Q.Find the scalar components and magnitude of the vector joining the points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$.

Showing the 12 most recent of 29 on this concept.

1. CBSE 2023Set 65/1/11 markMCQ
   Q.If $\vec{a} + \vec{b} = \hat{i}$ and $\vec{a} = 2\hat{i} - 2\hat{j} + 2\hat{k}$, then $|\vec{b}|$ equals : (A) $\sqrt{14}$ (B) 3 (C) $\sqrt{12}$ (D) $\sqrt{17}$

   ›Reveal solutionSolution

   $\vec{b} = \hat{i} - \vec{a} = -\hat{i} + 2\hat{j} - 2\hat{k}$, so $|\vec{b}| = 3$ — option (B).

   We are given $\vec{a} + \vec{b} = \hat{i}$ and $\vec{a} = 2\hat{i} - 2\hat{j} + 2\hat{k}$.

   Isolate $\vec{b}$:

   $\vec{b} = \hat{i} - \vec{a} = \hat{i} - (2\hat{i} - 2\hat{j} + 2\hat{k}) = -\hat{i} + 2\hat{j} - 2\hat{k}.$

   Compute the magnitude:

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

   ✓Final answer

   $|\vec{b}| = 3$ — option (B).

2. CBSE 2026Set 65/3/11 markMCQ
   Q.For any two vectors $\vec{a}$ and $\vec{b}$, which of the following statements is always true? (A) $\vec{a} \cdot \vec{b} \leq |\vec{a}| |\vec{b}|$ (B) $|\vec{a} + \vec{b}| \geq |\vec{a}| + |\vec{b}|$ (C) $|\vec{a} - \vec{b}| = |\vec{a}| - |\vec{b}|$ (D) $|\vec{a} \times \vec{b}| \geq |\vec{a}| |\vec{b}|$

   ›Reveal solutionSolution

   The dot product satisfies $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta$, and since $\cos\theta \leq 1$, we always have $\vec{a} \cdot \vec{b} \leq |\vec{a}| |\vec{b}|$. The correct option is (A).

   The key here is to recall the geometric definitions of the dot product and cross product, and the triangle inequality for vector addition. Each option claims an inequality or equality that must hold for any two vectors — so we test each against the general formulas.

   1. Option (A): $\vec{a} \cdot \vec{b} \leq |\vec{a}| |\vec{b}|$

   The dot product is defined as $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta$, where $\theta$ is the angle between the vectors. Since $\cos\theta$ ranges from $-1$ to $1$, the product $|\vec{a}| |\vec{b}| \cos\theta$ can be as large as $|\vec{a}| |\vec{b}|$ (when $\cos\theta = 1$) and as small as $-|\vec{a}| |\vec{b}|$ (when $\cos\theta = -1$). Therefore, $\vec{a} \cdot \vec{b} \leq |\vec{a}| |\vec{b}|$ is always true — the dot product never exceeds the product of magnitudes.

   Tip

   The inequality $\vec{a} \cdot \vec{b} \leq |\vec{a}| |\vec{b}|$ is actually a form of the Cauchy–Schwarz inequality, which holds for any inner product space.

   2. Option (B): $|\vec{a} + \vec{b}| \geq |\vec{a}| + |\vec{b}|$

   This is the reverse of the triangle inequality. The actual triangle inequality states $|\vec{a} + \vec{b}| \leq |\vec{a}| + |\vec{b}|$, with equality only when the vectors point in the same direction. So the given statement is false — for example, take $\vec{a} = (1,0)$ and $\vec{b} = (-1,0)$; then $|\vec{a} + \vec{b}| = 0$, which is not $\geq 2$.

   3. Option (C): $|\vec{a} - \vec{b}| = |\vec{a}| - |\vec{b}|$

   This would require the vectors to be parallel and pointing in the same direction, with $|\vec{a}| \geq |\vec{b}|$. In general, $|\vec{a} - \vec{b}|$ depends on the angle between them. For instance, if $\vec{a} = (1,0)$ and $\vec{b} = (0,1)$, then $|\vec{a} - \vec{b}| = \sqrt{2}$, but $|\vec{a}| - |\vec{b}| = 0$. So this is not always true.

