The projection vector of vector veca on vector vecb is: (A) (dfracveca·vecb|vecb|²)vecb (B) dfracveca·vecb|vec

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\vec{a}

CBSECBSE Class XII Board 2025MCQ· 1mImportance★★★★★

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Concept understanding — Vector Projection

Vector Projection

Picture a stick leaning in sunlight with the sun directly overhead: the shadow it casts on the ground is the projection of the stick onto the ground. The stick is your vector, the ground is the direction you project onto, and the shadow tells you how much of the stick lies along that direction.

That is the whole idea: projection answers "how much of this vector points in that particular direction?"

The Geometry

Take two vectors $\vec{a}$ and $\vec{b}$ . The projection of $\vec{a}$ onto $\vec{b}$ is a new vector that

lies along the line of $\vec{b}$ (parallel to $\vec{b}$ ), and
has length equal to how much of $\vec{a}$ points along $\vec{b}$ .

Note

The scalar projection is a number; the vector projection is a vector — same information, but the vector version also carries direction.

The Formula

For $\vec{b} \neq \vec{0}$ ,

\text{proj}_{\vec{b}}\,\vec{a} = \frac{\vec{a}\cdot\vec{b}}{\lVert\vec{b}\rVert^{2}}\,\vec{b}, \qquad \text{comp}_{\vec{b}}\,\vec{a} = \frac{\vec{a}\cdot\vec{b}}{\lVert\vec{b}\rVert}.

Why it works: $\vec{a}\cdot\vec{b}$ measures how much $\vec{a}$ "agrees" with $\vec{b}$ (positive if aligned, negative if opposed, zero if perpendicular). Dividing by $\lVert\vec{b}\rVert^{2}$ turns that into the signed length of the shadow relative to $\vec{b}$ , and multiplying by $\vec{b}$ places that length along $\vec{b}$ .

A Quick Example

Let $\vec{a}=(3,4)$ and $\vec{b}=(1,1)$ (the line $y=x$ ):

$\vec{a}\cdot\vec{b}=3+4=7$ , and $\lVert\vec{b}\rVert^{2}=2$
$\text{proj}_{\vec{b}}\,\vec{a}=\frac{7}{2}(1,1)=(3.5,\,3.5)$

The shadow sits exactly on the line $y=x$ .

Watch out

Do not write the projection as $\frac{\vec{a}\cdot\vec{b}}{\lVert\vec{b}\rVert}\vec{b}$ — that gives the right direction but the wrong length. The denominator must be $\lVert\vec{b}\rVert^{2}$ .

Why It Matters

Projection resolves a force into components along and across a surface, gives work done ( $W=\vec{F}\cdot\vec{d}$ is the scalar projection of force onto displacement), and splits any vector into a part parallel to a chosen direction plus a perpendicular part — the basis of orthogonal decomposition.

Vector Projection is formally introduced in the CBSE Class 12 Vector Algebra chapter, where the scalar and vector projection formulas using the dot product are standard NCERT content tested in board exams. "Projection of a vector on another vector formula" is a common search term, and the same idea reappears in JEE Main and NEET physics problems on resolving forces along a direction.

The projection vector of $\vec{a}$ on $\vec{b}$ is the component of $\vec{a}$ that lies along the direction of $\vec{b}$ , and it is given by the formula $\left(\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|^2}\right)\vec{b}$ .

When we talk about the projection of vector $\vec{a}$ onto vector $\vec{b}$ , we are essentially asking: "How much of vector $\vec{a}$ points in the same direction as vector $\vec{b}$ ?" Imagine shining a light perpendicular to vector $\vec{b}$ . The shadow of $\vec{a}$ cast on the line containing $\vec{b}$ is its projection.

This projection is itself a vector. It will always point in the same direction as $\vec{b}$ (or opposite, if the angle between $\vec{a}$ and $\vec{b}$ is obtuse). To define this vector, we need two things: its magnitude and its direction.

Determine the magnitude of the projection (Scalar Projection).

