Q.Find the projection of the vector $\hat{i}-\hat{j}$ on the vector $\hat{i}+\hat{j}$.

Showing the 12 most recent of 26 on this concept.

1. CBSE 2023Set 65/2/11 markMCQ
   Q.The projection of the vector $2\hat{i} + 3\hat{j}$ on the vector $3\hat{i} - 2\hat{j}$ is:

   (a) 0

   (b) 12

   (c) $\frac{12}{\sqrt{13}}$

   (d) $-\frac{12}{\sqrt{13}}$

   ›Reveal solutionSolution

   The projection of $\vec{a}=2\hat{i}+3\hat{j}$ onto $\vec{b}=3\hat{i}-2\hat{j}$ is found using the scalar projection formula $\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|}$. The dot product is $0$, so the projection is $0$ — option (a).

   The idea of projecting one vector onto another is like asking: how much of vector $\vec{a}$ points in the direction of vector $\vec{b}$?

   If you shine a light straight down onto $\vec{b}$, the shadow that $\vec{a}$ casts along $\vec{b}$ is its projection. That shadow can be positive (same direction), negative (opposite direction), or zero (perpendicular).

   The scalar projection (also called the component) of $\vec{a}$ onto $\vec{b}$ is given by:

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

   This formula works because the dot product measures how aligned two vectors are, and dividing by the length of $\vec{b}$ scales it to a unit direction.

   Now let’s apply it step by step.

   1. Identify the vectors

      $\vec{a} = 2\hat{i} + 3\hat{j}$

      $\vec{b} = 3\hat{i} - 2\hat{j}$

   2. Compute the dot product

      $\vec{a} \cdot \vec{b} = (2)(3) + (3)(-2) = 6 - 6 = 0$

      The dot product is zero — this already tells us the vectors are perpendicular.

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

      $|\vec{b}| = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}$

   4. Apply the projection formula

   $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{0}{\sqrt{13}} = 0$

   Watch out

   A common mistake is to forget the denominator and just pick the dot product (option b), or to compute the vector projection instead of the scalar projection. The scalar projection is a number, not a vector.

   Tip

   Whenever the dot product is zero, the projection is automatically zero — no need to compute the magnitude at all. This is a quick sanity check in exams.

   ✓Final answer

   The projection is $0$, so the correct option is (a).

2. 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

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

   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

   The projection is $\boxed{5}$, which matches option (B).

3. 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 $\vec{b}$.

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

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

   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

   The projection vector of vector $\vec{a}$ on vector $\vec{b}$ is $\boxed{\left(\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|^2}\right)\vec{b}}$.

4. CBSE 2020Set 65/1/11 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) $-\frac{2}{3}$ (D) $-\frac{3}{2}$

   ›Reveal solutionSolution

   The projection of $\vec{a}$ on $\vec{b}$ is zero when $\vec{a}$ is perpendicular to $\vec{b}$. Using the dot product condition $\vec{a} \cdot \vec{b} = 0$, we get $2 - 0 + 3\lambda = 0$, so $\lambda = -\frac{2}{3}$. The correct option is (C).

   The idea of projection is simple: it tells you how much of one vector lies along the direction of another. When the projection is zero, it means the two vectors are perpendicular — they have no "overlap" in that direction. Mathematically, the scalar projection of $\vec{a}$ on $\vec{b}$ is given by $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$. If this equals zero, then $\vec{a} \cdot \vec{b} = 0$ (provided $\vec{b}$ is not the zero vector, which it isn't here).

   So the entire problem reduces to one clean condition: the dot product must vanish.

   Let's work through it step by step.

   1. Write the vectors clearly.

      $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$

      $\vec{b} = 2\hat{i} + 0\hat{j} + \lambda\hat{k}$

      Notice that $\vec{b}$ has no $\hat{j}$ component — that's fine.

   2. Set up the dot product.

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

   3. Apply the zero-projection condition.

      Since projection is zero, $\vec{a} \cdot \vec{b} = 0$.

      So $2 + 3\lambda = 0$.

   4. Solve for $\lambda$.

      $3\lambda = -2$

      $\lambda = -\frac{2}{3}$.

   Watch out

   A common mistake is to forget that the projection formula involves the magnitude of $\vec{b}$ in the denominator. But since we only care when the projection is zero, the denominator doesn't matter — as long as $\vec{b} \neq \vec{0}$, which is true here because $2\hat{i} + \lambda\hat{k}$ is never the zero vector for any real $\lambda$. So the condition $\vec{a} \cdot \vec{b} = 0$ is both necessary and sufficient.

   Tip

   You could also think geometrically: projection zero means the vectors are orthogonal. So this is just a perpendicularity problem in disguise. No need to compute magnitudes at all.

   ✓Final answer

   The value of $\lambda$ is $\boxed{-\frac{2}{3}}$, which corresponds to option (C).