   4. Option (D): $|\vec{a} \times \vec{b}| \geq |\vec{a}| |\vec{b}|$

   The magnitude of the cross product is $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| |\sin\theta|$. Since $|\sin\theta| \leq 1$, we have $|\vec{a} \times \vec{b}| \leq |\vec{a}| |\vec{b}|$, not greater than or equal. So this is false — equality occurs only when $\sin\theta = 1$ (vectors perpendicular), but the inequality sign is reversed.

   Watch out

   A common mistake is to confuse the dot product inequality with the cross product one. Remember: dot uses $\cos\theta$ (bounded above by 1), cross uses $\sin\theta$ (also bounded above by 1), so both products are at most the product of magnitudes — but the dot product can be negative, while the cross product magnitude is always non-negative.

   Thus, only option (A) holds for all vectors.

   ✓Final answer

   The correct option is (A), because $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta \leq |\vec{a}| |\vec{b}|$ always.

3. CBSE 2020Set 65/2/11 markMCQ
   Q.If $|\vec{a}| = 4$ and $-3 \le \lambda \le 2$, then $|\lambda\vec{a}|$ lies in (A) $[0, 12]$ (B) $[2, 3]$ (C) $[8, 12]$ (D) $[-12, 8]$

   ›Reveal solutionSolution

   The magnitude of a scalar multiple is $|\lambda\vec{a}| = |\lambda|\,|\vec{a}|$, so we need the range of $|\lambda| \cdot 4$ for $\lambda \in [-3, 2]$. The smallest $|\lambda|$ is $0$ and the largest is $3$, giving the range $[0, 12]$. The correct option is (A).

   The key idea here is simple but easy to mess up if you rush. The magnitude of a vector is always non-negative — it's a length. When you multiply a vector by a scalar $\lambda$, the new vector's length is $|\lambda|$ times the original length. Notice the absolute value around $\lambda$: that's the crucial detail.

   If you forget that absolute value and just plug the endpoints $-3$ and $2$ directly into $\lambda \cdot 4$, you'd get $-12$ and $8$, which is option (D). But a length can never be negative, so that can't be right. The magnitude $|\lambda\vec{a}|$ is always $\ge 0$, and the question asks where it lies — meaning the set of all possible values it can take.

   Let's walk through it step by step.

   1. Write the magnitude formula. For any vector $\vec{a}$ and scalar $\lambda$,

   $|\lambda\vec{a}| = |\lambda| \, |\vec{a}|.$

   This is a standard property: scaling a vector scales its length by the absolute value of the scalar.

   2. Plug in the given length. We have $|\vec{a}| = 4$, so

   $|\lambda\vec{a}| = |\lambda| \cdot 4.$

   3. Find the range of $|\lambda|$.

      $\lambda$ can be any real number between $-3$ and $2$, inclusive.

      • The absolute value $|\lambda|$ is smallest when $\lambda = 0$, giving $|\lambda| = 0$.
      • The absolute value $|\lambda|$ is largest at the endpoint farthest from zero, which is $\lambda = -3$, giving $|\lambda| = 3$. So $|\lambda|$ ranges from $0$ to $3$.
      Watch out

      A common mistake is to think the maximum of $|\lambda|$ occurs at $\lambda = 2$ because $2$ is the largest number in $[-3, 2]$. But $|\lambda|$ measures distance from zero, not the number itself. The point $-3$ is farther from zero than $2$ is.

   4. Multiply by 4.

      Since $|\lambda\vec{a}| = 4|\lambda|$, the smallest value is $4 \times 0 = 0$ and the largest is $4 \times 3 = 12$.

      As $\lambda$ varies continuously over $[-3, 2]$, $|\lambda|$ takes every value between $0$ and $3$, so $|\lambda\vec{a}|$ takes every value between $0$ and $12$.

      Tip

      You can think of it this way: the function $f(\lambda) = |\lambda|$ on $[-3, 2]$ is V-shaped, hitting a minimum of $0$ at $\lambda = 0$ and a maximum of $3$ at $\lambda = -3$. Scaling by $4$ just stretches that V vertically.

