Q.Show that vectors $\vec{a} = 2\hat{i}+3\hat{j}+6\hat{k}$, $\vec{b} = 3\hat{i}-6\hat{j}+2\hat{k}$ and $\vec{c} = 6\hat{i}+2\hat{j}-3\hat{k}$ are mutually perpendicular.

Showing the 12 most recent of 15 on this concept.

1. CBSE 2026Set A1 markMCQ
   Q.$2\vec{j} \cdot (-3)\vec{k} =$

   (a) $6$

   (b) $-6$

   (c) $0$

   (d) $-6\vec{i}$

   ›Reveal solutionSolution

   Perpendicular unit vectors have zero dot product.

   $2\vec{j}\cdot(-3)\vec{k} = (2)(-3)\,(\vec{j}\cdot\vec{k}) = -6\,(\vec{j}\cdot\vec{k}).$

   Since $\vec{j}$ and $\vec{k}$ are mutually perpendicular, $\vec{j}\cdot\vec{k} = 0$, so the result is $0$.

   ✓Final answer

   (c) $0$.

2. CBSE 2025Set E1 markMCQ
   Q.$(7\vec{i} - 8\vec{j} + 9\vec{k}) \cdot (\vec{i} - \vec{j} + \vec{k}) =$

   (a) $25$

   (b) $24$

   (c) $23$

   (d) $22$

   ›Reveal solutionSolution

   The scalar (dot) product of the two vectors is $24$.

   The dot product multiplies corresponding components and adds:

   $(7\vec{i}-8\vec{j}+9\vec{k})\cdot(\vec{i}-\vec{j}+\vec{k})=7(1)+(-8)(-1)+9(1).$

   $=7+8+9=24.$

   ✓Final answer

   (B) $24$.

3. CBSE 2025Set E1 markMCQ
   Q.$(11\vec{i} + \vec{j} + \vec{k}) \cdot (\vec{i} + \vec{j} + 11\vec{k}) =$

   (a) $22$

   (b) $23$

   (c) $24$

   (d) $20$

   ›Reveal solutionSolution

   The dot product of the two vectors is $23$.

   Multiply corresponding components and add:

   $(11\vec{i}+\vec{j}+\vec{k})\cdot(\vec{i}+\vec{j}+11\vec{k})=11(1)+1(1)+1(11)=11+1+11=23.$

   ✓Final answer

   (B) $23$.

4. CBSE 2025Set E1 markMCQ
   Q.$(\vec{i} - 2\vec{j} + 5\vec{k}) \cdot (-2\vec{i} + 4\vec{j} + 2\vec{k}) =$

   (a) $20$

   (b) $18$

   (c) $0$

   (d) $4$

   ›Reveal solutionSolution

   The dot product is $0$, so the two vectors are perpendicular.

   $(\vec{i}-2\vec{j}+5\vec{k})\cdot(-2\vec{i}+4\vec{j}+2\vec{k})=1(-2)+(-2)(4)+5(2).$

   $=-2-8+10=0.$

   ✓Final answer

   (C) $0$.

5. CBSE 2025Set ANNUAL1 markMCQ
   Q.If $|\vec{a}| = 3, |\vec{b}| = 4, |\vec{c}| = 5$ and $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ then the angle between $\vec{a}$ and $\vec{b}$ is:

   (a) $60^\circ$

   (b) $0$

   (c) $45^\circ$

   (d) $90^\circ$

   ›Reveal solutionSolution

   From $\vec a+\vec b+\vec c=\vec 0$, we get $\vec a+\vec b=-\vec c$; squaring both sides and using the given magnitudes shows $\vec a\cdot\vec b=0$.

   Since $\vec a+\vec b+\vec c=\vec 0$, we have $\vec a+\vec b=-\vec c$.

   Taking magnitudes squared: $|\vec a+\vec b|^2=|\vec c|^2=25$.

   Expanding the left side: $|\vec a|^2+|\vec b|^2+2\vec a\cdot\vec b=9+16+2\vec a\cdot\vec b=25+2\vec a\cdot\vec b$.

   Setting equal to 25: $25+2\vec a\cdot\vec b=25\Rightarrow \vec a\cdot\vec b=0$.

   Since neither vector is zero, $\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta=0$ forces $\cos\theta=0$, so $\theta=90^\circ$.

   ✓Final answer

   The correct option is (d) $90^\circ$.

6. CBSE 2024Set ANNUAL1 mark
   Q.Evaluate the product $(3\vec{a} - 5\vec{b}) \cdot (2\vec{a} + 7\vec{b})$.

   ›Reveal solutionSolution

   Expand the dot product like a binomial product, using $\vec a\cdot\vec a=|\vec a|^2$, $\vec b\cdot\vec b=|\vec b|^2$, and $\vec a\cdot\vec b=\vec b\cdot\vec a$.

