Question 1
Sketch a 40 km displacement whose direction is 30° east of north.
Steps only
Draw a north arrow, then measure 30° towards the east.
Draw a 40 km arrow along that bearing.
Answer
A 40 km arrow pointing 30° east of north.
CLASS 12 · MATHEMATICS PART II · CHAPTER 10
Vector Algebra
Student-friendly short and explained solutions for vector operations, dot products, cross products, angles, projections and areas.
73 worked answersAll 4 exercise setsAll 19 end-of-chapter questions
What you will learn
- Represent vectors in component form and calculate magnitudes and directions.
- Use dot products to find angles, projections and perpendicularity.
- Use cross products to find perpendicular directions and areas.
73 worked answers · short steps first, explanations when you need them
Exercise 10.1
Textbook pages 342–343
Question 2
Classify: 10 kg; 2 m north-west; 40°; 40 W; 10⁻¹⁹ C; 20 m/s².
Steps only
A magnitude alone is scalar; a magnitude with direction is vector.
Mass, angle, power and charge are scalars; the directed length and acceleration are vectors.
Answer
10 kg scalar; 2 m north-west vector; 40° scalar; 40 W scalar; 10⁻¹⁹ C scalar; 20 m/s² vector.
Question 3
Classify time period, distance, force, velocity and work done.
Steps only
Time period and distance have magnitude only.
Force and velocity have magnitude and direction; work is scalar.
Answer
Time period—scalar; distance—scalar; force—vector; velocity—vector; work done—scalar.
Question 4
In the square diagram, identify the coinitial vectors, equal vectors, and collinear vectors that are not equal.
Steps only
The four arrows run around the square: top right, right side down, bottom left and left side up.
Opposite sides are parallel, but their arrows point in opposite directions.
Answer
Coinitial: none. Equal: none. Collinear but not equal: a and c; b and d.
Question 5
Decide whether each statement is true: a and −a are collinear; collinear vectors always have equal magnitudes; equal magnitudes imply collinearity; collinear vectors with equal magnitudes must be equal.
Steps only
A vector and its negative lie on the same line.
Collinearity fixes a line direction, not length; equal lengths do not fix direction.
Equal length and collinearity still allow opposite directions.
Answer
- (i) True.
- (ii) False.
- (iii) False.
- (iv) False.
Exercise 10.2
Textbook pages 354–355
Question 1
Find the magnitudes of a=i+j+k, b=2i−7j−3k and c=(i+j−k)/√3.
Steps only
Use |(x,y,z)|=√(x²+y²+z²).
For a, b and c, the squared magnitudes are 3, 62 and 3/3.
Answer
|a|=√3, |b|=√62, |c|=1.
Question 2
Give two different vectors with the same magnitude.
Steps only
Choose two perpendicular unit vectors.
Each has length 1, but their directions differ.
Answer
For example, i and j; both have magnitude 1.
Question 3
Give two different vectors pointing in the same direction.
Steps only
Take a nonzero vector and multiply it by a positive number other than 1.
Answer
For example, i and 2i.
Question 4
Find x and y if 2i+3j equals xi+yj.
Steps only
Equal vectors have equal corresponding components.
Match the i-components and then the j-components.
Answer
x=2 and y=3.
Question 5
Find the scalar and vector components from P=(2,1) to Q=(−5,7).
Steps only
Subtract the initial coordinates from the terminal coordinates: Q−P=(−7,6).
Answer
Scalar components: (−7,6). Vector: −7i+6j.
Question 6
Add a=i−2j+k, b=−2i+4j+5k and c=i−6j−7k.
Steps only
Add the i, j and k coefficients separately.
The totals are 1−2+1, −2+4−6 and 1+5−7.
Answer
a+b+c=−4j−k.
Question 7
Find the unit vector in the direction i+j+2k.
Steps only
The vector has length √(1²+1²+2²)=√6.
Divide every component by √6.
Answer
(i+j+2k)/√6.
Question 8
Find the unit vector from P=(1,2,3) to Q=(4,5,6).
Steps only
Q−P=(3,3,3), whose length is 3√3.
Divide (3,3,3) by 3√3.
Answer
(i+j+k)/√3.
Question 9
For a=2i−j+2k and b=−i+j−k, find the unit vector along a+b.
Steps only
a+b=i+k.
