MathsISC Class 12 · Vector Algebra

Projection of a Vector: Scalar and Vector Projection

Q: What is the difference between scalar and vector projection?

The scalar projection is a number, the signed length of the shadow of b along a: (a·b)/|a|. The vector projection is that number times the unit vector along a, so it is a vector pointing along a (or opposite to it if negative).

Q: Why can a projection be negative?

Because it equals |b| cos θ, and cos θ is negative when the angle between the vectors is obtuse. The shadow of b then falls on the opposite side, pointing backwards along a.

Q: Is the projection of a on b the same as the projection of b on a?

Not in general. Both use a·b in the numerator, but one divides by |b| and the other by |a|. They are equal only if |a| = |b| or a·b = 0.

Q: How are projections used with direction cosines?

The projection of a vector on a line with direction cosines l, m, n is l b₁ + m b₂ + n b₃. In particular, the components of a vector are its projections on the three axes.

Q: Is vector projection in the ISC 2027 syllabus?

Yes. The CISCE 2027 syllabus lists scalar and vector projection under Vector Algebra, and ISC 2025 asked a 4-mark case question involving the projection of one side of a triangle on another.

Updated 15 October 2026 · Fact-checked

The scalar projection of b on a is (a·b)/|a| = |b|cos θ, the signed length of the shadow of b along a. The vector projection is ((a·b)/|a|²) a, a vector along a. The scalar projection is negative when the angle between the vectors is obtuse, so do not take its modulus.

The orange shadow of b on a is |b| cos θ; when θ is obtuse it points backwards, so keep the negative sign and never take its modulus.

Illustration drawn for QuestionBuddy.

Understand Projection of a Vector

Picture it. Place $a$ and $b$ tail to tail. Drop a perpendicular from the tip of $b$ to the line of $a$ . The piece of that line from the common tail to the foot of the perpendicular is the **projection of $b$ on $a$ **. It is the part of $b$ that points along $a$ .

Scalar projection. Its signed length is
$proj_{a} b = ∣ b ∣ cos θ = \frac{a \cdot b}{∣ a ∣} = b \cdot \overset{a}{^} .$
Notice that you divide by the magnitude of the vector you are projecting onto. Mixing up "projection of $b$ on $a$ " with "projection of $a$ on $b$ " is the most common error; the ISC 2024 examiners noted candidates who used an incorrect formula for it.

The sign matters. If $θ$ is obtuse, $cos θ < 0$ and the scalar projection is negative: the shadow points backwards along $a$ . The ISC 2025 examiners noted candidates who took the absolute value of a negative projection and lost the mark. A projection of zero means the vectors are perpendicular.

Vector projection. The projection as a vector along $a$ is the scalar projection times the unit vector $\overset{a}{^}$ :
$proj_{a} b = (\frac{a \cdot b}{∣ a ∣}) \overset{a}{^} = \frac{a \cdot b}{∣ a ∣ ^{2}} a .$
The syllabus lists both scalar and vector projection.

Resolving a vector. Any vector $b$ can be split into a part parallel to $a$ and a part perpendicular to it:
$b = parallel \frac{a \cdot b}{∣ a ∣ ^{2}} a + perpendicular (b - \frac{a \cdot b}{∣ a ∣ ^{2}} a) .$
You can check the perpendicular part by showing its dot product with $a$ is zero.

Projections on the axes. The projection of $b = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$ on the x-axis is $b \cdot \hat{i} = b_{1}$ : the components of a vector are its projections on the axes. Similarly, the projection on a line with direction cosines $l, m, n$ is $l b_{1} + m b_{2} + n b_{3}$ .

With points. If the question gives position vectors of points, form the side vectors correctly: $A B = O B - O A$ . The 2025 examiners noted candidates who subtracted in the wrong order, which flips the sign of the projection.

Physical meaning. The component of a force along a direction of motion is a projection; so is the progress a hiker makes along a trail heading, or the useful part of a wind along a boat's course.