   5. Match with the options.

      The interval $[0, 12]$ is exactly option (A). Option (B) $[2, 3]$ is too narrow, option (C) $[8, 12]$ misses the smaller values, and option (D) $[-12, 8]$ includes negative numbers, which are impossible for a magnitude.

   ✓Final answer

   The correct option is (A) $[0, 12]$.

4. CBSE 2026Set 65/2/11 markMCQ
   Q.If $|\vec{a}| = 5$ and $-2 \le \lambda \le 1$, then the sum of greatest and the smallest value of $|\lambda \vec{a}|$ is (A) $-5$ (B) $5$ (C) $10$ (D) $15$

   ›Reveal solutionSolution

   Scalar multiplication scales a vector's magnitude by the absolute value of the scalar; since $|\lambda \vec{a}| = |\lambda| \cdot |\vec{a}|$, we find the extreme values of $|\lambda|$ over $[-2, 1]$ and multiply by $5$.

   When you multiply a vector by a scalar, the magnitude of the resulting vector depends only on the absolute value of that scalar. This is because direction reversals (negative scalars) don't affect length—they just flip the arrow. The fundamental property is:

   $|\lambda \vec{a}| = |\lambda| \cdot |\vec{a}|$

   Given $|\vec{a}| = 5$, we have $|\lambda \vec{a}| = 5|\lambda|$. The problem now reduces to finding the maximum and minimum values of $|\lambda|$ as $\lambda$ ranges over $[-2, 1]$.

   Finding the extreme values of $|\lambda|$:

   1. Understand the absolute value function on the interval. For $\lambda \in [-2, 1]$, the function $|\lambda|$ equals $-\lambda$ when $\lambda < 0$ and equals $\lambda$ when $\lambda \ge 0$. The graph is V-shaped with its vertex at $\lambda = 0$.

   2. Identify the minimum. The absolute value $|\lambda|$ is smallest at $\lambda = 0$, where $|\lambda| = 0$. Therefore, the minimum value of $|\lambda \vec{a}|$ is $5 \cdot 0 = 0$.

   3. Identify the maximum. Since $|\lambda|$ increases as we move away from zero in either direction, we check the endpoints of the interval:

      • At $\lambda = -2$: $|\lambda| = |-2| = 2$
      • At $\lambda = 1$: $|\lambda| = |1| = 1$

      The maximum occurs at $\lambda = -2$, giving $|\lambda| = 2$. Therefore, the maximum value of $|\lambda \vec{a}|$ is $5 \cdot 2 = 10$.

   Tip

   For any interval containing zero, the minimum of $|\lambda|$ is always $0$. The maximum is the larger of the absolute values of the endpoints.

   4. Compute the sum. The greatest value is $10$, the smallest is $0$, so their sum is $10 + 0 = 10$.

   ✓Final answer

   The sum of the greatest and smallest values is $\boxed{10}$, so the correct option is (C).

5. CBSE 2026Set CX1 mark
   Q.If for a unit vector $\vec{a}$, $(\vec{x}-\vec{a})\cdot(\vec{x}+\vec{a})=12$, then find $|\vec{x}|$.

   ›Reveal solutionSolution

   The dot product expands to $|\vec x|^{2}-|\vec a|^{2}=12$; with a unit $\vec a$ this gives $|\vec x|=\sqrt{13}$.

   Concept: For any vectors, $(\vec p-\vec q)\cdot(\vec p+\vec q)=\vec p\cdot\vec p-\vec q\cdot\vec q=|\vec p|^{2}-|\vec q|^{2}$ (the cross terms cancel).

   $(\vec{x}-\vec{a})\cdot(\vec{x}+\vec{a})=|\vec{x}|^{2}-|\vec{a}|^{2}=12.$

   Since $\vec a$ is a unit vector, $|\vec a|=1$, so

   $|\vec{x}|^{2}-1=12\;\Rightarrow\;|\vec{x}|^{2}=13\;\Rightarrow\;|\vec{x}|=\sqrt{13}.$

   ✓Final answer

   $|\vec{x}|=\sqrt{13}$.