Its length is √2, so divide by √2.
Answer
(i+k)/√2.
Question 10
Find a vector of length 8 pointing in the direction 5i−j+2k.
Steps only
The given vector has length √30.
Multiply its unit vector by 8.
Answer
(8/√30)(5i−j+2k).
Question 11
Show that 2i−3j+4k and −4i+6j−8k are collinear.
Steps only
The second vector is −2 times the first.
Answer
They are collinear, since one is a scalar multiple of the other.
Question 12
Find the direction cosines of i+2j+3k.
Steps only
Its length is √14.
Divide each component by √14.
Answer
(1/√14, 2/√14, 3/√14).
Question 13
Find the direction cosines from A=(1,2,−3) to B=(−1,−2,1).
Steps only
B−A=(−2,−4,4), with length 6.
Divide the three components by 6.
Answer
(−1/3, −2/3, 2/3).
Question 14
Show that i+j+k makes equal angles with the coordinate axes.
Steps only
Its length is √3.
Its direction cosines are (1/√3,1/√3,1/√3).
Answer
The three angles are equal because their cosines are equal.
Question 15
P and Q have position vectors i+2j−k and −i+j+k. Find R when PR:RQ=2:1, internally and externally.
Steps only
Internal: r=(p+2q)/3=(−1,4,1)/3.
External: r=(2q−p)/(2−1)=(−3,0,3).
Answer
Internal r=−(1/3)i+(4/3)j+(1/3)k; external r=−3i+3k.
Question 16
Find the midpoint of P=(2,3,4) and Q=(4,1,−2).
Steps only
Average corresponding coordinates: ((2+4)/2,(3+1)/2,(4−2)/2).
Answer
The midpoint position vector is 3i+2j+k.
Question 17
Check whether position vectors a=3i−4j−4k, b=2i−j+k and c=i−3j−5k give a right-angled triangle.
Steps only
AB=b−a=(−1,3,5), BC=c−b=(−1,−2,−6), CA=a−c=(2,−1,1).
Their squared lengths are 35, 41 and 6; 35+6=41.
Answer
They form a right-angled triangle; the right angle is at A.
Question 18
In triangle ABC, check which listed vector equation is not true.
Steps only
The triangle rule gives AB+BC=AC, while CA=−AC and CB=−BC.
Each of the four displayed options in the official PDF simplifies to AB+BC−AC=0.
Answer
As printed, there is no false option: all four equations are true. Options B and C appear duplicated in the official PDF, so the question likely contains a printing error.
Question 19
Which statements about two collinear vectors are incorrect?
Steps only
One vector is a scalar multiple of the other, so A is true and components are proportional.
Their lengths and orientations need not match, so B and D are not always true.
Answer
B, C and D are incorrect as general statements; A is true.
Exercise 10.3
Textbook pages 361–362
Question 1
Vectors a,b have lengths √3 and 2, and a·b=√6. Find their angle.
Steps only
cosθ=(a·b)/(|a||b|)=√6/(2√3)=1/√2.
Answer
θ=π/4 (45°).
Question 2
Find the angle between i−2j+3k and 3i−2j+k.
Steps only
Their dot product is 3+4+3=10; each length is √14.
Thus cosθ=10/14=5/7.
Answer
θ=cos⁻¹(5/7).
Question 3
Find the projection of i−j on i+j.
Steps only
(i−j)·(i+j)=1−1=0.
Answer
The scalar projection is 0; the vector projection is also the zero vector.
Question 4
Project i+3j+7k onto 7i−j+8k.
Steps only
The dot product is 7−3+56=60 and the second vector has length √114.
Scalar projection =60/√114; vector projection = (60/114)(7i−j+8k).
Answer
Scalar projection: 60/√114. Vector projection: (10/19)(7i−j+8k).
Question 5
Show that (2i+3j+6k)/7, (3i−6j+2k)/7 and (6i+2j−3k)/7 are unit and mutually perpendicular.
Steps only
Each numerator has squared length 49, so each vector has length 1.
Pairwise numerator dot products are 6−18+12=0, 12+6−18=0 and 18−12−6=0.
Answer
All three are unit vectors and every pair is perpendicular.
Question 6
Given (a+b)·(a−b)=8 and |a|=8|b|, find |a| and |b|.