Projection of a Vector: formulas to remember

Scalar projection of b on a: $\frac{a \cdot b}{∣ a ∣} = ∣ b ∣ cos θ = b \cdot \overset{a}{^}$; Divide by the length of the vector you project onto.
Vector projection of b on a: $\frac{a \cdot b}{∣ a ∣ ^{2}} a$; A vector along a.
Perpendicular component: $b - \frac{a \cdot b}{∣ a ∣ ^{2}} a$; Its dot product with a is zero.
Projection on an axis: $b \cdot \hat{i} = b_{1}$; Components are projections on the axes.
Projection on a line: $l b_{1} + m b_{2} + n b_{3}$; l, m, n are the direction cosines of the line.
Sign: $proj < 0 ⟺ θ > 9 0^{\circ}$; Do not take the modulus.

How to solve Projection of a Vector questions

Use these steps for projection questions, including those set with points.

1Read carefully which vector is projected and which is projected onto: "projection of $b$ on $a$ " divides by $∣ a ∣$ .
2If points are given, form the vectors as (terminal − initial).
3Compute the dot product $a \cdot b$ .
4Compute the magnitude of the vector being projected onto.
5Divide to get the scalar projection; keep its sign.
6For the vector projection, multiply the scalar projection by the unit vector, or use $\frac{a \cdot b}{∣ a ∣ ^{2}} a$ .
7For resolving, subtract the vector projection from $b$ and check the remainder is perpendicular to $a$ .

Projection of a Vector in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Component of a force along a direction

Suppose a support cable pulls on the anchor of a sloping beam with force $F = 2 \hat{i} + 3 \hat{j} - 6 \hat{k}$ kN. The beam runs in the direction of the unit vector $\overset{u}{^} = \frac{1}{3} (2 \hat{i} + 2 \hat{j} + \hat{k})$ .

Find the component of the force along the beam.

Try it first, then show the solutionSolution

Component $= F \cdot \overset{u}{^} = \frac{1}{3} (4 + 6 - 6) = \frac{4}{3}$ kN.
It is positive, so the cable pulls partly along the direction of the beam.

Answer: $\frac{4}{3} \approx 1.33$ kN along the beam.

Memory hook: Component along a direction = dot product with the unit vector in that direction.

Scalar projection as progress along a heading

Imagine a hiker whose displacement over an hour is $d = 3 \hat{i} + 4 \hat{j} + 0.5 \hat{k}$ km (east, north, up), while the trail she wants to follow heads north-east, along $\hat{i} + \hat{j}$ .

How far has she progressed along the trail direction?

Try it first, then show the solutionSolution

Projection of $d$ on $\hat{i} + \hat{j}$ : $\frac{3 + 4 + 0}{2} = \frac{7}{2}$ .
$\frac{7}{2} \approx 4.95$ km.

Answer: About 4.95 km along the trail direction.

Memory hook: Progress along a heading is a projection: dot with the heading, divide by its length.

Tricks to remember Projection of a Vector

Short memory aids. Say each one aloud once, then use it on a question the same day.

Divide by the floor
Projecting b onto a? a is the floor the shadow falls on: divide by |a|.
Use it when: Avoiding the wrong denominator.
Shadow can point back
An obtuse angle casts a backward shadow: the projection is negative.
Use it when: Not taking the modulus.
Scalar × unit = vector
Vector projection = scalar projection × â.
Use it when: Writing the vector projection.
Terminal minus initial
AB = OB − OA, always.
Use it when: Questions with position vectors.

Projection of a Vector: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: A scalar projection · 1 mark · MCQBoard level

The projection of $b = \hat{i} + 2 \hat{j} + 4 \hat{k}$ on $a = 2 \hat{i} - 2 \hat{j} + \hat{k}$ is:
(a) $\frac{2}{3}$
(b) $\frac{2}{21}$
(c) 2
(d) $- \frac{2}{3}$

Try it first, then show the solutionSolution

$a \cdot b = 2 - 4 + 4 = 2$ ; $∣ a ∣ = 3$ ; projection $= \frac{2}{3}$ .

