NCERT Solutions›Class 12 Mathematics›Chapter 10

📐

Chapter 10 · Class 12 Mathematics

Vector Algebra

5 exercises73 questions solved

Exercise 10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4 Exercise Misc

Exercise 10.1Basic Concepts of Vectors

Represent graphically a displacement of 40 km, 30° east of north.

Solution

Draw a vector from origin at angle 30° east of north (i.e., 60° from x-axis toward y-axis), with magnitude 40 km. The vector makes 60° with the east direction.

Classify the following measures as scalars and vectors: (i) 10 kg (ii) 2 metres north-west (iii) 40° (iv) 40 watt (v) 10⁻¹⁹ coulomb (vi) 20 m/s²

Solution

(i) Scalar (mass). (ii) Vector (displacement). (iii) Scalar (angle). (iv) Scalar (power). (v) Scalar (charge). (vi) Scalar (speed/acceleration magnitude without direction).

Classify as scalar/vector: (i) Time period (ii) Distance (iii) Force (iv) Velocity (v) Work done

Solution

(i) Scalar. (ii) Scalar. (iii) Vector. (iv) Vector. (v) Scalar.

In the figure, identify the following vectors: (i) Coinitial (ii) Equal (iii) Collinear but not equal

Solution

(i) Coinitial vectors share the same initial point. (ii) Equal vectors have same magnitude and direction. (iii) Collinear but not equal: same line of action but different magnitudes or opposite directions.

Answer true/false: (i) a and −a are collinear. (ii) Two collinear vectors are always equal in magnitude. (iii) Two vectors having same magnitude are collinear. (iv) Two collinear vectors having same magnitude are equal.

Solution

(i) True. (ii) False — they may have different magnitudes. (iii) False — equal magnitude doesn't imply collinear. (iv) False — they may have opposite directions.

Exercise 10.2Addition of Vectors and Multiplication by Scalar

Compute the magnitude of the vector a = î + ĵ + k̂.

Solution

|a| = √(1²+1²+1²) = √3.

Write two different vectors having same magnitude.

Solution

a = î + 2ĵ + 3k̂, b = 3î + 2ĵ + k̂. |a| = |b| = √14.

Write two different vectors having same direction.

Solution

a = î + ĵ + k̂, b = 2î + 2ĵ + 2k̂. Both point in direction (1,1,1).

Find the values of x and y so that 2î + 3ĵ = xî + yĵ.

Solution

Equating components: x = 2, y = 3.

Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (−5, 7).

Solution

Vector = (−5−2)î+(7−1)ĵ = −7î+6ĵ. Scalar components: −7, 6. Vector components: −7î, 6ĵ.

Find the sum of vectors a = î − 2ĵ + k̂, b = −2î + 4ĵ + 5k̂, c = î − 6ĵ − 7k̂.

Solution

a+b+c = (1−2+1)î+(−2+4−6)ĵ+(1+5−7)k̂ = 0î−4ĵ−1k̂ = −4ĵ−k̂.

Find the unit vector in the direction of vector PQ where P=(1,2,3) and Q=(4,5,6).

Solution

PQ = 3î+3ĵ+3k̂. |PQ|=3√3. Unit vector = (1/√3)(î+ĵ+k̂).

Find the unit vector in the direction of sum of vectors a = 2î+2ĵ−5k̂ and b = 2î+ĵ+3k̂.

Solution

a+b = 4î+3ĵ−2k̂. |a+b|=√(16+9+4)=√29. Unit vector = (4î+3ĵ−2k̂)/√29.

Write the direction cosines of vector −2î+ĵ−5k̂ and magnitude of a = 3î−4k̂.

Solution

|−2î+ĵ−5k̂|=√30. DCs: −2/√30, 1/√30, −5/√30. |3î−4k̂|=√(9+16)=5.

Q10

Find the vector joining P(2,3,0) and Q(−1,−2,−4).

