The angle between two vectors $\vec{a}$ and $\vec{b}$ with magnitudes $\sqrt{3}$ and 4, respectively, and $\vec{a}\cdot\vec{b}=2\sqrt{3}$ is (A) $\dfrac{\pi}{6}$ (B) $\dfrac{\pi}{3}$ (C) $\dfrac{\pi}{2}$ (D) $\dfrac{5\pi}{2}$

The angle between two vectors veca and vecb with magnitudes √(3) and 4, respectively, and veca·vecb=2√(3) is (

NCERT Exemplar · Q33

\vec{a}

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Concept understanding — Dot Product Angle

Finding the Angle Between Vectors

Suppose you have two arrows drawn from the same point. One question is unavoidable in geometry, physics, and mechanics: what is the angle between them? You could measure it with a protractor on paper, but that fails the moment the vectors live in 3D. The dot product gives you the angle by pure calculation.

The Core Idea

The scalar (dot) product of two vectors has two faces that describe the same number:

\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \qquad \text{(components)}

\vec{a} \cdot \vec{b} = |\vec{a}|\,|\vec{b}|\cos\theta \qquad \text{(geometry)}

The first is easy to compute from coordinates; the second hides the angle $\theta$ (with $0 \le \theta \le \pi$ ) between the vectors. Setting them equal and solving for $\cos\theta$ gives the master formula.

\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|\,|\vec{b}|}, \qquad \theta = \cos^{-1}\!\left(\frac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}\right)

Why It Works

Both vectors have a fixed length, so the only thing the dot product can "vary" with is how aligned they are. When they point the same way, $\cos\theta = 1$ and the dot product is as large as possible, $|\vec{a}||\vec{b}|$ . When they are perpendicular, $\cos\theta = 0$ and the dot product vanishes. When they point opposite ways, $\cos\theta = -1$ . Dividing $\vec{a}\cdot\vec{b}$ by the two lengths simply strips away the size information and leaves behind a pure measure of alignment — exactly $\cos\theta$ .

Note

The sign of the dot product tells you the type of angle at a glance: positive $\Rightarrow$ acute, zero $\Rightarrow$ right angle, negative $\Rightarrow$ obtuse.

Using the Formula

For $\vec{a} = \hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{b} = \hat{i} + 0\hat{j} + 0\hat{k}$ :

\vec{a}\cdot\vec{b} = 1, \quad |\vec{a}| = 3, \quad |\vec{b}| = 1

$\cos\theta = \frac{1}{3\cdot 1} = \frac{1}{3} \;\Rightarrow\; \theta = \cos^{-1}\tfrac{1}{3} \approx 70.5^\circ$ …

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The angle between two vectors is found using the dot product formula $\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos\theta$ . Substituting the given magnitudes and dot product gives $\cos\theta = \frac{1}{2}$ , so $\theta = \frac{\pi}{3}$ . The correct option is (B).

The dot product of two vectors isn't just a number — it encodes how much one vector "points along" the other. The formula $\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos\theta$ is the bridge between the algebraic product and the geometric angle $\theta$ . When you know the magnitudes and the dot product, you can solve directly for $\cos\theta$ , and then identify the angle from standard trigonometric values.

Here’s the step-by-step:

Write the dot product formula For any two vectors $\vec{a}$ and $\vec{b}$ , the dot product is

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

where $\theta$ is the angle between them, measured from $0$ to $\pi$ .

Plug in the given values We have $|\vec{a}| = \sqrt{3}$ , $|\vec{b}| = 4$ , and $\vec{a}\cdot\vec{b} = 2\sqrt{3}$ . Substituting:

2\sqrt{3} = (\sqrt{3})(4)\cos\theta

Simplify the right-hand side

2\sqrt{3} = 4\sqrt{3}\,\cos\theta

Solve for $\cos\theta$ Divide both sides by $4\sqrt{3}$ (since $\sqrt{3} \neq 0$ ):

\cos\theta = \frac{2\sqrt{3}}{4\sqrt{3}} = \frac{2}{4} = \frac{1}{2}

Identify the angle …

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Q.The angle between two vectors $\vec{a}$ and $\vec{b}$ with magnitudes $\sqrt{3}$ and 4, respectively, and $\vec{a}\cdot\vec{b}=2\sqrt{3}$ is
(A) $\dfrac{\pi}{6}$
(B) $\dfrac{\pi}{3}$
(C) $\dfrac{\pi}{2}$
(D) $\dfrac{5\pi}{2}$

Concept understanding — Dot Product Angle

Finding the Angle Between Vectors

The Core Idea

Why It Works

Using the Formula

Method: Angle Between Two Vectors from the Dot Product

Steps

Common Mistakes

Q.The angle between two vectors a⃗\vec{a}a and b⃗\vec{b}b with magnitudes 3\sqrt{3}3​ and 4, respectively, and a⃗⋅b⃗=23\vec{a}\cdot\vec{b}=2\sqrt{3}a⋅b=23​ is (A) π6\dfrac{\pi}{6}6π​ (B) π3\dfrac{\pi}{3}3π​ (C) π2\dfrac{\pi}{2}2π​ (D) 5π2\dfrac{5\pi}{2}25π​

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Q.The angle between two vectors $\vec{a}$ and $\vec{b}$ with magnitudes $\sqrt{3}$ and 4, respectively, and $\vec{a}\cdot\vec{b}=2\sqrt{3}$ is
(A) $\dfrac{\pi}{6}$
(B) $\dfrac{\pi}{3}$
(C) $\dfrac{\pi}{2}$
(D) $\dfrac{5\pi}{2}$

Unlock everything free for 14 days