Let $\theta$ be the angle between vectors $\vec{a}$ and $\vec{b}$ . Geometrically, if we drop a perpendicular from the tip of $\vec{a}$ onto the line containing $\vec{b}$ , the length of the segment formed on $\vec{b}$ is the magnitude of the projection. This length is given by $|\vec{a}|\cos\theta$ .

We know the dot product of two vectors is defined as:

\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta

From this, we can express $|\vec{a}|\cos\theta$ as:

|\vec{a}|\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}

This value, $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$, is called the **scalar projection** of $\vec{a}$ on $\vec{b}$. It tells us the "length" of the projection, including a sign that indicates whether it's in the same or opposite direction as $\vec{b}$.

> [!WARNING]
> A common mistake is to confuse the scalar projection with the vector projection. The scalar projection is a number (a scalar), while the vector projection is a vector. Option (B) in the question represents the scalar projection.

2. Determine the direction of the projection.

Since the projection vector lies along $\vec{b}$ , its direction must be the same as the direction of $\vec{b}$ . The unit vector in the direction of $\vec{b}$ is given by:

\hat{b} = \frac{\vec{b}}{|\vec{b}|}

Combine magnitude and direction to form the vector projection. To get the vector projection, we multiply its magnitude (the scalar projection) by its direction (the unit vector $\hat{b}$ ). Let $\text{proj}_{\vec{b}}\vec{a}$ denote the vector projection of $\vec{a}$ on $\vec{b}$ .

\text{proj}_{\vec{b}}\vec{a} = (\text{scalar projection}) \times (\text{unit vector in direction of } \vec{b})

\text{proj}_{\vec{b}}\vec{a} = \left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\right) \left(\frac{\vec{b}}{|\vec{b}|}\right)

Multiplying these terms, we get:

\text{proj}_{\vec{b}}\vec{a} = \left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right)\vec{b}

> [!FORMULA]
> The vector projection of $\vec{a}$ on $\vec{b}$ is given by:
> $$ \text{proj}_{\vec{b}}\vec{a} = \left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right)\vec{b} $$

Comparing this derived formula with the given options:

(A) $\left(\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|^2}\right)\vec{b}$

(B) $\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|}$ (This is the scalar projection)

(D) $\left(\dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|^2}\right)\vec{b}$ (Incorrect denominator)

The derived formula matches option (A).

✓Final answer

\vec{a}

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 26 on this concept.

CBSE 2026Set A1 markMCQ
$\vec{i} + 3\vec{j} + 7\vec{k}$
(a) $5$
(b) $25$
(c) $6$
(d) none of these
›Reveal solutionSolution

Scalar projection of $\vec{a}$ on $\vec{b}$ is $\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|}$ .

With $\vec{a} = \vec{i}+3\vec{j}+7\vec{k}$ and $\vec{b} = 2\vec{i}-3\vec{j}+6\vec{k}$ :

$\vec{a}\cdot\vec{b} = (1)(2) + (3)(-3) + (7)(6) = 2 - 9 + 42 = 35,$

$|\vec{b}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4+9+36} = \sqrt{49} = 7.$

Projection $= \dfrac{35}{7} = 5$ .

✓Final answer
$5$
CBSE 2026Set ANNUAL1 markMCQ
$\hat{i}-\hat{j}$
(a) 0
(b) 1
(c) -1
(d) 2
›Reveal solutionSolution

The two vectors are perpendicular, so the projection is $0$ .

The projection of vector $\vec u$ on vector $\vec v$ is

$\text{proj}_{\vec v}\vec u=\frac{\vec u\cdot\vec v}{|\vec v|}$

Here $\vec u=\hat i-\hat j$ and $\vec v=\hat i+\hat j$ .

Dot product:

$\vec u\cdot\vec v=(1)(1)+(-1)(1)=1-1=0$

Since the dot product is $0$ , the projection is

$\text{proj}_{\vec v}\vec u=\frac{0}{|\vec v|}=0$

(This makes geometric sense: $\hat i-\hat j$ and $\hat i+\hat j$ are perpendicular diagonals of the unit square, so one has zero projection on the other.)