Answer: (a) $\frac{2}{3}$ .

Example 2: Projection with a surd · 2 marks · Short answerBoard level

Find the projection of $a = 2 \hat{i} + 3 \hat{j} + 2 \hat{k}$ on $b = \hat{i} + 2 \hat{j} + \hat{k}$ .

Try it first, then show the solutionSolution

$a \cdot b = 2 + 6 + 2 = 10$ ; $∣ b ∣ = 6$ . (1 mark)
Projection of $a$ on $b$ $= \frac{10}{6} = \frac{5 6}{3}$ . (1 mark)

Answer: $\frac{5 6}{3}$ .

Example 3: Vector projection and the perpendicular part · 3 marks · Short answerHarder board level

Find the vector projection of $b = 3 \hat{i} + \hat{j} + 2 \hat{k}$ on $a = \hat{i} + \hat{j}$ , and the component of $b$ perpendicular to $a$ .

Try it first, then show the solutionSolution

$a \cdot b = 4$ , $∣ a ∣^{2} = 2$ . Vector projection $= \frac{4}{2} (\hat{i} + \hat{j}) = 2 \hat{i} + 2 \hat{j}$ . (1½ marks)
Perpendicular part $= b - (2 \hat{i} + 2 \hat{j}) = \hat{i} - \hat{j} + 2 \hat{k}$ . (1 mark)
Check: $(\hat{i} - \hat{j} + 2 \hat{k}) \cdot (\hat{i} + \hat{j}) = 0$ . (½ mark)

Answer: Parallel part $2 \hat{i} + 2 \hat{j}$ ; perpendicular part $\hat{i} - \hat{j} + 2 \hat{k}$ .

Example 4: A negative projection from points · 3 marks · Short answerHarder board level

For $A (1, 1, 1)$ , $B (2, 3, 0)$ and $C (0, 2, 3)$ , find the projection of $A B$ on $A C$ and interpret its sign.

Try it first, then show the solutionSolution

$A B = \hat{i} + 2 \hat{j} - \hat{k}$ , $A C = - \hat{i} + \hat{j} + 2 \hat{k}$ . (1 mark)
$A B \cdot A C = - 1 + 2 - 2 = - 1$ ; $∣ A C ∣ = 6$ . (1 mark)
Projection $= - \frac{1}{6}$ . It is negative, so angle $B A C$ is obtuse. (1 mark)

Answer: $- \frac{1}{6}$ ; angle $B A C$ is slightly more than $9 0^{\circ}$ .

Example 5: Equal projections · 2 marks · Short answerHarder board level

If the projection of $a$ on $b$ equals the projection of $b$ on $a$ , and $a \cdot b \neq = 0$ , show that $∣ a ∣ = ∣ b ∣$ .

Try it first, then show the solutionSolution

$\frac{a \cdot b}{∣ b ∣} = \frac{a \cdot b}{∣ a ∣}$ . (1 mark)
Divide by $a \cdot b \neq = 0$ : $\frac{1}{∣ b ∣} = \frac{1}{∣ a ∣}$ , so $∣ a ∣ = ∣ b ∣$ . (1 mark)

Answer: Shown.

Example 6: Projection and area from position vectors · 5 marks · Case-basedHOTS / competency

Suppose the position vectors of $A$ , $B$ , $C$ are $\hat{i} + 2 \hat{j} + 3 \hat{k}$ , $2 \hat{i} - \hat{j} + \hat{k}$ and $3 \hat{i} + \hat{j} - 2 \hat{k}$ .
(i) Find $A B$ and $B C$ . (1 mark)
(ii) Find the scalar projection of $A B$ on $B C$ . (1 mark)
(iii) Find the vector projection of $A B$ on $B C$ . (1 mark)
(iv) Find the area of triangle $A B C$ . (2 marks)