Steps only
The first expression simplifies to |a|²−|b|²=8.
Let |b|=t; then 64t²−t²=8, so t=√(8/63).
Answer
|b|=2√14/21 and |a|=16√14/21.
Question 7
Evaluate (3a−5b)·(2a+7b).
Steps only
Distribute the dot product across both brackets.
Use a·a=|a|², b·b=|b|² and a·b=b·a.
Answer
6|a|²+11(a·b)−35|b|².
Question 8
Vectors a and b have the same length, their angle is 60°, and a·b=1/2. Find their lengths.
Steps only
Let their common length be r. Then a·b=r²cos60°=r²/2.
Set r²/2=1/2 and use r≥0.
Answer
|a|=|b|=1.
Question 9
For a unit vector a, (x−a)·(x+a)=12. Find |x|.
Steps only
Expand: x·x+x·a−a·x−a·a=|x|²−1.
So |x|²−1=12.
Answer
|x|=√13.
Question 10
Let a=2i+2j+3k, b=−i+2j+k and c=3i+j. If a+λb is perpendicular to c, find λ.
Steps only
a·c=8 and b·c=−1.
(a+λb)·c=8−λ=0.
Answer
λ=8.
Question 11
Show |a|b+|b|a is perpendicular to |a|b−|b|a.
Steps only
Take the dot product of the two displayed vectors.
The equal squared-magnitude terms cancel, as do the two mixed terms.
Answer
Their dot product is 0, so the vectors are perpendicular.
Question 12
If a·a=0 and a·b=0, what can be concluded about b?
Steps only
Since a·a=|a|²=0, a must be the zero vector.
Then a·b=0 holds for every b.
Answer
Nothing specific about b follows; b may be any vector.
Question 13
Unit vectors a,b,c satisfy a+b+c=0. Find a·b+b·c+c·a.
Steps only
Square both sides: |a+b+c|²=0.
The three squared lengths sum to 3, leaving 3+2(a·b+b·c+c·a)=0.
Answer
a·b+b·c+c·a=−3/2.
Question 14
If either a or b is zero, then a·b=0. Give a counterexample to the converse.
Steps only
Choose two nonzero perpendicular vectors, such as i and j.
Their dot product is zero, though neither vector is zero.
Answer
The converse is false: a=i and b=j is a counterexample.
Question 15
For A=(1,2,3), B=(−1,0,0), C=(0,1,2), find ∠ABC.
Steps only
BA=(2,2,3), BC=(1,1,2); their dot product is 10.
Their lengths are √17 and √6, so cos∠ABC=10/√102.
Answer
∠ABC=cos⁻¹(10/√102).
Question 16
Show A=(1,2,7), B=(2,6,3), C=(3,10,−1) are collinear.
Steps only
B−A=(1,4,−4) and C−B=(1,4,−4).
Answer
The consecutive displacement vectors are equal, so A, B and C lie on one straight line.
Question 17
Show that the points with position vectors 2i−j+k, i−3j−5k and 3i−4j−4k form a right-angled triangle.
Steps only
Take these as A,B,C. Then BC=(2,−1,1) and CA=(1,−3,−5).
BC·CA=2+3−5=0.
Answer
The triangle is right-angled at C.
Question 18
A nonzero vector has magnitude a>0 and λ≠0. When is λa a unit vector?
Steps only
|λa|=|λ|a.
Set this equal to 1.
Answer
a=1/|λ| (choice D).
Exercise 10.4
Textbook pages 368–369
Question 1
Find |a×b| for a=i−7j+7k and b=3i−2j+2k.
Steps only
a×b=0i−19j+19k.
Its magnitude is √(19²+19²).
Answer
|a×b|=19√2.
Question 2
Find a unit vector perpendicular to both a=3i+2j+2k and b=i+2j−2k.
Steps only
a+b=(4,4,0), a−b=(2,0,4).
Their cross product is (16,−16,−8), parallel to (2,−2,−1), whose length is 3.
Answer
Either ±(2i−2j−k)/3.
Question 3
A unit vector makes angles π/3 with i and π/4 with j, and an acute angle θ with k. Find θ and its components.
Steps only
Its first two direction cosines are 1/2 and √2/2.
The squares of all three direction cosines add to 1, giving the third cosine 1/2 because θ is acute.
