î + ĵ + k̂, b⃗ = 2î + 3ĵ + k̂, c⃗ = 3î + 5ĵ − 2k̂.
Answer
In a parallelogram ABCD: d⃗ = a⃗ + c⃗ − b⃗ = (1+3−2)î + (1+5−3)ĵ + (1−2−1)k̂ = 2î + 3ĵ − 2k̂
Find the position vector of point R which divides the line joining P(2,−3,1) and Q(3,1,4) in ratio 2:1 internally and externally.
Answer
Internally: OR⃗ = (2·OQ⃗ + 1·OP⃗)/(2+1) = (2(3î+ĵ+4k̂) + (2î−3ĵ+k̂))/3 = (8î−ĵ+9k̂)/3 = (8/3)î − (1/3)ĵ + 3k̂
Externally (ratio 2:1): OR⃗ = (2·OQ⃗ − 1·OP⃗)/(2−1) = 2(3î+ĵ+4k̂) − (2î−3ĵ+k̂) = 4î + 5ĵ + 7k̂
Find the position vector of the midpoint M of the segment joining A(5,−1,2) and B(1,3,4). Show that A, B and M are collinear.
Answer
OM⃗ = ((5+1)/2)î + ((−1+3)/2)ĵ + ((2+4)/2)k̂ = 3î + ĵ + 3k̂
AM⃗ = OM⃗ − OA⃗ = −2î + 2ĵ + k̂, AB⃗ = −4î + 4ĵ + 2k̂ = 2·AM⃗. So A, M, B are collinear. ✓
Exercise 10.3
Find the angle between two vectors a⃗ and b⃗ with |a⃗| = √3, |b⃗| = 2, a⃗·b⃗ = √6.
Answer
cos θ = (a⃗·b⃗)/(|a⃗||b⃗|) = √6/(√3·2) = √6/(2√3) = √2/2
⇒ θ = π/4 (45°)
Find the angle between a⃗ = î − 2ĵ + 3k̂ and b⃗ = 3î − 2ĵ + k̂.
Answer
a⃗·b⃗ = 3+4+3 = 10. |a⃗| = √(1+4+9) = √14. |b⃗| = √(9+4+1) = √14.
cos θ = 10/14 = 5/7 → θ = cos⁻¹(5/7)
Find the projection of a⃗ on b⃗ where a⃗ = 2î + 3ĵ + 2k̂, b⃗ = î + 2ĵ + k̂.
Answer
Projection = (a⃗·b⃗)/|b⃗| = (2+6+2)/√6 = 10/√6 = 5√6/3
Find a⃗·b⃗ if |a⃗| = 3, |b⃗| = 4, and the angle between them is 60°.
Answer
a⃗·b⃗ = |a⃗||b⃗|cos 60° = 3·4·(1/2) = 6
Find λ so that a⃗ = 2î + λĵ + k̂ and b⃗ = î − 2ĵ + k̂ are perpendicular.
Answer
a⃗·b⃗ = 0
⇒ 2−2λ+1 = 0
⇒ 3 = 2λ
⇒ λ = 3/2
Show that (a⃗ − b⃗) × (a⃗ + b⃗) = 2(a⃗ × b⃗)... actually Q6 is: Show |a⃗|b⃗ + |b⃗|a⃗ is perpendicular to |a⃗|b⃗ − |b⃗|a⃗ for any non-zero vectors a⃗ and b⃗.
Answer
Let p⃗ = |a⃗|b⃗ + |b⃗|a⃗ and q⃗ = |a⃗|b⃗ − |b⃗|a⃗.
p⃗·q⃗ = (|a⃗|b⃗ + |b⃗|a⃗)·(|a⃗|b⃗ − |b⃗|a⃗) = |a⃗|²|b⃗|² − |b⃗|²|a⃗|² = 0. So p⃗ ⊥ q⃗. ✓
Evaluate the product (3a⃗ − 5b⃗)·(2a⃗ + 7b⃗).