Try it first, then show the solutionSolution

(i) $A B = \hat{i} - 3 \hat{j} - 2 \hat{k}$ ; $B C = \hat{i} + 2 \hat{j} - 3 \hat{k}$ .
(ii) $A B \cdot B C = 1 - 6 + 6 = 1$ ; $∣ B C ∣ = 14$ ; projection $= \frac{1}{14}$ .
(iii) $\frac{1}{14} (\hat{i} + 2 \hat{j} - 3 \hat{k})$ .
(iv) $A B \times B C = (9 + 4) \hat{i} - (- 3 + 2) \hat{j} + (2 + 3) \hat{k} = 13 \hat{i} + \hat{j} + 5 \hat{k}$ , of magnitude $195$ .
Area $= \frac{1}{2} 195 \approx 6.98$ sq units.

Answer: (ii) $\frac{1}{14}$ ; (iii) $\frac{1}{14} (\hat{i} + 2 \hat{j} - 3 \hat{k})$ ; (iv) $\frac{195}{2}$ sq units.

Where students lose marks in Projection of a Vector

The slips examiners see most often, and what to write instead.

Mistake: Wrong denominator
The ISC 2024 examiners noted candidates using an incorrect formula for the projection of b on a.
Wrong working
Projection of $b$ on $a$ $= \frac{a \cdot b}{∣ b ∣}$ .
Right working
$= \frac{a \cdot b}{∣ a ∣}$ .
Do this: Divide by the magnitude of the vector you project onto.
Usually costs: 1 mark.
Mistake: Taking the modulus of a negative projection
The 2025 examiners noted this; a scalar projection carries a sign.
Do this: Keep the minus sign and, if asked, interpret it as an obtuse angle.
Usually costs: 1 mark.
Mistake: Subtracting position vectors in the wrong order
Writing AB = OA − OB reverses the vector and flips the sign of the projection.
Do this: Terminal minus initial: AB = OB − OA.
Usually costs: 1 to 2 marks in a case question.
Mistake: Confusing scalar and vector projection
Giving a number when a vector is asked (or vice versa) loses the mark.
Do this: Vector projection = scalar projection × unit vector along a.
Usually costs: 1 mark.

Exam tips

Write the formula with the correct denominator before substituting; it secures the method mark.
Leave surds in exact form (for example $\frac{5 6}{3}$ ) unless a decimal is asked for.
In a case question, label each vector ( $A B$ , $B C$ ) and keep them for later parts such as area.
Check a perpendicular component by dotting it with the original direction: the result must be 0.
If the projection is zero, say "the vectors are perpendicular".

