Q15

1 markMCQSection A

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

Q: If $|\vec{a}| = 8$, $|\vec{b}| = 3$ and $|\vec{a} \times \vec{b}| = 12$, then the value of $|\vec{a} \cdot \vec{b}|$ is

**Correct option: (C) $12\sqrt{3}$** From $|a \times b| = |a||b|\sin\theta$: 12 = $8 \times 3 \times \sin\theta = 24\sin\theta$, so $\sin\theta = \frac{1}{2}$, giving $\cos\theta = \pm\frac{\sqrt{3}}{2}$. Then $|a \cdot b| = |a||b||\cos\theta| = 8 \times 3 \times \frac{\sqrt{3}}{2}$ = $12\sqrt{3}$.

Vector Algebra

Relation between Dot Product and Cross Product Magnitudes

Options

(A)

6\sqrt{3}

(B)

8\sqrt{3}

(C)

12\sqrt{3}

(D)

3\sqrt{12}

Official Answer

Correct option: (C) $12\sqrt{3}$

From $|a \times b| = |a||b|\sin\theta$ : 12 = $8 \times 3 \times \sin\theta = 24\sin\theta$ , so $\sin\theta = \frac{1}{2}$ , giving $\cos\theta = \pm\frac{\sqrt{3}}{2}$ . Then $|a \cdot b| = |a||b||\cos\theta| = 8 \times 3 \times \frac{\sqrt{3}}{2}$ = $12\sqrt{3}$ .

dot productcross productangle between vectorssine cosine identityvector magnitudevector algebra class 12

Marking Scheme

11 mark: awarded only for selecting option (C) $12\sqrt{3}$ ; no partial credit.

Hint

Use $|a \times b| = |a||b|\sin\theta$ to find sin(theta), then get cos(theta) from sin-squared plus cos-squared equals 1, and substitute into $|a \cdot b| = |a||b|\cos\theta$ .

Quick Oral Answer

From $|a \times b|=|a||b|\sin\theta$ , $\sin\theta=\frac{1}{2}$ , so $\cos\theta=\frac{\sqrt{3}}{2}$ ; then $|a \cdot b|=|a||b|\cos\theta=24\times\frac{\sqrt{3}}{2}=12\sqrt{3}$ .

Analysis & Explanation

Concept: $|a \times b| = |a||b|\sin\theta$ and $a \cdot b = |a||b|\cos\theta$ , where theta is the angle between a and b.

Given $|a|=8$ , $|b|=3$ , $|a \times b|=12$ : using $|a \times b|=|a||b|\sin\theta$ , $12 = 24\sin\theta$ , so $\sin\theta = \frac{1}{2}$ , meaning theta = 30 degrees or 150 degrees. Using the Pythagorean identity, $\cos\theta = \pm \frac{\sqrt{3}}{2}$ in either case (positive for 30 degrees, negative for 150 degrees, but its magnitude is the same). So $|a \cdot b| = |a||b||\cos\theta| = 24 \times \frac{\sqrt{3}}{2} = 12\sqrt{3}$ .

Why the distractors fail:

(A) 6 sqrt(3) comes from using half of |a||b| (=12 instead of 24) in the final multiplication, an arithmetic slip.
(B) 8 sqrt(3) mistakenly uses only |a| (=8) times sqrt(3) instead of the full product |a||b|cos(theta).
(D) 3 sqrt(12) is a mathematically equal-looking but poorly simplified/incorrect combination (3 sqrt(12) = 6 sqrt(3), not 12 sqrt(3)), showing a common error of not simplifying surds correctly or mixing up which values multiply which.

Exam tip: Always find sin(theta) first from the cross product magnitude, get cos(theta) using sin-squared plus cos-squared equals 1, and only then substitute into the dot product formula — mixing up |a| and |b| values is the most common error here.

Common Mistakes

1Using only one of $|a|$ or $|b|$ instead of their full product $|a||b|$ when computing the final dot product magnitude.
2Forgetting that $\sin\theta = \frac{1}{2}$ gives two possible angles (30 degrees and 150 degrees), though $|\cos\theta|$ turns out the same in both cases.
3Simplification errors with surds, e.g., mixing up $3\sqrt{12}$ and $12\sqrt{3}$ , which are not equal.

Interesting Facts

Lagrange's identity, (a.b) squared plus |a x b| squared equals |a| squared times |b| squared, directly connects the dot and cross products and can be used as a shortcut here without separately finding theta.

This relation is the vector analogue of the trigonometric identity sin-squared plus cos-squared equals 1, extended to describe how much two vectors 'align' versus 'rotate' relative to each other.

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Frequently Asked Questions

Is there a shortcut avoiding finding theta explicitly?

Yes — Lagrange's identity (a.b) squared plus |a x b| squared equals |a| squared times |b| squared lets you directly compute $(a \cdot b)^2 = 576 - 144 = 432$ , so $|a \cdot b| = \sqrt{432} = 12\sqrt{3}$ , without separately computing sin(theta) or cos(theta).

Why are there two possible angles but only one final answer?

$\sin\theta=\frac{1}{2}$ gives theta=30 degrees or 150 degrees, and although cos(theta) differs in sign between these, the question asks for $|a \cdot b|$ (the magnitude), which uses $|\cos\theta|$ , making both cases give the same final numerical answer.