Answer
θ=π/3; vector=(1/2)i+(√2/2)j+(1/2)k.
Question 4
Show (a−b)×(a+b)=2(a×b).
Steps only
Expand the left side using distributivity.
a×a and b×b are zero, while b×a=−a×b.
Answer
(a−b)×(a+b)=a×b−b×a=2a×b.
Question 5
Find λ,μ if (2i+6j+27k)×(i+λj+μk)=0.
Steps only
A zero cross product means the nonzero vectors are parallel.
Match the component ratios: 2/1=6/λ=27/μ.
Answer
λ=3 and μ=27/2.
Question 6
Given a·b=0 and a×b=0, what follows about a and b?
Steps only
For nonzero vectors, a zero cross product makes them parallel.
Nonzero perpendicular parallel vectors cannot exist.
Answer
At least one of a,b is the zero vector.
Question 7
For component vectors a,b,c, show a×(b+c)=a×b+a×c.
Steps only
Write b+c by adding corresponding components.
Expand the determinant formula for the cross product; each component separates into the two required terms.
Answer
a×(b+c)=a×b+a×c.
Question 8
If either a or b is zero then a×b=0. Is the converse true? Justify.
Steps only
Take a=i and b=2i.
Both are nonzero, but they are parallel, so their cross product is zero.
Answer
No. For example, i×2i=0 although neither vector is zero.
Question 9
Find the area of triangle A=(1,1,2), B=(2,3,5), C=(1,5,5).
Steps only
AB=(1,2,3), AC=(0,4,3).
AB×AC=(−6,−3,4), of length √61.
Answer
Area=√61/2 square units.
Question 10
Find the area for adjacent side vectors a=i−j+3k and b=2i−7j+k.
Steps only
a×b=(20,5,−5).
Its length is √(400+25+25)=15√2.
Answer
Area=15√2 square units.
Question 11
If |a|=3, |b|=√2/3 and a×b is a unit vector, choose the angle between a and b.
Steps only
1=|a×b|=|a||b|sinθ=√2 sinθ.
Thus sinθ=1/√2; among the given choices, π/4 fits.
Answer
π/4 (choice B).
Question 12
Find the area of the rectangle with vertices A=(−1,1/2,4), B=(1,1/2,4), C=(1,−1/2,4), D=(−1,−1/2,4).
Steps only
AB has length 2 and BC has length 1.
Adjacent sides are perpendicular, so area=2×1.
Answer
Area=2 square units (choice C).
Miscellaneous Exercise on Chapter 10
Textbook pages 372–373
Question 1
Write a unit vector in the XY-plane at 30° from the positive x-axis.
Steps only
Use (cosθ)i+(sinθ)j for a unit vector in the plane.
At 30°, cosθ=√3/2 and sinθ=1/2.
Answer
(√3/2)i+(1/2)j.
Question 2
Give the components and magnitude of the vector from P=(x₁,y₁,z₁) to Q=(x₂,y₂,z₂).
Steps only
Subtract P’s coordinates from Q’s coordinates.
Apply the three-dimensional distance formula to those differences.
Answer
Components: (x₂−x₁,y₂−y₁,z₂−z₁); magnitude=√((x₂−x₁)²+(y₂−y₁)²+(z₂−z₁)²).
Question 3
A girl walks 4 km west, then 3 km at 30° east of north. Find her displacement from the start.
Steps only
The second walk has components 1.5 km east and (3√3/2) km north.
Adding the first walk gives (−5/2,3√3/2) km; its length is √13 km.
Answer
Displacement=(−5/2)i+(3√3/2)j km; magnitude=√13 km, about 46.1° north of west.
Question 4
If a=b+c, must |a|=|b|+|c|? Explain.
Steps only
Take b=i and c=j. Then a=i+j.
|a|=√2, while |b|+|c|=2.
Answer
No. For b=i,c=j, the lengths are √2 and 2, so they are unequal.
Question 5
Find x so x(i+j+k) is a unit vector.
Steps only
Its magnitude is |x|√3.
Set |x|√3=1.
Answer
x=±1/√3.
Question 6
Find a vector of length 5 parallel to the resultant of 2i+3j−k and i−2j+k.
Steps only
The resultant is 3i+j, of length √10.
Multiply its unit vector by 5.