6. CBSE 2026Set A1 markMCQ
   Q.$|3\vec{i} + 4\vec{j} + 7\vec{k}| =$

   (a) $\sqrt{14}$

   (b) $\sqrt{74}$

   (c) $\sqrt{61}$

   (d) $\sqrt{94}$

   ›Reveal solutionSolution

   Magnitude of $a\vec{i}+b\vec{j}+c\vec{k}$ is $\sqrt{a^2+b^2+c^2}$.

   $|3\vec{i} + 4\vec{j} + 7\vec{k}| = \sqrt{3^2 + 4^2 + 7^2} = \sqrt{9 + 16 + 49} = \sqrt{74}.$

   ✓Final answer

   (b) $\sqrt{74}$.

7. CBSE 2026Set ANNUAL1 markMCQ
   Q.The magnitude of the vector $\dfrac{1}{\sqrt{3}}\hat{i}+\dfrac{1}{\sqrt{3}}\hat{j}+\dfrac{1}{\sqrt{3}}\hat{k}$ is

   (a) 0

   (b) 3

   (c) 1

   (d) -1

   ›Reveal solutionSolution

   Use the magnitude formula $|\vec v|=\sqrt{v_x^2+v_y^2+v_z^2}$ on the given vector.

   Given $\vec v = \dfrac{1}{\sqrt3}\hat{i}+\dfrac{1}{\sqrt3}\hat{j}+\dfrac{1}{\sqrt3}\hat{k}$.

   $|\vec v| = \sqrt{\left(\frac{1}{\sqrt3}\right)^2 + \left(\frac{1}{\sqrt3}\right)^2 + \left(\frac{1}{\sqrt3}\right)^2}$

   $= \sqrt{\frac13+\frac13+\frac13} = \sqrt{\frac33} = \sqrt1 = 1$

   ✓Final answer

   (c) 1

8. CBSE 2025Set ANNUAL1 markMCQ
   Q.Assertion (A): Points $A(2\hat i - \hat j + \hat k)$, $B(\hat i - 3\hat j - 5\hat k)$ and $C(3\hat i - 4\hat j - 4\hat k)$ are the vertices of a right angled triangle. Reason (R): In triangle ABC, $|\overrightarrow{AB}|^2 = |\overrightarrow{BC}|^2 + |\overrightarrow{AC}|^2$. Choose the correct option:

   (a) Both A and R are correct and R is the correct explanation of A.

   (b) Both A and R are correct but R is not the correct explanation of A.

   (c) A is correct but R is incorrect.

   (d) Both A and R are incorrect.

   ›Reveal solutionSolution

   Compute the three squared side-lengths using position vectors and check the Pythagoras relation.

   $A(2,-1,1)$, $B(1,-3,-5)$, $C(3,-4,-4)$.

   $\overrightarrow{AB}=B-A=(-1,-2,-6)$, so $|\overrightarrow{AB}|^2 = 1+4+36 = 41$.

   $\overrightarrow{BC}=C-B=(2,-1,1)$, so $|\overrightarrow{BC}|^2 = 4+1+1 = 6$.

   $\overrightarrow{AC}=C-A=(1,-3,-5)$, so $|\overrightarrow{AC}|^2 = 1+9+25 = 35$.

   Check: $|\overrightarrow{BC}|^2+|\overrightarrow{AC}|^2 = 6+35 = 41 = |\overrightarrow{AB}|^2$.

   So the relation holds — by the converse of Pythagoras' theorem, triangle ABC is right-angled at $C$. Hence A is true, R is true, and R (the Pythagoras relation) is precisely why A holds.

   ✓Final answer

   (i) Both A and R are correct and R is the correct explanation of A.

9. CBSE 2025Set E1 markMCQ
   Q.$|\vec{i} - \vec{j} - 3\vec{k}| =$

   (a) $11$

   (b) $\sqrt{11}$

   (c) $\sqrt{7}$

   (d) $\sqrt{10}$

   ›Reveal solutionSolution

   The magnitude of $\vec{i}-\vec{j}-3\vec{k}$ is $\sqrt{11}$.

