Q.If $|\vec{a}|=10$, $|\vec{b}|=2$ and $\vec{a}\cdot\vec{b}=12$, then value of $|\vec{a}\times\vec{b}|$ is
(A) $5$
(B) $10$
(C) $14$
(D) $16$

We use the identity linking dot and cross product magnitudes:

$|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$.

Substitute the given values:

$|\vec{a}|^2 = 100$, $|\vec{b}|^2 = 4$, and $(\vec{a} \cdot \vec{b})^2 = 144$.

So $|\vec{a} \times \vec{b}|^2 = (100)(4) - 144 = 400 - 144 = 256$. …

The key here is that the dot product and cross product magnitudes are not independent — they come from the same angle between the vectors. If you know the lengths of two vectors and their dot product, you already know everything about the angle, including its sine. And the magnitude of the cross product is simply $|\vec{a}||\vec{b}|\sin\theta$.

So instead of finding $\theta$ explicitly (which would involve an inverse cosine and then a sine), we use a direct algebraic relationship that avoids angles altogether.

1. Recall the fundamental identity For any two vectors $\vec{a}$ and $\vec{b}$, the dot product is

and the magnitude of the cross product is

where $\theta$ is the angle between them.

2. Square both and add Notice that $\sin^2\theta + \cos^2\theta = 1$. So if we square both expressions and add, we get:

This gives the clean formula:

3. Plug in the given numbers $|\vec{a}| = 10$, $|\vec{b}| = 2$, and $\vec{a}\cdot\vec{b} = 12$. So:

4. Take the square root …

Method: Linking dot and cross magnitudes with the Lagrange identity

Use this whenever you are given $|\vec{a}|$, $|\vec{b}|$ and $\vec{a}\cdot\vec{b}$ and asked for $|\vec{a}\times\vec{b}|$ (or vice versa) — without knowing the angle.

Steps

Step 1: Recognise both come from the same angle

$\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos\theta$ and $|\vec{a}\times\vec{b}|=|\vec{a}||\vec{b}|\sin\theta$ share one $\theta$, so they are not independent.

Step 2: Square, add, and use $\sin^2\theta+\cos^2\theta=1$

This eliminates the angle entirely and gives the identity:

$|\vec{a}\times\vec{b}|^2=|\vec{a}|^2|\vec{b}|^2-(\vec{a}\cdot\vec{b})^2$ …

Common Mistakes

Mistake 1: Using the dot-product value directly as the cross-product magnitude

Why it's wrong: the cross product uses $\sin\theta$, not $\cos\theta$; reporting $12$ ignores the identity entirely. Correct approach: apply $|\vec a\times\vec b|^2=|\vec a|^2|\vec b|^2-(\vec a\cdot\vec b)^2=400-144=256$, so $|\vec a\times\vec b|=16$.

Mistake 2: Forgetting to square before subtracting …