Q.Find the sine of the angle between the vectors $\vec{a}=3\hat{i}+\hat{j}+2\hat{k}$ and $\vec{b}=2\hat{i}-2\hat{j}+4\hat{k}$.

Concept: Dot Product Angle — we use the dot product to find $\cos\theta$, then $\sin\theta$ from the identity $\sin^2\theta = 1 - \cos^2\theta$.

Step 1: Compute dot product and magnitudes.

$\vec{a} \cdot \vec{b} = (3)(2) + (1)(-2) + (2)(4) = 6 - 2 + 8 = 12$

$|\vec{a}| = \sqrt{3^2 + 1^2 + 2^2} = \sqrt{14}$

$|\vec{b}| = \sqrt{2^2 + (-2)^2 + 4^2} = \sqrt{24} = 2\sqrt{6}$

Step 2: Find $\cos\theta$.

$\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{12}{\sqrt{14} \cdot 2\sqrt{6}} = \frac{12}{2\sqrt{84}} = \frac{6}{\sqrt{84}} = \frac{6}{2\sqrt{21}} = \frac{3}{\sqrt{21}}$ …

The most direct way to find the sine of the angle between two vectors is through the cross product. The magnitude of the cross product is $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta$, where $\theta$ is the angle between them. So if we can compute the cross product magnitude and the individual magnitudes, we can isolate $\sin\theta$ without ever needing to find $\theta$ itself.

This is often more convenient than using the dot product to find $\cos\theta$ and then converting to sine, because the cross product gives us $\sin\theta$ directly — no sign ambiguity for acute vs obtuse angles (since we take the magnitude).

Let’s work through it step by step.

1. Write the vectors in component form

   $\vec{a} = 3\hat{i} + 1\hat{j} + 2\hat{k}$

   $\vec{b} = 2\hat{i} - 2\hat{j} + 4\hat{k}$

2. Compute the cross product $\vec{a} \times \vec{b}$

   Use the determinant method:

Expand:

3. Find the magnitude of the cross product

4. Find the magnitudes of $\vec{a}$ and $\vec{b}$

5. Use the cross product relation to find $\sin\theta$ From $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta$, we get:

Simplify the denominator: $\sqrt{14} \cdot \sqrt{6} = \sqrt{84} = 2\sqrt{21}$, so: …

Method: Sine of the angle between vectors via the cross product

Use this whenever the sine of the angle between two vectors is wanted.

Steps

Step 1: Choose the cross-product route.

Since $|\vec{a}\times\vec{b}|=|\vec{a}|\,|\vec{b}|\sin\theta$, the cross product gives $\sin\theta$ directly:

(The alternative is $\cos\theta$ from the dot product then $\sin\theta=\sqrt{1-\cos^2\theta}$ — both are valid.)

Step 2: Compute the cross product and all magnitudes. …

Common Mistakes

Mistake 1: Using the dot product to find the sine of the angle.

Why it's wrong: the dot product yields $\cos\theta$, not $\sin\theta$; sine comes from the cross-product magnitude, $\sin\theta=\dfrac{|\vec{a}\times\vec{b}|}{|\vec{a}||\vec{b}|}$. Correct approach: compute $\vec{a}\times\vec{b}$, or find $\cos\theta$ first and use $\sin\theta=\sqrt{1-\cos^2\theta}$.

Mistake 2: Forgetting to take the magnitude of the cross product. …