   $(3\vec a - 5\vec b)\cdot(2\vec a + 7\vec b)$

   $= 3\vec a\cdot2\vec a + 3\vec a\cdot7\vec b - 5\vec b\cdot2\vec a - 5\vec b\cdot7\vec b$

   $= 6(\vec a\cdot\vec a) + 21(\vec a\cdot\vec b) - 10(\vec b\cdot\vec a) - 35(\vec b\cdot\vec b)$

   Since $\vec a\cdot\vec b = \vec b\cdot\vec a$:

   $= 6|\vec a|^2 + (21-10)(\vec a\cdot\vec b) - 35|\vec b|^2 = 6|\vec a|^2 + 11(\vec a\cdot\vec b) - 35|\vec b|^2$

   ✓Final answer

   $6|\vec a|^2 + 11(\vec a\cdot\vec b) - 35|\vec b|^2$

7. CBSE 2024Set ANNUAL1 markMCQ
   Q.For vectors a⃗ and b⃗, |a⃗| = √3, |b⃗| = 2 and a⃗.b⃗ = √6, angle between a⃗ and b⃗ is

   (a) π/2

   (b) π/6

   (c) π/3

   (d) π/4

   ›Reveal solutionSolution

   Use $\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta$ and solve for $\theta$.

   Given $|\vec a| = \sqrt3$, $|\vec b| = 2$, $\vec a\cdot\vec b = \sqrt6$. Using $\vec a \cdot \vec b = |\vec a||\vec b|\cos\theta$:

   $\cos\theta = \frac{\sqrt6}{\sqrt3\cdot 2} = \frac{\sqrt6}{2\sqrt3} = \frac{\sqrt2}{2} = \frac{1}{\sqrt2}$

   So $\theta = \pi/4$.

   ✓Final answer

   $\pi/4$ — option (d).

8. CBSE 2023Set E1 markMCQ
   Q.$(\vec{j}-2\vec{i})\cdot(\vec{k}+3\vec{i}-\vec{j}) =$

   (a) $0$

   (b) $-6$

   (c) $-7$

   (d) $8$

   ›Reveal solutionSolution

   $(\vec{j}-2\vec{i})\cdot(\vec{k}+3\vec{i}-\vec{j})=-7$.

   Write first vector as $(-2,1,0)$ and second as $(3,-1,1)$ in $(\vec i,\vec j,\vec k)$ components:

   $(-2)(3)+(1)(-1)+(0)(1)=-6-1+0=-7.$

   ✓Final answer

   (C) $-7$.

9. CBSE 2023Set E1 markMCQ
   Q.$3\vec{k}\cdot(13\vec{i}-7\vec{k}) =$

   (a) $39$

   (b) $0$

   (c) $-21$

   (d) $18$

   ›Reveal solutionSolution

   Use $\vec{k}\cdot\vec{i}=0$, $\vec{k}\cdot\vec{k}=1$; the expression reduces to $3\times(-7)=-21$.

   Compute $3\vec{k}\cdot(13\vec{i}-7\vec{k})=3\big(13(\vec{k}\cdot\vec{i})-7(\vec{k}\cdot\vec{k})\big)$.

   The unit vectors are mutually orthogonal, so $\vec{k}\cdot\vec{i}=0$ and $\vec{k}\cdot\vec{k}=1$.

   Thus $3\big(13(0)-7(1)\big)=3(-7)=-21$.

   ✓Final answer

   (c) $-21$.

10. CBSE 2023Set E1 markMCQ
    Q.$(2\vec{i}-3\vec{j})\cdot(\vec{i}+\vec{k}) =$

    (a) $2$

    (b) $-1$

    (c) $3$

    (d) $0$

    ›Reveal solutionSolution

    Only the $\vec{i}\cdot\vec{i}$ term survives, giving $2\times1=2$.

    Compute $(2\vec{i}-3\vec{j})\cdot(\vec{i}+\vec{k})$ using $\vec{i}\cdot\vec{i}=1$ and all other unlike dot products $=0$.

    $=2(\vec{i}\cdot\vec{i})+2(\vec{i}\cdot\vec{k})-3(\vec{j}\cdot\vec{i})-3(\vec{j}\cdot\vec{k})=2(1)+0-0-0=2$.

    ✓Final answer

    (a) $2$.

11. CBSE 2023Set E1 markMCQ
    Q.$\vec{k}\cdot(\vec{i}+\vec{j}) =$

    (a) $0$

    (b) $1$

    (c) $2$

    (d) $-1$

    ›Reveal solutionSolution

    $\vec{k}$ is orthogonal to both $\vec{i}$ and $\vec{j}$, so the dot product is $0$.

    $\vec{k}\cdot(\vec{i}+\vec{j})=\vec{k}\cdot\vec{i}+\vec{k}\cdot\vec{j}$.

    The standard unit vectors are mutually perpendicular, so $\vec{k}\cdot\vec{i}=0$ and $\vec{k}\cdot\vec{j}=0$.

    Hence the sum is $0+0=0$.

    ✓Final answer

    (a) $0$.

12. CBSE 2023Set E1 markMCQ
    Q.$(\vec{i}-\vec{j}+\vec{k})\cdot(7\vec{i}-8\vec{j}+9\vec{k}) =$

    (a) $22$

    (b) $23$

    (c) $24$

    (d) $25$

    ›Reveal solutionSolution

    The dot product multiplies corresponding components and adds: $7+8+9=24$.

    The dot product of $a_1\vec{i}+a_2\vec{j}+a_3\vec{k}$ and $b_1\vec{i}+b_2\vec{j}+b_3\vec{k}$ is $a_1b_1+a_2b_2+a_3b_3$.

    $(\vec{i}-\vec{j}+\vec{k})\cdot(7\vec{i}-8\vec{j}+9\vec{k})=(1)(7)+(-1)(-8)+(1)(9)=7+8+9=24.$

    ✓Final answer

    (c) $24$.