Answer
= 6|a⃗|² + 21a⃗·b⃗ − 10a⃗·b⃗ − 35|b⃗|² = 6|a⃗|² + 11(a⃗·b⃗) − 35|b⃗|²
Find |x⃗| if (x⃗ − a⃗)·(x⃗ + a⃗) = 12, |a⃗| = 5.
Answer
|x⃗|² − |a⃗|² = 12
⇒ |x⃗|² = 12 + 25 = 37
⇒ |x⃗| = √37
For a⃗ = 2î + 2ĵ + 3k̂, b⃗ = −î + 2ĵ + k̂, c⃗ = 3î + ĵ, find λ such that a⃗ + λb⃗ is perpendicular to c⃗.
Answer
a⃗ + λb⃗ = (2−λ)î + (2+2λ)ĵ + (3+λ)k̂. For ⊥ to c⃗ = 3î + ĵ:
(a⃗+λb⃗)·c⃗ = 3(2−λ) + (2+2λ) = 6−3λ+2+2λ = 8−λ = 0
⇒ λ = 8
Show that |a⃗|b⃗ + |b⃗|a⃗ bisects the angle between a⃗ and b⃗ for non-zero vectors.
Answer
Unit vectors along a⃗ and b⃗ are â = a⃗/|a⃗| and b̂ = b⃗/|b⃗|. The bisector direction is â + b̂ = a⃗/|a⃗| + b⃗/|b⃗| = (|b⃗|a⃗ + |a⃗|b⃗)/(|a⃗||b⃗|). So |a⃗|b⃗ + |b⃗|a⃗ is in the direction of â + b̂, which bisects the angle. ✓
Find the value of λ for which a⃗ = λ(î + ĵ + k̂) is a unit vector.
Answer
|a⃗|² = λ²(1+1+1) = 3λ² = 1
⇒ λ = ±1/√3.
⇒ λ = ±1/√3
Find a vector of magnitude √2 making equal angles with î, ĵ, k̂.
Answer
Let the vector be a(î+ĵ+k̂). Magnitude = a√3 = √2
⇒ a = √(2/3). Vector = √(2/3)(î+ĵ+k̂)
a⃗ = 2î − ĵ + k̂, b⃗ = î + 2ĵ − k̂, c⃗ = î + ĵ − 2k̂. Find λ, μ such that c⃗ = λa⃗ + μb⃗... actually: show a⃗, b⃗, c⃗ form a triangle and find (2a⃗ − b⃗).(a⃗ + b⃗ + c⃗)...
Answer
Actual Q13: a⃗·b⃗ = |a⃗||b⃗|, show that a⃗ = λb⃗.
Ans. a⃗·b⃗ = |a⃗||b⃗|cos θ = |a⃗||b⃗| means cos θ = 1, so θ = 0, so a⃗ and b⃗ are parallel. Hence a⃗ = λb⃗ for some scalar λ. ✓
If a⃗, b⃗, c⃗ are mutually perpendicular unit vectors, find |a⃗ + b⃗ + c⃗|.
Answer
|a⃗+b⃗+c⃗|² = |a⃗|² + |b⃗|² + |c⃗|² + 2(a⃗·b⃗ + b⃗·c⃗ + c⃗·a⃗) = 1+1+1+0 = 3
⇒ |a⃗+b⃗+c⃗| = √3
Show that the points A(2,3,4), B(−1,−2,1), C(5,8,7) are collinear.
Answer
AB⃗ = −3î − 5ĵ − 3k̂, BC⃗ = 6î + 10ĵ + 6k̂ = −2AB⃗. Since BC⃗ = −2AB⃗, vectors are parallel and B is common. So A, B, C are collinear. ✓
Show that a⃗(b⃗ + c⃗) + b⃗(c⃗ + a⃗) + c⃗(a⃗ + b⃗) = 0.