Solution

PQ = (−1−2)î+(−2−3)ĵ+(−4−0)k̂ = −3î−5ĵ−4k̂.

Q11

Find the direction cosines of the vector joining A(1,2,−3) and B(−1,−2,1), directed from A to B.

Solution

AB = −2î−4ĵ+4k̂. |AB|=√(4+16+16)=6. DCs: −1/3, −2/3, 2/3.

Q12

Show that vector î+ĵ+k̂ is equally inclined to the positive directions of coordinate axes.

Solution

a=î+ĵ+k̂. DCs: cosα=cosβ=cosγ=1/√3. All equal → equally inclined.

Q13

Find a vector of magnitude 5 units parallel to the resultant of a = 2î+3ĵ−k̂ and b = î−2ĵ+k̂.

Solution

a+b = 3î+ĵ. |a+b|=√10. Unit vector=3î+ĵ)/√10. Required: 5(3î+ĵ)/√10 = (15î+5ĵ)/√10 = (3√10î+√10ĵ)/2.

Q14

If a = î+ĵ+k̂, b = 2î−ĵ+3k̂, c = î−2ĵ+k̂, find unit vector parallel to 2a−b+3c.

Solution

2a=2î+2ĵ+2k̂, −b=−2î+ĵ−3k̂, 3c=3î−6ĵ+3k̂. Sum=3î−3ĵ+2k̂. |Sum|=√(9+9+4)=√22. Unit vector=(3î−3ĵ+2k̂)/√22.

Q15

If a = î+ĵ, b = ĵ+k̂, c = k̂+î, find unit vector in direction of a+b+c.

Solution

a+b+c=2î+2ĵ+2k̂. Unit vector=(î+ĵ+k̂)/√3.

Q16

Show that points A(1,2,7), B(2,6,3), C(3,10,−1) are collinear.

Solution

AB=î+4ĵ−4k̂, BC=î+4ĵ−4k̂. AB=BC → collinear.

Q17

Show that the vectors 2î−ĵ+k̂, î−3ĵ−5k̂, 3î−4ĵ−4k̂ form the vertices of a right triangle.

Solution

Let a=2î−ĵ+k̂, b=î−3ĵ−5k̂, c=3î−4ĵ−4k̂. AB=b−a=−î−2ĵ−6k̂, BC=c−b=2î−ĵ+k̂, CA=a−c=−î+3ĵ+5k̂. |AB|²=1+4+36=41, |BC|²=4+1+1=6, |CA|²=1+9+25=35. 6+35=41=|AB|² → right angle at B.

Q18

If a is a nonzero vector of magnitude a, what is unit vector â?

Solution

â = a/|a| = a/a (where a = |a|).

Q19

Find the value of x for which x(î+ĵ+k̂) is a unit vector.

Solution

|x(î+ĵ+k̂)| = |x|√3 = 1 → x = ±1/√3.

Exercise 10.3Product of Two Vectors — Dot Product

Find the angle between a = î−2ĵ+3k̂ and b = 3î−2ĵ+k̂.

Solution

a·b = 3+4+3 = 10. |a|=√14, |b|=√14. cosθ=10/14=5/7. θ=cos⁻¹(5/7).

If a = î−ĵ+k̂ and b = î+ĵ−k̂, find a·b and the angle between them.

Solution

a·b=1−1−1=−1. |a|=|b|=√3. cosθ=−1/3. θ=cos⁻¹(−1/3).

Find a·b if a = 3î−ĵ+2k̂, b = 2î+3ĵ−k̂.

Solution

a·b = 6−3−2 = 1.

If a·a = 0 and a·b = 0, what can you conclude about b?

Solution

a·a=0 → a=0 (zero vector). Then a·b=0 regardless of b, so b can be anything.

Find projection of a = 2î−3ĵ+6k̂ on b = î+2ĵ+2k̂.

Solution

Projection = a·b/|b| = (2−6+12)/3 = 8/3.

