Use a⃗·b⃗ = |a||b|cos θ to find the angle between two vectors.

Dot Product of Two Vectors — Find Angle Between Them

Question

Two vectors a⃗ and b⃗ are given as:

\vec{a} = 2\hat{i} + 2\hat{j} - \hat{k}, \quad \vec{b} = 6\hat{i} - 3\hat{j} + 2\hat{k}

Find the angle θ between them.

Solution — Step by Step

Multiply corresponding components and add them up:

\vec{a} \cdot \vec{b} = (2)(6) + (2)(-3) + (-1)(2) = 12 - 6 - 2 = 4

We square each component, add, and take the square root. Why? Because magnitude is the 3D version of the Pythagorean theorem.

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

Same process for the second vector:

|\vec{b}| = \sqrt{6^2 + (-3)^2 + 2^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7

The formula connects dot product to angle. Rearrange it to isolate cos θ:

\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{4}{3 \times 7} = \frac{4}{21}

Take the inverse cosine to get the angle:

\theta = \cos^{-1}\!\left(\frac{4}{21}\right)

That’s the answer — leave it in this form unless the exam asks for a decimal approximation.

Answer: $\theta = \cos^{-1}\!\left(\dfrac{4}{21}\right)$

Why This Works

The dot product formula $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$ comes directly from the geometric definition of the dot product. When two vectors point in the same direction, the angle is 0° and cos θ = 1, so the dot product equals the product of magnitudes — maximum value. When they’re perpendicular, cos 90° = 0, so the dot product is zero.

This formula is reversible: if you know the components, you can compute the dot product algebraically (Step 1). If you know the magnitudes, you can compute it geometrically. Equating both sides gives you cos θ — and then the angle.

In board and JEE problems, the magnitudes often come out to clean numbers (like 3 and 7 here). That’s your signal that the question is designed to be solved this way — always check if the magnitudes simplify before reaching for a calculator.

Alternative Method — Using the Vector Components Directly

This method is faster if the question already gives you unit vectors or simple components.

We can verify using the formula for angle in terms of direction cosines. The direction cosines of a⃗ are:

l_a = \frac{2}{3},\quad m_a = \frac{2}{3},\quad n_a = \frac{-1}{3}

And for b⃗:

l_b = \frac{6}{7},\quad m_b = \frac{-3}{7},\quad n_b = \frac{2}{7}

Then:

\cos\theta = l_a l_b + m_a m_b + n_a n_b = \frac{12}{21} + \frac{-6}{21} + \frac{-2}{21} = \frac{4}{21}

Same result. Both methods are valid — use whichever the question setup makes easier.

Common Mistake

The most common error is forgetting the negative sign on a component when computing the dot product. In this problem, (-1)(2) = -2, not +2. Students in a hurry write 12 − 6 + 2 = 8, then get cos θ = 8/21 — wrong answer, wrong angle. Slow down on the signs.

A related mistake: computing |a⃗| as 2 + 2 + 1 = 5 instead of √(4 + 4 + 1) = 3. Magnitudes are always square roots of the sum of squares, never just the sum of components.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using the Vector Components Directly

Common Mistake

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