Answer
Expanding: a⃗·b⃗ + a⃗·c⃗ + b⃗·c⃗ + b⃗·a⃗ + c⃗·a⃗ + c⃗·b⃗ = 2(a⃗·b⃗ + b⃗·c⃗ + c⃗·a⃗) ≠ 0 in general. Actually the Q is a⃗×(b⃗+c⃗) + b⃗×(c⃗+a⃗) + c⃗×(a⃗+b⃗) = 0 (cross product).
= a⃗×b⃗ + a⃗×c⃗ + b⃗×c⃗ + b⃗×a⃗ + c⃗×a⃗ + c⃗×b⃗ = (a⃗×b⃗ − a⃗×b⃗) + (b⃗×c⃗ − b⃗×c⃗) + (c⃗×a⃗ − c⃗×a⃗) = 0 ✓
Find the area of triangle with vertices A(1,1,2), B(2,3,5), C(1,5,5).
Answer
AB⃗ = î+2ĵ+3k̂, AC⃗ = 0î+4ĵ+3k̂. AB⃗×AC⃗ = |î ĵ k̂; 1 2 3; 0 4 3| = î(6−12) − ĵ(3−0) + k̂(4−0) = −6î − 3ĵ + 4k̂.
|AB⃗×AC⃗| = √(36+9+16) = √61. Area = (1/2)√61 = √61/2
Find the area of parallelogram with adjacent sides a⃗ = 3î + ĵ + 4k̂, b⃗ = î − ĵ + k̂.
Answer
a⃗×b⃗ = |î ĵ k̂; 3 1 4; 1 −1 1| = î(1+4) − ĵ(3−4) + k̂(−3−1) = 5î + ĵ − 4k̂.
Area = |a⃗×b⃗| = √(25+1+16) = √42
Exercise 10.4
Q1. Find a⃗×b⃗ for
(i) a⃗ = î − ĵ + k̂, b⃗ = î + ĵ + k̂
Ans.(i) |î ĵ k̂; 1 −1 1; 1 1 1| = î(−1−1) − ĵ(1−1) + k̂(1+1) = −2î + 0ĵ + 2k̂ = −2î + 2k̂
(ii) a⃗ = 2î + 3ĵ + k̂, b⃗ = î + 2ĵ − k̂
Ans.(ii) = î(−3−2) − ĵ(−2−1) + k̂(4−3) = −5î + 3ĵ + k̂
Find a unit vector perpendicular to each of a⃗+b⃗ and a⃗−b⃗ where a⃗ = î+ĵ+k̂, b⃗ = î+2ĵ+3k̂.
Answer
a⃗+b⃗ = 2î+3ĵ+4k̂, a⃗−b⃗ = 0î−ĵ−2k̂.
(a⃗+b⃗)×(a⃗−b⃗) = |î ĵ k̂; 2 3 4; 0 −1 −2| = î(−6+4) − ĵ(−4−0) + k̂(−2−0) = −2î+4ĵ−2k̂.
Magnitude = √(4+16+4) = √24 = 2√6. Unit vector = (−î+2ĵ−k̂)/√6
If two adjacent sides of a triangle are a⃗ = −2î+4ĵ+5k̂ and b⃗ = î−2ĵ−3k̂, find the unit normal to the plane of the triangle and area.
Answer
a⃗×b⃗ = |î ĵ k̂; −2 4 5; 1 −2 −3| = î(−12+10) − ĵ(6−5) + k̂(4−4) = −2î−ĵ.
|a⃗×b⃗| = √(4+1) = √5. Unit normal = (−2î−ĵ)/√5. Area = √5/2.
Show that (a⃗−b⃗)×(a⃗+b⃗) = 2(a⃗×b⃗).
Answer
(a⃗−b⃗)×(a⃗+b⃗) = a⃗×a⃗ + a⃗×b⃗ − b⃗×a⃗ − b⃗×b⃗ = 0 + a⃗×b⃗ + a⃗×b⃗ − 0 = 2(a⃗×b⃗) ✓
Find λ and μ if (2î+6ĵ+27k̂)×(î+λĵ+μk̂) = 0⃗.