✓Final answer
$0$
CBSE 2026Set ANNUAL1 markMCQ
$\vec{a} = 2\hat{i}+3\hat{j}+2\hat{k}$
(a) $\dfrac{3}{2}$
(b) $\dfrac{10}{\sqrt{6}}$
(c) $\dfrac{\sqrt{6}}{10}$
(d) None of these
›Reveal solutionSolution

The projection of $\vec a$ on $\vec b$ is $\dfrac{\vec a\cdot\vec b}{|\vec b|}$ .

$\vec a\cdot\vec b = (2)(1)+(3)(2)+(2)(1) = 2+6+2=10$

$|\vec b| = \sqrt{1^2+2^2+1^2} = \sqrt{6}$

Projection $= \dfrac{10}{\sqrt{6}}$ .

✓Final answer
$\dfrac{10}{\sqrt6}$
CBSE 2026Set ANNUAL1 mark
$Q. If a⃗ = 2î + 3ĵ + 5k̂ and b⃗ = 3î + ĵ + k̂, then find the projection of a⃗ on b⃗.$
›Reveal solutionSolution

Projection of $\vec a$ on $\vec b$ is $\dfrac{\vec a\cdot \vec b}{|\vec b|}$ .

$\vec a\cdot \vec b = 2(3)+3(1)+5(1) = 6+3+5=14$

$|\vec b| = \sqrt{3^2+1^2+1^2} = \sqrt{11}$

Projection $= \dfrac{14}{\sqrt{11}}$

✓Final answer
$\vec a$
CBSE 2026Set ANNUAL1 markMCQ
$3\hat{i}-\hat{j}-2\hat{k}$
(a) $\dfrac{7}{\sqrt{14}}$
(b) $\dfrac{7}{14}$
(c) $\dfrac{6}{13}$
(d) $\dfrac{7}{2}$
›Reveal solutionSolution

$\vec a\cdot\vec b=7$ , $|\vec b|=\sqrt{14}$ , so projection $=\tfrac{7}{\sqrt{14}}$ .

With $\vec a=3\hat i-\hat j-2\hat k$ and $\vec b=\hat i+2\hat j-3\hat k$ :

$\vec a\cdot\vec b=(3)(1)+(-1)(2)+(-2)(-3)=3-2+6=7.$

$|\vec b|=\sqrt{1^{2}+2^{2}+(-3)^{2}}=\sqrt{14}.$

Scalar projection $=\dfrac{\vec a\cdot\vec b}{|\vec b|}=\dfrac{7}{\sqrt{14}}.$

✓Final answer
$\dfrac{7}{\sqrt{14}}$
CBSE 2025Set 65/2/11 markMCQ
$\vec{a}$
›Reveal solutionSolution
The projection vector of $\vec{a}$ on $\vec{b}$ is the component of $\vec{a}$ that lies along the direction of $\vec{b}$ , and it is given by the formula $\left(\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|^2}\right)\vec{b}$ .

When we talk about the projection of vector $\vec{a}$ onto vector $\vec{b}$ , we are essentially asking: "How much of vector $\vec{a}$ points in the same direction as vector $\vec{b}$ ?" Imagine shining a light perpendicular to vector $\vec{b}$ . The shadow of $\vec{a}$ cast on the line containing $\vec{b}$ is its projection.

This projection is itself a vector. It will always point in the same direction as $\vec{b}$ (or opposite, if the angle between $\vec{a}$ and $\vec{b}$ is obtuse). To define this vector, we need two things: its magnitude and its direction.

Determine the magnitude of the projection (Scalar Projection).

Let $\theta$ be the angle between vectors $\vec{a}$ and $\vec{b}$ . Geometrically, if we drop a perpendicular from the tip of $\vec{a}$ onto the line containing $\vec{b}$ , the length of the segment formed on $\vec{b}$ is the magnitude of the projection. This length is given by $|\vec{a}|\cos\theta$ .

We know the dot product of two vectors is defined as:

$\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$
From this, we can express $|\vec{a}|\cos\theta$ as:
$|\vec{a}|\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$
This value, $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$, is called the **scalar projection** of $\vec{a}$ on $\vec{b}$. It tells us the "length" of the projection, including a sign that indicates whether it's in the same or opposite direction as $\vec{b}$. > [!WARNING] > A common mistake is to confuse the scalar projection with the vector projection. The scalar projection is a number (a scalar), while the vector projection is a vector. Option (B) in the question represents the scalar projection.