Practice questions on Projection of a Vector

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The projection of $\hat{i} + \hat{j} + \hat{k}$ on $\hat{j}$ is:
(a) 0
(b) 1
(c) $3$
(d) $\frac{1}{3}$
Hint
Projections on the axes are the components.
Answer
(b) 1.
Q2 · 2 marks · Short answerBoard level
Find the projection of $4 \hat{i} - \hat{j} + 2 \hat{k}$ on $2 \hat{i} + 3 \hat{j} + 6 \hat{k}$ .
Hint
$∣2 \hat{i} + 3 \hat{j} + 6 \hat{k} ∣ = 7$ .
Answer
$\frac{8 - 3 + 12}{7} = \frac{17}{7}$ .
Q3 · 2 marks · Short answerHarder board level
Find $λ$ if the projection of $λ \hat{i} + \hat{j} + 4 \hat{k}$ on $2 \hat{i} + 6 \hat{j} + 3 \hat{k}$ is 4 units.
Hint
Set $\frac{2 λ + 6 + 12}{7} = 4$ .
Answer
$2 λ + 18 = 28$ , so $λ = 5$ .
Q4 · 3 marks · Short answerHarder board level
Write $b = 2 \hat{i} - \hat{j} + 3 \hat{k}$ as the sum of a vector parallel to $a = \hat{i} + \hat{j} + \hat{k}$ and a vector perpendicular to $a$ .
Hint
Parallel part $= \frac{a \cdot b}{∣ a ∣ ^{2}} a$ .
Answer
Parallel: $\frac{4}{3} (\hat{i} + \hat{j} + \hat{k})$ . Perpendicular: $\frac{2}{3} \hat{i} - \frac{7}{3} \hat{j} + \frac{5}{3} \hat{k}$ (its dot product with $a$ is $\frac{2 - 7 + 5}{3} = 0$ ).
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The projection of one vector on another can be negative.
Reason (R): The projection of $b$ on $a$ is $∣ b ∣ cos θ$ , and $cos θ < 0$ for obtuse $θ$ .
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Think of an obtuse angle.
Answer
(a).
Q6 · 3 marks · Short answerHOTS / competency
If $\overset{a}{^}$ is a unit vector, $∣ b ∣ = 4$ , and the projection of $b$ on $\overset{a}{^}$ is $23$ , find the angle between them and $∣ \overset{a}{^} \times b ∣$ .
Hint
Projection $= ∣ b ∣ cos θ$ .
Answer
$4 cos θ = 23 \Rightarrow cos θ = \frac{3}{2}$ , so $θ = 3 0^{\circ}$ . $∣ \overset{a}{^} \times b ∣ = 1 \times 4 \times sin 3 0^{\circ} = 2$ .
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose the wind velocity over a lake is $w = 6 \hat{i} + 8 \hat{j}$ m/s, and a sailing boat is heading in the direction $\hat{i} + \hat{j}$ .
(i) Find the speed of the wind. (1 mark)
(ii) Find the component of the wind along the boat's heading. (1 mark)
(iii) Find the size of the component across the heading. (1 mark)
(iv) In which direction should the boat head for the whole wind to act along its heading? (1 mark)
Hint
Use Pythagoras for (iii).
Answer
(i) 10 m/s. (ii) $\frac{6 + 8}{2} = 72 \approx 9.90$ m/s. (iii) $100 - 98 = 2 \approx 1.41$ m/s. (iv) Along $w$ itself, that is in the direction $\frac{3}{5} \hat{i} + \frac{4}{5} \hat{j}$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State the scalar projection of $b$ on $a$ .Show answer
$\frac{a \cdot b}{∣ a ∣}$ .
2.State the vector projection of $b$ on $a$ .Show answer
$\frac{a \cdot b}{∣ a ∣ ^{2}} a$ .
3.What does a negative projection tell you?Show answer
The angle between the vectors is obtuse.
4.Find the projection of $3 \hat{i} - 4 \hat{k}$ on $\hat{k}$ .Show answer
$- 4$ .
5.What is the projection of $b$ on $a$ if they are perpendicular?Show answer
0.
6.How do you form $A B$ from position vectors?Show answer
$O B - O A$ .

Projection of a Vector: frequently asked questions

What is the difference between scalar and vector projection?

The scalar projection is a number, the signed length of the shadow of b along a: (a·b)/|a|. The vector projection is that number times the unit vector along a, so it is a vector pointing along a (or opposite to it if negative).

Why can a projection be negative?

Because it equals |b| cos θ, and cos θ is negative when the angle between the vectors is obtuse. The shadow of b then falls on the opposite side, pointing backwards along a.

Is the projection of a on b the same as the projection of b on a?

Not in general. Both use a·b in the numerator, but one divides by |b| and the other by |a|. They are equal only if |a| = |b| or a·b = 0.

How are projections used with direction cosines?

The projection of a vector on a line with direction cosines l, m, n is l b₁ + m b₂ + n b₃. In particular, the components of a vector are its projections on the three axes.

Is vector projection in the ISC 2027 syllabus?

Yes. The CISCE 2027 syllabus lists scalar and vector projection under Vector Algebra, and ISC 2025 asked a 4-mark case question involving the projection of one side of a triangle on another.