Answer
The vectors are parallel: 2/1 = 6/λ = 27/μ
⇒ λ = 3, μ = 27/2.
⇒ λ = 3, μ = 27/2
Given vectors a⃗ = 3î+4ĵ+5k̂ and b⃗ = 2î+ĵ−4k̂. Find a⃗×b⃗ and verify that it is perpendicular to both a⃗ and b⃗.
Answer
a⃗×b⃗ = |î ĵ k̂; 3 4 5; 2 1 −4| = î(−16−5) − ĵ(−12−10) + k̂(3−8) = −21î+22ĵ−5k̂
Verify: (−21)(3)+(22)(4)+(−5)(5) = −63+88−25 = 0 ✓. (−21)(2)+(22)(1)+(−5)(−4) = −42+22+20 = 0 ✓
Let a⃗ = î+4ĵ+2k̂, b⃗ = 3î−2ĵ+7k̂, c⃗ = 2î−ĵ+4k̂. Find a vector d⃗ perpendicular to both a⃗ and b⃗ with c⃗·d⃗ = 18.
Answer
d⃗ = λ(a⃗×b⃗). a⃗×b⃗ = |î ĵ k̂; 1 4 2; 3 −2 7| = î(28+4) − ĵ(7−6) + k̂(−2−12) = 32î−ĵ−14k̂.
c⃗·d⃗ = λ(c⃗·(a⃗×b⃗)) = λ(64+1−56) = 9λ = 18
⇒ λ = 2.
d⃗ = 64î − 2ĵ − 28k̂
If |a⃗| = √26, |b⃗| = 7, |a⃗×b⃗| = 35, find a⃗·b⃗.
Answer
(a⃗·b⃗)² + |a⃗×b⃗|² = |a⃗|²|b⃗|²
⇒ (a⃗·b⃗)² = 26×49 − 1225 = 1274 − 1225 = 49
⇒ a⃗·b⃗ = ±7
Find the area of triangle with vertices A(1,1,2), B(2,3,5), C(1,5,5). (Same as Q17 of Ex 10.3)
Answer
Area = (1/2)|AB⃗×AC⃗| = √61/2 (computed above)
Find the area of parallelogram ABCD where diagonals are AC⃗ = 3î+ĵ−2k̂ and BD⃗ = î−3ĵ+4k̂.
Answer
Area = (1/2)|AC⃗×BD⃗|. AC⃗×BD⃗ = |î ĵ k̂; 3 1 −2; 1 −3 4| = î(4−6)−ĵ(12+2)+k̂(−9−1) = −2î−14ĵ−10k̂.
|AC⃗×BD⃗| = √(4+196+100) = √300 = 10√3. Area = 5√3
Show that (a⃗+b⃗)×(a⃗−b⃗) = 2(b⃗×a⃗)... actually same as Q4. Prove a⃗×(b⃗+c⃗) + b⃗×(c⃗+a⃗) + c⃗×(a⃗+b⃗) = 0⃗.
Answer
Already shown in Ex 10.3 Q16. ✓
If a⃗·a⃗ = 0 and a⃗·b⃗ = 0, then what can we conclude about b⃗?
Answer
a⃗·a⃗ = |a⃗|² = 0
⇒ a⃗ = 0⃗ (zero vector). Then a⃗·b⃗ = 0⃗·b⃗ = 0 for any b⃗. So b⃗ can be any vector; no conclusion about b⃗ is possible.
Miscellaneous Exercise on Chapter 10
Write the projection of b⃗+c⃗ on a⃗, where a⃗ = 2î−2ĵ+k̂, b⃗ = î+2ĵ−2k̂, c⃗ = 2î−ĵ+4k̂.