   For a vector $a\vec{i}+b\vec{j}+c\vec{k}$ the magnitude is $\sqrt{a^2+b^2+c^2}$.

   Here $a=1,\ b=-1,\ c=-3$, so

   $|\vec{i}-\vec{j}-3\vec{k}|=\sqrt{1^2+(-1)^2+(-3)^2}=\sqrt{1+1+9}=\sqrt{11}.$

   ✓Final answer

   (B) $\sqrt{11}$.

10. CBSE 2025Set E1 markMCQ
    Q.$(4\vec{i} + 3\vec{j})^2 =$

    (a) $7$

    (b) $19$

    (c) $25$

    (d) $49$

    ›Reveal solutionSolution

    The square of a vector means its dot product with itself, i.e. $|\vec{a}|^2$, which is $25$.

    For any vector $\vec{a}$, $\vec{a}^2=\vec{a}\cdot\vec{a}=|\vec{a}|^2$.

    Here $\vec{a}=4\vec{i}+3\vec{j}$, so

    $\vec{a}^2=4^2+3^2=16+9=25.$

    ✓Final answer

    (C) $25$.

11. CBSE 2024Set 65/3/11 markMCQ
    Q.If $|\vec{a}| = 2$ and $-3 \le k \le 2$, then $|k\vec{a}| \in$: (A) $[-6,\ 4]$ (B) $[0,\ 4]$ (C) $[4,\ 6]$ (D) $[0,\ 6]$

    ›Reveal solutionSolution

    The magnitude of a scalar multiple is the absolute value of the scalar times the magnitude of the vector; since $|k|$ ranges from $0$ to $3$ when $-3 \le k \le 2$, we have $|k\vec{a}| \in [0,\,6]$.

    The key concept here is how scalar multiplication affects vector magnitude. When you multiply a vector by a scalar, the magnitude scales by the absolute value of that scalar—the direction may flip (if the scalar is negative), but magnitude is always non-negative.

    The fundamental property is:

    $|k\vec{a}| = |k| \cdot |\vec{a}|$

    This tells us that to find the range of $|k\vec{a}|$, we need to find the range of $|k|$ and multiply by the fixed magnitude $|\vec{a}| = 2$.

    Now let's trace through the reasoning:

    1. Identify the range of the scalar $k$.

       We're given $-3 \le k \le 2$.

    2. Find the range of $|k|$.

       The absolute value function $|k|$ measures distance from zero. On the interval $[-3,\,2]$:

       • At $k = 0$, we have $|k| = 0$ (the minimum).
       • At $k = -3$, we have $|k| = 3$.
       • At $k = 2$, we have $|k| = 2$.

       The maximum value of $|k|$ occurs at the endpoint farthest from zero, which is $k = -3$, giving $|k| = 3$.

       Therefore, $|k| \in [0,\,3]$.

    3. Compute the range of $|k\vec{a}|$.

       Using the scaling property:

    $|k\vec{a}| = |k| \cdot |\vec{a}| = |k| \cdot 2 = 2|k|$

    Since $|k| \in [0,\,3]$, multiplying through by $2$ gives:

    $2|k| \in [0,\,6]$

    Watch out

    A common mistake is to think $|k\vec{a}|$ ranges from $|-3| \cdot 2 = 6$ down to $|2| \cdot 2 = 4$, forgetting that $k$ can be zero. The magnitude $|k\vec{a}|$ achieves its minimum when $k=0$, not at the endpoints of the $k$-interval.

    ✓Final answer

    The correct option is (D) $[0,\,6]$.

12. CBSE 2024Set D1 markMCQ
    Q.$|\vec{i} - \vec{j} - \vec{k}| =$

    (a) $\sqrt{3}$

    (b) $3$

    (c) $\sqrt{2}$

    (d) $2$

    ›Reveal solutionSolution

    Magnitude = square root of sum of squares of components.

    $|\vec{i} - \vec{j} - \vec{k}| = \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{3}.$

    ✓Final answer

    (A) $\sqrt{3}$