2. Determine the direction of the projection.

Since the projection vector lies along $\vec{b}$ , its direction must be the same as the direction of $\vec{b}$ . The unit vector in the direction of $\vec{b}$ is given by:
$\hat{b} = \frac{\vec{b}}{|\vec{b}|}$

Combine magnitude and direction to form the vector projection. To get the vector projection, we multiply its magnitude (the scalar projection) by its direction (the unit vector $\hat{b}$ ). Let $\text{proj}_{\vec{b}}\vec{a}$ denote the vector projection of $\vec{a}$ on $\vec{b}$ .

$\text{proj}_{\vec{b}}\vec{a} = (\text{scalar projection}) \times (\text{unit vector in direction of } \vec{b})$ $\text{proj}_{\vec{b}}\vec{a} = \left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\right) \left(\frac{\vec{b}}{|\vec{b}|}\right)$
Multiplying these terms, we get:
$\text{proj}_{\vec{b}}\vec{a} = \left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right)\vec{b}$
> [!FORMULA] > The vector projection of $\vec{a}$ on $\vec{b}$ is given by: > $$ \text{proj}_{\vec{b}}\vec{a} = \left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right)\vec{b} $$

Comparing this derived formula with the given options:

(A) $\left(\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|^2}\right)\vec{b}$

(B) $\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|}$ (This is the scalar projection)

(C) $\dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|}$ (This would be the scalar projection of $\vec{b}$ on $\vec{a}$ )

(D) $\left(\dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|^2}\right)\vec{b}$ (Incorrect denominator)

The derived formula matches option (A).

✓Final answer
$\vec{a}$
CBSE 2025Set X11 mark
$\hat{i} + \hat{j}$
›Reveal solutionSolution

The vectors are perpendicular ( $\vec{a}\cdot\vec{b}=0$ ), so the projection is $0$ .

Concept: The (scalar) projection of $\vec{a}$ along $\vec{b}$ is $\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|}$ .

Step 1 — Dot product.

With $\vec{a} = \hat{i}+\hat{j}$ and $\vec{b} = \hat{i}-\hat{j}$ ,
$\vec{a}\cdot\vec{b} = (1)(1) + (1)(-1) = 0.$
Step 2 — Magnitude of $\vec{b}$ .
$|\vec{b}| = \sqrt{1^2 + (-1)^2} = \sqrt{2}.$
Step 3 — Projection.
$\text{Projection} = \frac{\vec{a}\cdot\vec{b}}{|\vec{b}|} = \frac{0}{\sqrt{2}} = 0.$
✓Final answer
$0$
CBSE 2025Set ANNUAL1 markMCQ
$X$
(i) 12
(ii) 13
(iii) 4
(iv) 3
›Reveal solutionSolution

The length of a line segment equals the square root of the sum of the squares of its projections on the three axes.

If a line segment has projections $a,b,c$ on the $X,Y,Z$ axes, its length is $\sqrt{a^2+b^2+c^2}$ .

Here $a=12,\ b=4,\ c=3$ :

$L = \sqrt{12^2+4^2+3^2} = \sqrt{144+16+9} = \sqrt{169} = 13$

✓Final answer
$(ii) 13.$
CBSE 2025Set ANNUAL1 mark
$\hat{i}+2\hat{j}$
›Reveal solutionSolution

The projection of a vector on the $x$ -axis is simply its $\hat{i}$ -component.

For $\vec{a}=\hat{i}+2\hat{j}$ , projection on $x$ -axis $=\vec{a}\cdot\hat{i}=1$ .

✓Final answer
$1$
CBSE 2025Set ANNUAL1 markMCQ
$\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$
(a) $0$
(b) $1$
(c) $-\dfrac{2}{3}$
(d) $-\dfrac{3}{2}$
›Reveal solutionSolution

A zero projection means the dot product a·b is zero; solve for λ.