Answer
b⃗+c⃗ = 3î+ĵ+2k̂. Projection on a⃗ = (b⃗+c⃗)·a⃗/|a⃗| = (6−2+2)/√(4+4+1) = 6/3 = 2
Show that a⃗×b⃗ = a⃗×c⃗ does NOT imply b⃗ = c⃗.
Answer
a⃗×b⃗ = a⃗×c⃗
⇒ a⃗×(b⃗−c⃗) = 0⃗. This means either a⃗ = 0⃗ or (b⃗−c⃗) is parallel to a⃗. So b⃗−c⃗ = λa⃗ for some scalar λ, i.e., b⃗ = c⃗ + λa⃗ ≠ c⃗ in general. ✓
Find |a⃗|, |b⃗| if |a⃗| = 2|b⃗| and (a⃗+b⃗)·(a⃗−b⃗) = 12.
Answer
|a⃗|² − |b⃗|² = 12
⇒ 4|b⃗|² − |b⃗|² = 3|b⃗|² = 12
⇒ |b⃗|² = 4
⇒ |b⃗| = 2, |a⃗| = 4.
Show that for any two vectors a⃗ and b⃗: |a⃗+b⃗|² + |a⃗−b⃗|² = 2(|a⃗|² + |b⃗|²).
Answer
|a⃗+b⃗|² = |a⃗|²+2a⃗·b⃗+|b⃗|². |a⃗−b⃗|² = |a⃗|²−2a⃗·b⃗+|b⃗|². Sum = 2|a⃗|²+2|b⃗|² = 2(|a⃗|²+|b⃗|²) ✓
Show that (a⃗·b⃗)² + |a⃗×b⃗|² = |a⃗|²|b⃗|².
Answer
a⃗·b⃗ = |a⃗||b⃗|cos θ
⇒ (a⃗·b⃗)² = |a⃗|²|b⃗|²cos²θ. |a⃗×b⃗| = |a⃗||b⃗|sin θ
⇒ |a⃗×b⃗|² = |a⃗|²|b⃗|²sin²θ. Sum = |a⃗|²|b⃗|²(cos²θ+sin²θ) = |a⃗|²|b⃗|² ✓
Given a⃗ = î+ĵ+k̂, b⃗ = 2î−ĵ+3k̂, c⃗ = î−2ĵ+k̂. Find 2a⃗−b⃗+3c⃗.
Answer
2a⃗ = 2î+2ĵ+2k̂, −b⃗ = −2î+ĵ−3k̂, 3c⃗ = 3î−6ĵ+3k̂. Sum = 3î−3ĵ+2k̂
If two vectors are such that a⃗+b⃗+c⃗ = 0⃗ and |a⃗| = 3, |b⃗| = 5, |c⃗| = 7, find the angle between a⃗ and b⃗.
Answer
c⃗ = −(a⃗+b⃗). |c⃗|² = |a⃗+b⃗|² = |a⃗|²+2a⃗·b⃗+|b⃗|²
⇒ 49 = 9+2a⃗·b⃗+25
⇒ a⃗·b⃗ = 15/2.
cos θ = (a⃗·b⃗)/(|a⃗||b⃗|) = (15/2)/(15) = 1/2
⇒ θ = 60° (π/3)
Show that the vector î+ĵ+k̂ makes equal angles with the coordinate axes.
Answer
The direction cosines are l = m = n = 1/√3. Since all are equal, the vector makes equal angles with each coordinate axis. The angle α = cos⁻¹(1/√3) is the same with each axis. ✓
For any three vectors a⃗, b⃗, c⃗, show that a⃗·(b⃗+c⃗) + b⃗·(c⃗+a⃗) + c⃗·(a⃗+b⃗) = 0... Actually prove: if a⃗ ⊥ b⃗ then |a⃗+b⃗| = |a⃗−b⃗|.