Find the projection of b+c on a where a=2î−2ĵ+k̂, b=î+2ĵ−2k̂, c=2î−ĵ+4k̂.

Solution

b+c=3î+ĵ+2k̂. (b+c)·a=(6−2+2)=6. |a|=3. Projection=6/3=2.

Show that a = 2î+3ĵ+4k̂ and b = −4î+2ĵ+... are perpendicular. Actually show 3î+2ĵ+9k̂ and î+2ĵ−k̂ are perpendicular.

Solution

a·b=3·1+2·2+9·(−1)=3+4−9=−2 ≠0... Standard problem: a=î+2ĵ+3k̂, b=3î+2ĵ−k̂: a·b=3+4−3=4. Try given: if a·b=0 then perpendicular.

Find |a−b| given |a|=2, |b|=3, a·b=4.

Solution

|a−b|²=|a|²−2a·b+|b|²=4−8+9=5. |a−b|=√5.

If |a+b|=|a−b|, show a⊥b.

Solution

|a+b|²=|a|²+2a·b+|b|². |a−b|²=|a|²−2a·b+|b|². Setting equal: 4a·b=0 → a·b=0 → a⊥b.

Q10

The scalar product of the vector î+ĵ+k̂ with a unit vector along the sum of 2î+4ĵ−5k̂ and λî+2ĵ+3k̂ is 1. Find λ.

Solution

Sum=(2+λ)î+6ĵ−2k̂. |Sum|=√((2+λ)²+40). Unit vector dot (î+ĵ+k̂)=1: [(2+λ)+6−2]/√((2+λ)²+40)=1. (λ+6)²=(λ+2)²+40. λ²+12λ+36=λ²+4λ+44+... → 8λ=8 → λ=1.

Q11

If a, b, c are mutually perpendicular unit vectors, then |a+b+c|=?

Solution

|a+b+c|²=|a|²+|b|²+|c|²+2(a·b+b·c+c·a)=1+1+1+0=3. |a+b+c|=√3.

Q12

Show that |a|b + |b|a is perpendicular to |a|b − |b|a for any nonzero a, b.

Solution

(|a|b+|b|a)·(|a|b−|b|a)=|a|²|b|²−|b|²|a|²=0. ✓

Q13

If a·a=0 and a·b=0, what can you conclude?

Solution

a·a=|a|²=0 → a=0. Then a·b=0 holds trivially for any b.

Q14

If a, b, c are such that a+b+c=0, then find a·b+b·c+c·a.

Solution

|a+b+c|²=0. |a|²+|b|²+|c|²+2(a·b+b·c+c·a)=0. If |a|=1,|b|=1,|c|=1: 3+2(a·b+b·c+c·a)=0 → sum=−3/2. General: sum=−(|a|²+|b|²+|c|²)/2.

Q15

If the vertices of a triangle are A(1,1,2), B(2,3,5), C(1,5,5), find the angles of the triangle.

Solution

AB=î+2ĵ+3k̂, AC=0î+4ĵ+3k̂, BC=−î+2ĵ+0k̂. cosA=AB·AC/(|AB||AC|)=(0+8+9)/(√14·5)=17/5√14. cosB=BA·BC/|BA||BC|=(−1−4+0)/(√14·√5)=−5/√70. cosC=CA·CB/(|CA||CB|)=(0−8+0)/(5·√5)=−8/5√5. Angles=cos⁻¹ of above.

Q16

Show that for the triangle with vertices A, B, C: a·(b−c)+b·(c−a)+c·(a−b)=0.

Solution

= a·b−a·c+b·c−b·a+c·a−c·b = (a·b−b·a)+(b·c−c·b)+(c·a−a·c)=0 (dot product is commutative). ✓

Q17

If a is unit vector and (x−a)·(x+a)=8, find |x|.

Solution

(x−a)·(x+a)=|x|²−|a|²=|x|²−1=8. |x|²=9. |x|=3.

Q18

Show that the vectors 2î−3ĵ+4k̂ and −4î+6ĵ−8k̂ are collinear.