$\vec a = \hat i - 2\hat j + 3\hat k$ , $\vec b = 2\hat i + 0\hat j + \lambda\hat k$

The projection of $\vec a$ on $\vec b$ is $\dfrac{\vec a\cdot\vec b}{|\vec b|}$ . This is zero exactly when $\vec a \cdot \vec b = 0$ (since $|\vec b|\ne0$ ).

$\vec a \cdot \vec b = (1)(2) + (-2)(0) + (3)(\lambda) = 2 + 3\lambda$

Set equal to zero: $2+3\lambda = 0 \Rightarrow \lambda = -\dfrac{2}{3}$

✓Final answer
$-\dfrac{2}{3}$
CBSE 2025Set ANNUAL1 markMCQ
$Q. The projection of the vector a⃗ = 2î + 3ĵ + 2k̂ on the vector b⃗ = î + 2ĵ + k̂ is :$
(a) 10/√6
(b) −10/√6
(c) 5/√6
(d) −5/√6
›Reveal solutionSolution

Projection of $\vec a$ on $\vec b$ is $\dfrac{\vec a\cdot \vec b}{|\vec b|}$ .

Given $\vec a = 2\hat i+3\hat j+2\hat k$ and $\vec b=\hat i+2\hat j+\hat k$ :

$\vec a\cdot\vec b = 2(1)+3(2)+2(1) = 2+6+2=10.$

$|\vec b| = \sqrt{1^2+2^2+1^2}=\sqrt6.$

Projection of $\vec a$ on $\vec b$ :

$\frac{\vec a\cdot\vec b}{|\vec b|} = \frac{10}{\sqrt6}.$

✓Final answer
$\dfrac{10}{\sqrt6}$
CBSE 20241 markMCQ
$\vec{a} = 2\hat{i} - 2\hat{j} + \hat{k}$
›Reveal solutionSolution
The projection of a vector onto another is the scalar component along that direction, found by the dot product divided by the magnitude of the reference vector. Here, the projection of $(\vec{c} - \vec{b})$ along $\vec{a}$ equals 5, which corresponds to option (B).

The idea of projection is simple: if you shine a light straight down onto a line, the shadow a vector casts on that line is its projection. The scalar projection (what we want here) is just the length of that shadow — how much of one vector points in the direction of another. It’s given by the formula:
$\text{Projection of } \vec{u} \text{ along } \vec{v} = \frac{\vec{u} \cdot \vec{v}}{|\vec{v}|}$
This is a scalar (could be positive or negative), telling you the signed magnitude of the component of $\vec{u}$ in the direction of $\vec{v}$ . No unit vector needed — just the dot product divided by the length of the reference vector.

Let’s apply this step by step.

Find the vector we’re projecting: $\vec{c} - \vec{b}$ $\vec{c} = 2\hat{i} - \hat{j} + 4\hat{k}$ $\vec{b} = \hat{i} + 2\hat{j} - 3\hat{k}$ Subtract component-wise:

$\vec{c} - \vec{b} = (2-1)\hat{i} + (-1-2)\hat{j} + (4 - (-3))\hat{k} = \hat{i} - 3\hat{j} + 7\hat{k}$

Compute the dot product of $(\vec{c} - \vec{b})$ with $\vec{a}$ $\vec{a} = 2\hat{i} - 2\hat{j} + \hat{k}$

$(\vec{c} - \vec{b}) \cdot \vec{a} = (1)(2) + (-3)(-2) + (7)(1) = 2 + 6 + 7 = 15$

Find the magnitude of $\vec{a}$

$|\vec{a}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$

Apply the projection formula

$\text{Projection} = \frac{(\vec{c} - \vec{b}) \cdot \vec{a}}{|\vec{a}|} = \frac{15}{3} = 5$
Watch out

A common mistake is to forget the denominator and just take the dot product as the projection. The dot product alone gives 15, but that’s not the projection — it’s the numerator. Always divide by the magnitude of the reference vector.

Tip

Notice that the projection is a scalar, not a vector. If the problem had asked for the vector projection, you’d multiply this scalar by the unit vector in the direction of $\vec{a}$ . But here, “projection … along $\vec{a}$ ” in Indian exam language means the scalar projection.

✓Final answer
$\boxed{5}$

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