Q.For any vector $\vec{a}$, prove that $|\vec{a}\times\hat{i}|^2+|\vec{a}\times\hat{j}|^2+|\vec{a}\times\hat{k}|^2=2|\vec{a}|^2$. OR For any vector $\vec{a}$, prove that $\vec{a}=(\vec{a}.\hat{i})\hat{i}+(\vec{a}.\hat{j})\hat{j}+(\vec{a}.\hat{k})\hat{k}$.

Showing the 12 most recent of 29 on this concept.

1. 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: …

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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. …

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3. 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).

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

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

   …

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

   …

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

   …

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7. 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 …

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8. 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 $\vec{a}$, $\vec{a}^2=\vec{a}\cdot\vec{a}=|\vec{a}|^2$.

   …

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9. 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, $|k| \in [0,\,3]$. …

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10. 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.

    …

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11. 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 $\vec{b}$:

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

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12. CBSE 2023Set E1 markMCQ
    Q.$|\vec{i}-2\vec{j}+2\vec{k}| =$

    (a) $3$

    (b) $6$

    (c) $7$

    (d) $5$

    ›Reveal solutionSolution

    The magnitude is $\sqrt{1^2+(-2)^2+2^2}=3$.

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

    Here $a=1,\ b=-2,\ c=2$, so …

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