Solution

−4î+6ĵ−8k̂ = −2(2î−3ĵ+4k̂). One is scalar multiple of other → collinear.

Exercise 10.4Cross Product of Two Vectors

Find |a × b| if a = î−7ĵ+7k̂ and b = 3î−2ĵ+2k̂.

Solution

a×b=|î ĵ k̂; 1 −7 7; 3 −2 2|=(−14+14)î−(2−21)ĵ+(−2+21)k̂=0î+19ĵ+19k̂. |a×b|=√(0+361+361)=19√2.

Find a unit vector perpendicular to each of a = 4î+3ĵ+2k̂ and b = 3î+2ĵ+k̂.

Solution

a×b=|î ĵ k̂; 4 3 2; 3 2 1|=(3−4)î−(4−6)ĵ+(8−9)k̂=−î+2ĵ−k̂. |a×b|=√6. Unit vector=±(−î+2ĵ−k̂)/√6.

If a unit vector â makes angles π/3 with î, π/4 with ĵ and acute angle θ with k̂, find θ and hence the components.

Solution

cos²(π/3)+cos²(π/4)+cos²θ=1. 1/4+1/2+cos²θ=1. cos²θ=1/4. θ=π/3. â=(1/2)î+(1/√2)ĵ+(1/2)k̂.

Show that (a−b)×(a+b)=2(a×b).

Solution

(a−b)×(a+b)=a×a+a×b−b×a−b×b=0+a×b+a×b+0=2(a×b). ✓ (since −b×a=a×b)

Find the area of triangle with vertices A(1,1,2), B(2,3,5), C(1,5,5).

Solution

AB=î+2ĵ+3k̂, AC=0î+4ĵ+3k̂. AB×AC=|î ĵ k̂; 1 2 3; 0 4 3|=(6−12)î−(3−0)ĵ+(4−0)k̂=−6î−3ĵ+4k̂. |AB×AC|=√(36+9+16)=√61. Area=√61/2 sq. units.

Find the area of the parallelogram whose adjacent sides are a = î−ĵ+3k̂ and b = 2î−7ĵ+k̂.

Solution

a×b=|î ĵ k̂; 1 −1 3; 2 −7 1|=(−1+21)î−(1−6)ĵ+(−7+2)k̂=20î+5ĵ−5k̂. |a×b|=√(400+25+25)=√450=15√2. Area=15√2 sq. units.

Let a = î+4ĵ+2k̂, b = 3î−2ĵ+7k̂, c = 2î−ĵ+4k̂. Find a vector p perpendicular to both a and b and p·c = 18.

Solution

a×b=|î ĵ k̂; 1 4 2; 3 −2 7|=(28+4)î−(7−6)ĵ+(−2−12)k̂=32î−ĵ−14k̂. Any scalar multiple: p=λ(32î−ĵ−14k̂). p·c=λ(64+1−56)=9λ=18→λ=2. p=64î−2ĵ−28k̂.

If a·b = a·c and a×b = a×c, a≠0, then show b = c.

Solution

a·(b−c)=0 and a×(b−c)=0. Let d=b−c. a·d=0 → a⊥d. a×d=0 → a∥d. A vector perpendicular and parallel to a must be 0. d=0 → b=c. ✓

Find the area of the triangle with vertices A(1,2,3), B(2,−1,4), C(4,5,−1).

Solution

AB=î−3ĵ+k̂, AC=3î+3ĵ−4k̂. AB×AC=|î ĵ k̂; 1 −3 1; 3 3 −4|=(12−3)î−(−4−3)ĵ+(3+9)k̂=9î+7ĵ+12k̂. |AB×AC|=√(81+49+144)=√274. Area=√274/2.

Q10

Find the unit vector perpendicular to the plane of vectors a = 2î+6ĵ−3k̂ and b = 4î+3ĵ+k̂.

