Let the vectors $\vec{a}$ and $\vec{b}$ be such that $|\vec{a}|=3$ and $|\vec{b}|=\dfrac{\sqrt{2}}{3}$, then $\vec{a}\times\vec{b}$ is a unit vector, if the angle between $\vec{a}$ and $\vec{b}$ is (A) $\pi/6$ (B) $\pi/4$ (C) $\pi/3$ (D) $\pi/2$

Let the vectors veca and vecb be such that |veca|=3 and |vecb|=dfrac√(2)3, then veca×vecb is a unit vector, if

Exercise 10.4 · Q11

\vec{a}

Mizoram MbseTextbookSubjective· 1mImportance★★★★★

Appeared in past exams:KEAM 2025· Set eng-2025-0429· 4mexactGUJCET 2023· Set 09· 1mexactGUJCET 2022· Set 08· 1mexactGUJCET 2021· Set 15· 1mexact

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Concept understanding — Cross Product Area

Area from the Cross Product

The cross product $\vec{a}\times\vec{b}$ of two vectors in 3D is itself a vector, and the most useful thing about its magnitude is that it measures area.

Place the two vectors tail-to-tail. They span a parallelogram. The magnitude of their cross product is exactly the area of that parallelogram:

\text{Area of parallelogram} = |\vec{a}\times\vec{b}| = |\vec{a}|\,|\vec{b}|\,\sin\theta

where $\theta$ is the angle between them.

Why sine, not cosine

The area of a parallelogram is base $\times$ height. Take $|\vec{a}|$ as the base. The height is the part of $\vec{b}$ perpendicular to $\vec{a}$ , namely $|\vec{b}|\sin\theta$ . Multiplying gives $|\vec{a}||\vec{b}|\sin\theta$ — precisely $|\vec{a}\times\vec{b}|$ . The dot product uses $\cos\theta$ (overlap along); the cross product uses $\sin\theta$ (spread across), and "across" is what builds area.

Area of a triangle

A triangle with adjacent sides $\vec{a}$ and $\vec{b}$ is half that parallelogram:

\text{Area of triangle} = \tfrac{1}{2}\,|\vec{a}\times\vec{b}|

For a triangle with vertices $A, B, C$ , take $\vec{a}=\overrightarrow{AB}$ and $\vec{b}=\overrightarrow{AC}$ .

A quick example

For $\vec{a}=\hat{i}+2\hat{j}$ and $\vec{b}=3\hat{i}+\hat{j}$ ,

$\vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&2&0\\3&1&0\end{vmatrix} = (1\cdot 1 - 2\cdot 3)\hat{k} = -5\hat{k}.$ …

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$|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta = 1$ gives $\sqrt{2}\sin\theta = 1$ , so $\sin\theta = \tfrac{1}{\sqrt{2}}$ and $\theta = \tfrac{\pi}{4}$ — option (B).

The idea

The length of a cross product is $|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta$ , where $\theta$ is the angle between the vectors. Saying " $\vec{a}\times\vec{b}$ is a unit vector" simply pins that length to $1$ , turning the problem into a short trigonometric equation.

Step-by-step

1. Write the unit-vector condition.

|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta = 1.

2. Put in the given magnitudes. With $|\vec{a}|=3$ and $|\vec{b}|=\dfrac{\sqrt{2}}{3}$ , the $3$ 's cancel:

3\cdot\frac{\sqrt{2}}{3}\cdot\sin\theta = \sqrt{2}\,\sin\theta = 1.

3. Solve for $\sin\theta$ .

$\sin\theta = \frac{1}{\sqrt{2}}.$ …

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Q.Let the vectors $\vec{a}$ and $\vec{b}$ be such that $|\vec{a}|=3$ and $|\vec{b}|=\dfrac{\sqrt{2}}{3}$ , then $\vec{a}\times\vec{b}$ is a unit vector, if the angle between $\vec{a}$ and $\vec{b}$ is (A) $\pi/6$ (B) $\pi/4$ (C) $\pi/3$ (D) $\pi/2$

Concept understanding — Cross Product Area

Area from the Cross Product

Why sine, not cosine

Area of a triangle

A quick example

The idea

Step-by-step

Method: Finding the Angle from a Cross-Product Magnitude Condition

Steps

Common Mistakes

The idea

Step-by-step

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