Two statements are given, one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true. Assertion (A): Projection of $\vec{a}$ on $\vec{b}$ is same as projection of $\vec{b}$ on $\vec{a}$. Reason (R): Angle between $\vec{a}$ and $\vec{b}$ is same as angle between $\vec{b}$ and $\vec{a}$ numerically.

Two statements are given, one labelled Assertion (A) and the other labelled Reason (R). Select the correct ans

Question

\vec{a}

CBSECBSE Class XII Board 2024MCQ· 1mImportance★★★★★

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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 $y=x$ . …

Start your 14-day free trial to unlock the full solution →

The projection of one vector onto another is not symmetric — it depends on which vector is being projected. The angle between two vectors is symmetric, but that alone does not make the projections equal. Assertion (A) is false; Reason (R) is true.

The key idea here is the definition of vector projection. The projection of $\vec{a}$ on $\vec{b}$ is the component of $\vec{a}$ along the direction of $\vec{b}$ . It is given by:

\text{proj}_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}

Notice the denominator uses the magnitude of the vector onto which you are projecting. Similarly, the projection of $\vec{b}$ on $\vec{a}$ is:

\text{proj}_{\vec{a}} \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|}

These two are clearly different unless $|\vec{a}| = |\vec{b}|$ . So the assertion that they are always the same is false.

Reason (R) states that the angle between $\vec{a}$ and $\vec{b}$ is the same as the angle between $\vec{b}$ and $\vec{a}$ numerically. This is true — angle is a symmetric property: $\theta(\vec{a}, \vec{b}) = \theta(\vec{b}, \vec{a})$ , since the dot product is commutative and the magnitudes are positive.

Now let's go step by step.

Write the projection formulas explicitly. Projection of $\vec{a}$ on $\vec{b}$ :

p_1 = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = |\vec{a}| \cos \theta

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

p_2 = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|} = |\vec{b}| \cos \theta

Here $\theta$ is the angle between them.

Compare $p_1$ and $p_2$ .

For $p_1 = p_2$ , we need $|\vec{a}| \cos \theta = |\vec{b}| \cos \theta$ .

If $\cos \theta \neq 0$ , this forces $|\vec{a}| = |\vec{b}|$ . If $\cos \theta = 0$ , both projections are zero, so they are equal — but that is a special case, not generally true.

So the assertion is false in general.
Examine Reason (R). …

Start your 14-day free trial to unlock the full solution →

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

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

Start your 14-day free trial to see the full solution →
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

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

…

Start your 14-day free trial to see the full solution →
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 $\dfrac{\vec a\cdot\vec b}{|\vec b|}$ .

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

…

Start your 14-day free trial to see the full solution →
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 $\dfrac{\vec a\cdot \vec b}{|\vec b|}$ .

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

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

…

Start your 14-day free trial to see the full solution →
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 $\vec b=\hat i+2\hat j-3\hat k$ :

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

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

…

Start your 14-day free trial to see the full solution →
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 $\vec{b}$ . …
Start your 14-day free trial to see the full solution →
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 $\vec{b}$ . …
Start your 14-day free trial to see the full solution →
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 $\sqrt{a^2+b^2+c^2}$ .

…

Start your 14-day free trial to see the full solution →
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.

…

Start your 14-day free trial to see the full solution →
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 $|\vec b|\ne0$ ).

…

Start your 14-day free trial to see the full solution →
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 $\vec b=\hat i+2\hat j+\hat k$ :

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

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

…

Start your 14-day free trial to see the full solution →
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 …
Start your 14-day free trial to see the full solution →

🎓Unlock everything free for 14 days

Unlock everything free for 14 days