5. CBSE 2023Set 65/3/11 markMCQ
   Q.$\vec{a}$ and $\vec{b}$ are two non-zero vectors such that the projection of $\vec{a}$ on $\vec{b}$ is $0$. The angle between $\vec{a}$ and $\vec{b}$ is :

   (a) $\frac{\pi}{2}$

   (b) $\pi$

   (c) $\frac{\pi}{4}$

   (d) $0$

   ›Reveal solutionSolution

   The projection of vector $\vec{a}$ on vector $\vec{b}$ is given by $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$. If this projection is $0$, it implies $\vec{a} \cdot \vec{b} = 0$, which means the vectors are perpendicular, so the angle between them is $\frac{\pi}{2}$.

   Understanding vector projection is crucial here. Geometrically, the projection of vector $\vec{a}$ onto vector $\vec{b}$ is the length of the "shadow" that $\vec{a}$ casts on $\vec{b}$ when a light source is directly above $\vec{a}$ and perpendicular to $\vec{b}$. More precisely, it's the scalar component of $\vec{a}$ in the direction of $\vec{b}$.

   If this projection is $0$, it means there is no "shadow" of $\vec{a}$ along the direction of $\vec{b}$. This can only happen if $\vec{a}$ is perpendicular to $\vec{b}$. Think about it: if you shine a light directly down on a vertical pole, its shadow on the ground is zero. This is the core intuition.

   Let's work through the steps using the mathematical definition.

   1. Recall the formula for the scalar projection of $\vec{a}$ on $\vec{b}$.

      The scalar projection of vector $\vec{a}$ onto vector $\vec{b}$, often denoted as $\text{proj}_{\vec{b}} \vec{a}$ or $\text{comp}_{\vec{b}} \vec{a}$, is given by:

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

      Here, $\vec{a} \cdot \vec{b}$ is the dot product of $\vec{a}$ and $\vec{b}$, and $|\vec{b}|$ is the magnitude of $\vec{b}$.

   2. Apply the given condition.

      The problem states that the projection of $\vec{a}$ on $\vec{b}$ is $0$. So, we set the formula equal to $0$:

      $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = 0$

   3. Interpret the result of the equation.

      For a fraction to be zero, its numerator must be zero, provided the denominator is non-zero. The problem states that $\vec{b}$ is a non-zero vector, which means its magnitude $|\vec{b}| \neq 0$.

      Therefore, we must have:

      $\vec{a} \cdot \vec{b} = 0$

   4. Relate the dot product to the angle between vectors.

      The dot product of two non-zero vectors $\vec{a}$ and $\vec{b}$ is also defined in terms of their magnitudes and the angle $\theta$ between them:

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

      Substituting $\vec{a} \cdot \vec{b} = 0$ into this formula, we get:

      $|\vec{a}| |\vec{b}| \cos \theta = 0$

   5. Solve for the angle $\theta$.

      The problem states that both $\vec{a}$ and $\vec{b}$ are non-zero vectors. This means their magnitudes $|\vec{a}| \neq 0$ and $|\vec{b}| \neq 0$.

      For the product $|\vec{a}| |\vec{b}| \cos \theta$ to be zero, and knowing that $|\vec{a}|$ and $|\vec{b}|$ are non-zero, it must be that $\cos \theta = 0$.

      The angle $\theta$ between two vectors is conventionally taken to be in the range $[0, \pi]$. Within this range, the value of $\theta$ for which $\cos \theta = 0$ is $\frac{\pi}{2}$.

      Important

      If the dot product of two non-zero vectors is zero, the vectors are perpendicular (orthogonal) to each other, and the angle between them is $\frac{\pi}{2}$.

   The angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{2}$. This corresponds to option (a).

   ✓Final answer

   The angle between $\vec{a}$ and $\vec{b}$ is $\boxed{\frac{\pi}{2}}$.

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

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

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

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

   ✓Final answer

   (a) $5$.

7. 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

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

   (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

   The correct option is (a) $0$.

8. 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 $\dfrac{\vec a\cdot\vec b}{|\vec b|}$.

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

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

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

   ✓Final answer

   (b) $\dfrac{10}{\sqrt6}$.

9. 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

   Projection of $\vec a$ on $\vec b$ is $\dfrac{14}{\sqrt{11}}$.

10. 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 $\vec b=\hat i+2\hat j-3\hat k$:

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

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

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

    ✓Final answer

    $\dfrac{7}{\sqrt{14}}$ — option (A).

11. 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 $\vec{b}$.

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

    Step 3 — Projection.

    $\text{Projection} = \frac{\vec{a}\cdot\vec{b}}{|\vec{b}|} = \frac{0}{\sqrt{2}} = 0.$

    ✓Final answer

    $0$

12. 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 $\sqrt{a^2+b^2+c^2}$.

    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.