Answer
(5/√10)(3i+j)=(3√10/2)i+(√10/2)j.
Question 7
For a=i+j+k, b=2i−j+3k and c=i−2j+k, find a unit vector parallel to 2a−b+3c.
Steps only
2a−b+3c=3i−3j+2k.
Its length is √22.
Answer
(3i−3j+2k)/√22.
Question 8
Show A=(1,−2,−8), B=(5,0,−2), C=(11,3,7) are collinear and find the ratio in which B divides AC.
Steps only
AB=(4,2,6), BC=(6,3,9)=(3/2)AB.
Therefore AB:BC=2:3.
Answer
A, B, C are collinear and B divides AC internally in the ratio 2:3.
Question 9
P,Q have position vectors a+b and a−3b. Find R dividing PQ externally in ratio 1:2 and show P is midpoint of RQ.
Steps only
With PR:RQ=1:2, the external formula is r=(q−2p)/(1−2)=2p−q.
Thus r=a+5b, and (r+q)/2=(a+5b+a−3b)/2=a+b=p.
Answer
r=a+5b; P is the midpoint of RQ.
Question 10
Adjacent vectors are 2i−4j+5k and i−2j−3k. Find the unit vector along their diagonal and the area.
Steps only
The diagonal is their sum 3i−6j+2k, with length 7.
Their cross product is (22,−11,0), with length 11√5.
Answer
Diagonal unit vector=(3i−6j+2k)/7; area=11√5 square units.
Question 11
Give the direction cosines of a vector equally inclined to the three positive coordinate axes.
Steps only
Equal inclinations give equal direction cosines l=m=n.
Since l²+m²+n²=1, each is 1/√3.
Answer
(1/√3,1/√3,1/√3). For directions equally inclined to the unoriented axes, the common sign may also be negative.
Question 12
Find d perpendicular to a=i+4j+2k and b=3i−2j+7k, given c=2i−j+4k and c·d=15.
Steps only
A×b=(32,−1,−14), so d=t(32,−1,−14).
c·(a×b)=64+1−56=9; hence 9t=15 and t=5/3.
Answer
d=(160/3)i−(5/3)j−(70/3)k.
Question 13
The dot product of i+j+k with the unit vector along (2i+4j−5k)+(λi+2j+3k) is 1. Find λ.
Steps only
The sum is (λ+2,6,−2), with length √((λ+2)²+40).
The dot product condition is (λ+6)/√((λ+2)²+40)=1; squaring gives λ=1, which satisfies the positive sign.
Answer
λ=1.
Question 14
If a,b,c are mutually perpendicular and have equal nonzero magnitudes, show a+b+c is equally inclined to them.
Steps only
(a+b+c)·a=|a|² because a·b=a·c=0.
The same result holds for dot products with b and c.
Answer
The vector a+b+c makes equal angles with a,b,c.
Question 15
For nonzero a,b, prove |a+b|=|a−b| exactly when a and b are perpendicular.
Steps only
Square both sides and expand: |a|²+2a·b+|b|²=|a|²−2a·b+|b|².
This is equivalent to a·b=0.
Answer
The lengths are equal if and only if a·b=0, which is equivalent to perpendicularity.
Question 16
For the angle θ between two nonzero vectors, when is a·b≥0?
Steps only
a·b=|a||b|cosθ and the magnitudes are positive.
Thus the sign is determined by cosθ.
Answer
0≤θ≤π/2 (choice B).
Question 17
For unit vectors a,b with angle θ, when is a+b a unit vector?
Steps only
|a+b|²=|a|²+|b|²+2a·b=2+2cosθ.
Set this equal to 1; cosθ=−1/2.
Answer
θ=2π/3 (choice D).
Question 18
Evaluate i·(j×k)+j·(i×k)+k·(i×j).
Steps only
j×k=i, i×k=−j and i×j=k.
The three dot products are 1, −1 and 1.
Answer
The value is 1 (choice C).
Question 19
If θ is the angle between nonzero vectors a,b and |a·b|=|a×b|, find θ from the choices.
Steps only
Cancel |a||b| to obtain |cosθ|=sinθ.
Both π/4 and 3π/4 satisfy the equation; among the listed choices, only π/4 appears.
Answer
π/4 (choice B). Mathematically, 3π/4 also satisfies the condition.