CBSE Board Examination 2026 · Section A · 1 mark

Q15: $|\vec a\cdot\vec b|$ from $|\vec a|$ ⁠, $|\vec b|$ and $|\vec a\times\vec b|$

The question

If $|\vec a| = 8$ ⁠, $|\vec b| = 3$ and $|\vec a \times \vec b| = 12$ ⁠, then the value of $|\vec a \cdot \vec b|$

(A) $6\sqrt{3}$

(B) $8\sqrt{3}$

(D) $3\sqrt{12}$

Exam answer

Using the identity

\begin{aligned} &(\vec a \cdot \vec b)^2 + |\vec a \times \vec b|^2 \\ &= |\vec a|^2|\vec b|^2, \end{aligned}

Why?

The identity comes from the angle $\theta$ between the vectors:

\begin{aligned} \vec a \cdot \vec b &= |\vec a||\vec b|\cos\theta \\ |\vec a \times \vec b| &= |\vec a||\vec b|\sin\theta \end{aligned}

Square both and add. Since $\cos^2\theta + \sin^2\theta = 1$ ⁠, the right sides add up to $|\vec a|^2|\vec b|^2$ ⁠.

\begin{aligned} &(\vec a \cdot \vec b)^2 + |\vec a \times \vec b|^2 \\ &= |\vec a|^2|\vec b|^2\cos^2\theta \\ &\quad + |\vec a|^2|\vec b|^2\sin^2\theta \\ &= |\vec a|^2|\vec b|^2(\cos^2\theta + \sin^2\theta) \\ &= |\vec a|^2|\vec b|^2(1) \end{aligned}

\begin{aligned} (\vec a \cdot \vec b)^2 &= 64 \times 9 - 144 \\ &= 576 - 144 = 432 \end{aligned}

Why?

$|\vec a|^2 = 8^2 = 64$ ⁠, $|\vec b|^2 = 3^2 = 9$ and $|\vec a \times \vec b|^2 = 12^2 = 144$ ⁠. The 144 moves to the other side.

\begin{aligned} (\vec a \cdot \vec b)^2 + 144 &= 64 \times 9 \\ (\vec a \cdot \vec b)^2 + 144 &= 576 \\ (\vec a \cdot \vec b)^2 &= 576 - 144 \\ &= 432 \end{aligned}

$\therefore\ |\vec a \cdot \vec b| = \sqrt{432} = 12\sqrt{3}$

Why?

$432 = 144 \times 3$ and $\sqrt{144} = 12$ ⁠, so $\sqrt{432} = 12\sqrt{3}$ ⁠.

\begin{aligned} \sqrt{432} &= \sqrt{144 \times 3} \\ &= \sqrt{144} \times \sqrt{3} \\ &= 12\sqrt{3} \end{aligned}

Check with the angle:

\begin{aligned} \sin\theta &= \frac{12}{8 \times 3} = \frac{12}{24} = \frac{1}{2} \\ \cos^2\theta &= 1 - \left(\tfrac{1}{2}\right)^2 = \frac{3}{4} \\ |\cos\theta| &= \frac{\sqrt{3}}{2} \\ |\vec a \cdot \vec b| &= 24 \cdot \frac{\sqrt{3}}{2} = 12\sqrt{3} \end{aligned}

Watch out: $3\sqrt{12} = 6\sqrt{3}$ ⁠, the same as (A), so neither is right.

Answer: (C) $12\sqrt{3}$

Where this comes from

Chapter 10, Vector Algebra: Scalar and vector products.

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Vector Algebra in past papers: question 3 of 48 · see them all

Q15: ∣a⃗⋅b⃗∣|\vec a\cdot\vec b|∣a⋅b∣ from ∣a⃗∣|\vec a|∣a∣⁠, ∣b⃗∣|\vec b|∣b∣ and ∣a⃗×b⃗∣|\vec a\times\vec b|∣a×b∣