Solution

a×b=|î ĵ k̂; 2 6 −3; 4 3 1|=(6+9)î−(2+12)ĵ+(6−24)k̂=15î−14ĵ−18k̂. |a×b|=√(225+196+324)=√745. Unit vector=(15î−14ĵ−18k̂)/√745.

Q11

Find the sine of angle between a = î+2ĵ+2k̂ and b = 3î+2ĵ+6k̂.

Solution

a×b=|î ĵ k̂; 1 2 2; 3 2 6|=(12−4)î−(6−6)ĵ+(2−6)k̂=8î+0ĵ−4k̂. |a×b|=√(64+16)=4√5. |a|=3,|b|=7. sinθ=4√5/21.

Q12

If a = 2î−ĵ+k̂, b = î+ĵ−2k̂, c = î+3ĵ−k̂, find λ such that a is perpendicular to λb+c.

Solution

λb+c=(λ+1)î+(λ+3)ĵ+(−2λ−1)k̂. a·(λb+c)=2(λ+1)−(λ+3)+1(−2λ−1)=2λ+2−λ−3−2λ−1=−λ−2=0. λ=−2.

Exercise MiscMiscellaneous Exercise

Write the position vector of a point dividing PQ internally in ratio 1:2 where P=(1,2,3) and Q=(4,5,6).

Solution

r = (1·Q+2·P)/3 = (4+2,5+4,6+6)/3 = (2î+3ĵ+4k̂).

If p=3î−2ĵ+6k̂, find |p|.

Solution

|p|=√(9+4+36)=√49=7.

If a and b are two vectors such that |a+b|=|a|, show b is perpendicular to 2a+b.

Solution

|a+b|²=|a|². |a|²+2a·b+|b|²=|a|². 2a·b+|b|²=0. (2a+b)·b=2a·b+|b|²=0. ✓

If a = î+ĵ+k̂ and b = ĵ−k̂, find c such that a×c=b and a·c=3.

Solution

Let c=xî+yĵ+zk̂. a·c=x+y+z=3. a×c=|î ĵ k̂; 1 1 1; x y z|=(z−y)î−(z−x)ĵ+(y−x)k̂=0î+ĵ−k̂. So z−y=0→y=z. −(z−x)=1→x−z=1. y−x=−1. From y=z and x=z+1: x+y+z=z+1+z+z=3z+1=3→z=2/3, y=2/3, x=5/3. c=(5/3)î+(2/3)ĵ+(2/3)k̂.

The value of λ for which i×(2i+λj)×k is 0.

Solution

i×(2i+λj) = 2(i×i)+λ(i×j) = 0+λk̂ = λk̂. λk̂×k̂ = 0 for any λ (since k̂×k̂=0). So this is 0 for all λ. Or: if question is i×(λj)×k: λ(i×j)×k=λk̂×k̂=0. Still all λ work.

Let a, b, c be position vectors of vertices A, B, C of triangle. Find a vector expression for the centroid.

Solution

Centroid G = (a+b+c)/3.

If a×b = c×d and a×c = b×d, show (a−d) is parallel to (b−c), where a≠d, b≠c.

Solution

a×b−a×c=c×d−b×d. a×(b−c)=(c−b)×d=−(b−c)×d. a×(b−c)+(b−c)×d=0. (a−d)×(b−c)=0. Hence (a−d)∥(b−c). ✓

Find the position vector of point R which divides the line joining two points P and Q (position vectors 2a+b and a−3b respectively) externally in ratio 1:2.

Solution

R=(1·(a−3b)−2·(2a+b))/(1−2)=(a−3b−4a−2b)/(−1)=(−3a−5b)/(−1)=3a+5b.

If a+b+c=0, show a×b = b×c = c×a.

Solution

c=−a−b. b×c=b×(−a−b)=−b×a−b×b=−b×a=a×b. c×a=(−a−b)×a=−a×a−b×a=b×a... wait: c×a=−b×a=a×b. So a×b=b×c=c×a. ✓

Q10

If |a|=√26, |b|=7 and |a×b|=35, find a·b.

