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

Showing the 12 most recent of 26 on this concept.

1. CBSE 2026Set A1 markMCQ
   Q.The projection of the vector $\vec{i} + 3\vec{j} + 7\vec{k}$ on the vector $2\vec{i} - 3\vec{j} + 6\vec{k}$ is

   (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 →
2. CBSE 2026Set ANNUAL1 markMCQ
   Q.Write the projection of the vector $\hat{i}-\hat{j}$ on the vector $\hat{i}+\hat{j}$.

   (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 →
3. CBSE 2026Set ANNUAL1 markMCQ
   Q.Find the projection of vector $\vec{a} = 2\hat{i}+3\hat{j}+2\hat{k}$ on the vector $\vec{b} = \hat{i}+2\hat{j}+\hat{k}$.

   (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 →
4. 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 →
5. CBSE 2026Set ANNUAL1 markMCQ
   Q.The scalar projection of the vector $3\hat{i}-\hat{j}-2\hat{k}$ on the vector $\hat{i}+2\hat{j}-3\hat{k}$ is

   (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 →
6. CBSE 2025Set 65/2/11 markMCQ
   Q.The projection vector of vector $\vec{a}$ on vector $\vec{b}$ is: (A) $\left(\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|^2}\right)\vec{b}$ (B) $\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|}$ (C) $\dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|}$ (D) $\left(\dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|^2}\right)\vec{b}$

   ›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.

   1. 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:

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

   3. 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 →
7. CBSE 2025Set X11 mark
   Q.The projection of vector $\hat{i} + \hat{j}$ along the vector $\hat{i} - \hat{j}$ is __________.

   ›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 →
8. CBSE 2025Set ANNUAL1 markMCQ
   Q.The projections of a line segment on $X$-axis, $Y$-axis and $Z$-axis are 12, 4 and 3 respectively. What is the length of the line segment?

   (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 →
9. CBSE 2025Set ANNUAL1 mark
   Q.The projection of the vector $\hat{i}+2\hat{j}$ on $x$-axis is _____.

   ›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 →
10. CBSE 2025Set ANNUAL1 markMCQ
    Q.If the projection of $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$ on $\vec{b} = 2\hat{i} + \lambda\hat{k}$ is zero, then the value of $\lambda$ is

    (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 →
11. 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 →
12. CBSE 20241 markMCQ
    Q.If $\vec{a} = 2\hat{i} - 2\hat{j} + \hat{k}$, $\vec{b} = \hat{i} + 2\hat{j} - 3\hat{k}$ and $\vec{c} = 2\hat{i} - \hat{j} + 4\hat{k}$, then the projection of $(\vec{c} - \vec{b})$ along $\vec{a}$ is: (A) 15 (B) 5 (C) $\frac{2}{3}$ (D) 1

    ›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.

    1. 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:

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

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

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

    3. Find the magnitude of $\vec{a}$

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

    4. Apply the projection formula …

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