Answer
If a⃗·b⃗ = 0: |a⃗+b⃗|² = |a⃗|²+2a⃗·b⃗+|b⃗|² = |a⃗|²+|b⃗|² and |a⃗−b⃗|² = |a⃗|²−2a⃗·b⃗+|b⃗|² = |a⃗|²+|b⃗|². Equal. ✓
Show that the direction cosines of a vector equally inclined to the axes are 1/√3, 1/√3, 1/√3.
Answer
Let l = m = n. Then l²+m²+n² = 1
⇒ 3l² = 1
⇒ l = 1/√3. So direction cosines are 1/√3, 1/√3, 1/√3. ✓
Find the vector such that a⃗ = î−ĵ+k̂ and a⃗+b⃗+c⃗ = î+ĵ+k̂ where b⃗ = 2î−ĵ+2k̂. Find c⃗.
Answer
c⃗ = (î+ĵ+k̂) − a⃗ − b⃗ = (î+ĵ+k̂) − (î−ĵ+k̂) − (2î−ĵ+2k̂) = −2î+3ĵ−2k̂.
⇒ c⃗ = −2î+3ĵ−2k̂
If a⃗, b⃗, c⃗ are mutually perpendicular unit vectors, show that a⃗+b⃗+c⃗ cannot equal zero unless each is zero.
Answer
If a⃗+b⃗+c⃗ = 0⃗, then |a⃗+b⃗+c⃗|² = |a⃗|²+|b⃗|²+|c⃗|²+2(a⃗·b⃗+b⃗·c⃗+c⃗·a⃗) = 1+1+1+0 = 3 ≠ 0. Contradiction. ✓
If the vertices of triangle are A(1,1,2), B(2,3,5), C(1,5,5), find the interior angle at vertex B.
Answer
BA⃗ = −î−2ĵ−3k̂, BC⃗ = −î+2ĵ. BA⃗·BC⃗ = 1−4+0 = −3. |BA⃗| = √14, |BC⃗| = √5.
cos B = −3/√70 → B = cos⁻¹(−3/√70)
Find the sine of the angle between a⃗ = 3î+ĵ+2k̂ and b⃗ = 2î−2ĵ+4k̂.
Answer
a⃗×b⃗ = |î ĵ k̂; 3 1 2; 2 −2 4| = î(4+4)−ĵ(12−4)+k̂(−6−2) = 8î−8ĵ−8k̂.
|a⃗×b⃗| = 8√3. |a⃗| = √14, |b⃗| = √24 = 2√6. sin θ = 8√3/(√14·2√6) = 8√3/(2√84) = 8√3/(4√21) = 2√3/√21 = 2/√7.
→ sin θ = 2/√7
If a⃗ = 2î+2ĵ+3k̂, b⃗ = −î+2ĵ+k̂, c⃗ = 3î+ĵ, find a vector coplanar with b⃗ and c⃗ perpendicular to a⃗... Actually Q15 is the MCQ section starting:
Answer
MCQ: The value of î·(ĵ×k̂) + ĵ·(î×k̂) + k̂·(î×ĵ)
Answer
î·(ĵ×k̂) = î·î = 1. ĵ·(î×k̂) = ĵ·(−ĵ) = −1. k̂·(î×ĵ) = k̂·k̂ = 1. Sum = 1−1+1 = 1. Answer: (C) 1
MCQ: If θ is the angle between two vectors a⃗ and b⃗, then a⃗·b⃗ ≥ 0 only when:
Answer
a⃗·b⃗ = |a⃗||b⃗|cos θ ≥ 0 when cos θ ≥ 0, i.e., 0 ≤ θ ≤ π/2. Answer: (B) 0 ≤ θ ≤ π/2
MCQ: Let a⃗ and b⃗ be two unit vectors with angle θ between them. |a⃗+b⃗| = 1 when:
Answer
|a⃗+b⃗|² = 1+1+2cos θ = 2+2cos θ = 1
⇒ cos θ = −1/2
⇒ θ = 2π/3. Answer: (B) θ = 2π/3
MCQ: Area of rectangle whose adjacent sides are a⃗ = î+2k̂ and b⃗ = î+2ĵ+k̂:
Answer
Area = |a⃗×b⃗|. a⃗×b⃗ = |î ĵ k̂; 1 0 2; 1 2 1| = î(0−4)−ĵ(1−2)+k̂(2−0) = −4î+ĵ+2k̂.
|a⃗×b⃗| = √(16+1+4) = √21. Answer: (C) √21
MCQ: If θ is angle between unit vectors a⃗ and b⃗, |a⃗−b⃗| = sin(θ/2). True/False?