Solution

|a×b|²+|a·b|²=|a|²|b|². 1225+|a·b|²=26·49=1274. |a·b|²=49. a·b=±7.

Q11

Find a unit vector perpendicular to each of a+b and a−b where a=î+ĵ+k̂, b=î+2ĵ+3k̂.

Solution

a+b=2î+3ĵ+4k̂, a−b=0î−ĵ−2k̂. (a+b)×(a−b)=|î ĵ k̂; 2 3 4; 0 −1 −2|=(−6+4)î−(−4−0)ĵ+(−2−0)k̂=−2î+4ĵ−2k̂. Magnitude=√(4+16+4)=√24=2√6. Unit vector=(−î+2ĵ−k̂)/√6.

Q12

Let a = î+4ĵ+2k̂, b = 3î−2ĵ+7k̂, c = 2î−ĵ+4k̂. Find a vector d perpendicular to both a and b, and c·d=15.

Solution

Same structure as 10.4 Q7 above. a×b=32î−ĵ−14k̂. d=λ(32î−ĵ−14k̂). c·d=λ(64+1−56)=9λ=15→λ=5/3. d=(5/3)(32î−ĵ−14k̂)=(160î−5ĵ−70k̂)/3.

Q13

Find the area of triangle with vertices using cross product: A(1,2,−1), B(−1,1,2), C(1,−1,0).

Solution

AB=−2î−ĵ+3k̂, AC=0î−3ĵ+k̂. AB×AC=|î ĵ k̂; −2 −1 3; 0 −3 1|=(−1+9)î−(−2−0)ĵ+(6−0)k̂=8î+2ĵ+6k̂. Area=√(64+4+36)/2=√104/2=√26 sq. units.

Q14

Show that a·(b×c) = (a×b)·c (scalar triple product equality).

Solution

Both equal the signed volume of the parallelepiped spanned by a,b,c. It is a determinant property: [a,b,c]=[b,c,a]=[c,a,b]. ✓

Q15

Find the vector from origin to the centroid of triangle ABC where A=(3,−1,2), B=(1,−1,−3), C=(4,−3,1).

Solution

Centroid=(3+1+4,−1−1−3,2−3+1)/3=(8/3,−5/3,0)=(8/3)î−(5/3)ĵ.

Q16

If a,b are two vectors |a|=1, |b|=2, a·b=0; find |(a×b)×a|².

Solution

|(a×b)×a|²=|a×b|²|a|²sin²90°... Use identity: (a×b)×a=b(a·a)−a(a·b)=b·1−a·0=b. |(a×b)×a|²=|b|²=4.

Q17

Find the unit vector perpendicular to plane ABC where A=(1,−1,2), B=(2,0,−1), C=(0,2,1).

Solution

AB=î+ĵ−3k̂, AC=−î+3ĵ−k̂. AB×AC=|î ĵ k̂; 1 1 −3; −1 3 −1|=(−1+9)î−(−1−3)ĵ+(3+1)k̂=8î+4ĵ+4k̂. |...| = 4√6. Unit vector=(2î+ĵ+k̂)/√6.

Q18

Find the value of [a×b a×c] where a,b,c are non-coplanar.

Solution

[a×b a×c d]... Let d=a×c. [a×b, a, a×c]= ... This requires specific context. Generally the scalar triple product of three cross products involves determinants of dot products.

Q19

If a×b = b×c ≠ 0, show a+c = λb for some scalar λ.

Solution

a×b = b×c → a×b−b×c = 0 → a×b+c×b = 0 → (a+c)×b = 0. Since b≠0 and (a+c)×b=0, (a+c) is parallel to b → a+c=λb for some scalar λ. ✓

More chapters

← All chapters: Class 12 Mathematics

Vector Algebra

Appearing for the May Phase 2 Board Exam? Practice with AI-ranked questions.

SST

SST