The value of $\theta\in\left(0,\dfrac\pi2\right)$ for which the vectors $\vec a=(\sin\theta)\hat i+(\cos\theta)\hat j$ and $\vec b=-\sqrt3\hat i+\hat j-2\hat k$ are perpendicular, is equal to (1) $\dfrac\pi3$ (2) $\dfrac\pi6$ (3) $\dfrac\pi4$ (4) $\dfrac\pi2$

The value of θ∈(0,dfracπ2) for which the vectors vec a=(sinθ)hat i+(cosθ)hat j and vec b=-√3hat i+hat j-2hat k

Exercise 8.5 · Q17

\theta\in\left(0,\dfrac\pi2\right)

(1)

\dfrac\pi3

(2)

\dfrac\pi6

(3)

\dfrac\pi4

(4)

\dfrac\pi2

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Concept understanding — Scalar (Dot) Product

For non-zero vectors $\vec a,\vec b$ with included angle $\theta$ ( $0\le\theta\le\pi$ ), the scalar (dot) product is the number $\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta.$

Geometric meaning (projection). $\vec a\cdot\vec b=|\vec a|\times(\text{projection of }\vec b\text{ on }\vec a)$ , and the projection of $\vec b$ on $\vec a$ is $\dfrac{\vec a\cdot\vec b}{|\vec a|}$ (symmetrically, projection of $\vec a$ on $\vec b$ is $\dfrac{\vec a\cdot\vec b}{|\vec b|}$ ).

Core properties.

Commutative: $\vec a\cdot\vec b=\vec b\cdot\vec a$ .
Sign follows the angle: positive for $0\le\theta<\pi/2$ , zero at $\theta=\pi/2$ , negative for $\pi/2<\theta\le\pi$ . In particular $\vec a\cdot\vec b=0 \iff \vec a=\vec 0$ or $\vec b=\vec 0$ or $\vec a\perp\vec b$ — for two non-zero vectors, $\vec a\cdot\vec b=0$ is exactly the perpendicularity test.
$\vec a\cdot\vec a=|\vec a|^2$ (often written $a^2$ ), so $|\vec a|=\sqrt{\vec a\cdot\vec a}$ .
$\hat i\cdot\hat i=\hat j\cdot\hat j=\hat k\cdot\hat k=1$ and $\hat i\cdot\hat j=\hat j\cdot\hat k=\hat k\cdot\hat i=0$ (they're mutually perpendicular unit vectors).
Distributive: $\vec a\cdot(\vec b+\vec c)=\vec a\cdot\vec b+\vec a\cdot\vec c$ , and likewise for subtraction and for the right factor. …

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Step 1. $\vec a=\sin\theta\,\hat i+\cos\theta\,\hat j=(\sin\theta,\cos\theta,0)$ , $\vec b=-\sqrt3\hat i+\hat j-2\hat k=(-\sqrt3,1,-2)$ .

Step 2. $\vec a\cdot\vec b=-\sqrt3\sin\theta+\cos\theta+0=0$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 15 on this concept.

CBSE 2026Set A1 markMCQ
$2\vec{j} \cdot (-3)\vec{k} =$
(a) $6$
(b) $-6$
(c) $0$
(d) $-6\vec{i}$
›Reveal solutionSolution

Perpendicular unit vectors have zero dot product.

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

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

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CBSE 2025Set E1 markMCQ
$(11\vec{i} + \vec{j} + \vec{k}) \cdot (\vec{i} + \vec{j} + 11\vec{k}) =$
(a) $22$
(b) $23$
(c) $24$
(d) $20$
›Reveal solutionSolution

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

Multiply corresponding components and add: …

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CBSE 2025Set E1 markMCQ
$(\vec{i} - 2\vec{j} + 5\vec{k}) \cdot (-2\vec{i} + 4\vec{j} + 2\vec{k}) =$
(a) $20$
(b) $18$
(c) $0$
(d) $4$
›Reveal solutionSolution

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

…

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CBSE 2025Set ANNUAL1 markMCQ
$|\vec{a}| = 3, |\vec{b}| = 4, |\vec{c}| = 5$
(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: $|\vec a|^2+|\vec b|^2+2\vec a\cdot\vec b=9+16+2\vec a\cdot\vec b=25+2\vec a\cdot\vec b$ . …

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CBSE 2024Set ANNUAL1 mark
$(3\vec{a} - 5\vec{b}) \cdot (2\vec{a} + 7\vec{b})$
›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 $\vec a\cdot\vec b=\vec b\cdot\vec a$ .

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

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

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

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. For vectors a⃗ and b⃗, |a⃗| = \sqrt3, |b⃗| = 2 and a⃗.b⃗ = \sqrt6, angle between a⃗ and b⃗ is$
(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 $\vec a \cdot \vec b = |\vec a||\vec b|\cos\theta$ : …

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CBSE 2023Set E1 markMCQ
$(\vec{j}-2\vec{i})\cdot(\vec{k}+3\vec{i}-\vec{j}) =$
(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 $(\vec i,\vec j,\vec k)$ compone …

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

…

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CBSE 2023Set E1 markMCQ
$(2\vec{i}-3\vec{j})\cdot(\vec{i}+\vec{k}) =$
(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 $=0$ .

…

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CBSE 2023Set E1 markMCQ
$\vec{k}\cdot(\vec{i}+\vec{j}) =$
(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 $0$ .

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

…

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CBSE 2023Set E1 markMCQ
$(\vec{i}-\vec{j}+\vec{k})\cdot(7\vec{i}-8\vec{j}+9\vec{k}) =$
(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 $a_1b_1+a_2b_2+a_3b_3$ .

…

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