Answer
|a⃗−b⃗|² = 1+1−2cos θ = 2(1−cos θ) = 4sin²(θ/2)
⇒ |a⃗−b⃗| = 2sin(θ/2) ≠ sin(θ/2). False; the correct value is 2sin(θ/2).
Important Extra Questions
If a⃗ = î+ĵ, b⃗ = ĵ+k̂, find a vector coplanar with a⃗ and b⃗ that is also perpendicular to b⃗.
Answer
Let d⃗ = αa⃗ + βb⃗ = α(î+ĵ) + β(ĵ+k̂) = αî + (α+β)ĵ + βk̂.
For d⃗·b⃗ = 0: d⃗·(ĵ+k̂) = (α+β)+β = α+2β = 0
⇒ α = −2β.
Take β = 1: α = −2. d⃗ = −2î + (−2+1)ĵ + k̂ = −2î − ĵ + k̂
Find a unit vector perpendicular to the plane of the triangle with vertices A(1,−1,2), B(2,0,−1), C(0,2,1).
Answer
AB⃗ = î+ĵ−3k̂, AC⃗ = −î+3ĵ−k̂.
AB⃗×AC⃗ = |î ĵ k̂; 1 1 −3; −1 3 −1| = î(−1+9)−ĵ(−1−3)+k̂(3+1) = 8î+4ĵ+4k̂ = 4(2î+ĵ+k̂).
|8î+4ĵ+4k̂| = 4√6. Unit vector = (2î+ĵ+k̂)/√6
For three vectors a⃗, b⃗, c⃗ prove that a⃗·(b⃗×c⃗) = (a⃗×b⃗)·c⃗ (scalar triple product).
Answer
Let a⃗ = (a₁,a₂,a₃), b⃗ = (b₁,b₂,b₃), c⃗ = (c₁,c₂,c₃). Then b⃗×c⃗ has components (b₂c₃−b₃c₂, b₃c₁−b₁c₃, b₁c₂−b₂c₁). a⃗·(b⃗×c⃗) = a₁(b₂c₃−b₃c₂)+a₂(b₃c₁−b₁c₃)+a₃(b₁c₂−b₂c₁) = det[a⃗ b⃗ c⃗]. Similarly (a⃗×b⃗)·c⃗ = same determinant. ✓
If |a⃗+b⃗| = |a⃗−b⃗|, show that a⃗ ⊥ b⃗.
Answer
|a⃗+b⃗|² = |a⃗−b⃗|²
⇒ |a⃗|²+2a⃗·b⃗+|b⃗|² = |a⃗|²−2a⃗·b⃗+|b⃗|²
⇒ 4a⃗·b⃗ = 0
⇒ a⃗·b⃗ = 0
⇒ a⃗ ⊥ b⃗. ✓
The position vectors of A and B are a⃗ = 2î−ĵ+k̂ and b⃗ = î+ĵ−k̂ respectively. Find the position vector of the point C that divides AB externally in the ratio 1:2.
Answer
External division (1:2): OC⃗ = (1·b⃗ − 2·a⃗)/(1−2) = (b⃗ − 2a⃗)/(−1) = 2a⃗ − b⃗.
= 2(2î−ĵ+k̂) − (î+ĵ−k̂) = (4−1)î + (−2−1)ĵ + (2+1)k̂ = 3î − 3ĵ + 3k̂