Q.Find $|\vec{a}\times\vec{b}|$, if $\vec{a}=2\hat{i}+\hat{j}+3\hat{k}$ and $\vec{b}=3\hat{i}+5\hat{j}-2\hat{k}$.

Key idea: compute $\vec{a}\times\vec{b}$ with the determinant, then take its length.

Step 1 — Determinant.

Step 2 — Expand.

Step 3 — Magnitude.

To find $|\vec{a}\times\vec{b}|$ we could use $|\vec{a}||\vec{b}|\sin\theta$, but we are not given the angle $\theta$. It is far quicker to compute the cross-product vector directly from the components and then measure its length.

1. Set up the determinant

With $\vec{a} = 2\hat{i} + \hat{j} + 3\hat{k}$ and $\vec{b} = 3\hat{i} + 5\hat{j} - 2\hat{k}$:

2. Expand along the top row

Remember the middle term carries a minus sign:

Evaluate each bracket:

• $\hat{i}$: $-2 - 15 = -17$
• $\hat{j}$: $-(-4 - 9) = -(-13) = 13$
• $\hat{k}$: $10 - 3 = 7$

So

3. Take the magnitude

Since $507 = 3 \times 169 = 3 \times 13^2$,

4. Quick sanity check

The cross product should be perpendicular to both $\vec{a}$ and $\vec{b}$. Indeed $(-17)(2) + 13(1) + 7(3) = -34 + 13 + 21 = 0$ and $(-17)(3) + 13(5) + 7(-2) = -51 + 65 - 14 = 0$. Both check out.

Method: Magnitude of a cross product from components

When the angle between the vectors is not given, do not use $|\vec{a}||\vec{b}|\sin\theta$ — instead compute the cross-product vector from components with a determinant, then take its length.

Steps

Step 1: Set up the determinant

Unit vectors on the top row, $\vec{a}$'s components on the second, $\vec{b}$'s on the third:

Step 2: Expand along the top row — mind the middle sign

The $\hat{j}$ term carries a minus sign; this is the single most common slip.

Step 3: Take the magnitude

Step 4 (quick check): confirm perpendicularity

The result should satisfy $(\vec{a}\times\vec{b})\cdot\vec{a} = 0$ and $(\vec{a}\times\vec{b})\cdot\vec{b} = 0$; a fast dot product catches an arithmetic error before you commit to the magnitude.

Common Mistakes

Mistake 1: Forgetting the minus sign on the $\hat{j}$ component.

Why it's wrong: the cofactor expansion alternates signs $+,-,+$, so the middle term is $-\hat{j}(a_1b_3-a_3b_1)$; keeping it positive gives the wrong vector (and usually the wrong magnitude). Correct approach: always write the $\hat{j}$ term with its leading minus, then simplify.

Mistake 2: Using $|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|$.

Why it's wrong: the magnitude is $|\vec{a}||\vec{b}|\sin\theta$, which equals $|\vec{a}||\vec{b}|$ only if the vectors are perpendicular. Correct approach: compute the cross-product vector's own length via the square root of its squared components.

Mistake 3: Taking the magnitude of only part of the result.

Why it's wrong: all three components must be squared and summed; skipping the zero-looking or negative ones understates the magnitude. Correct approach: square every component (signs vanish under squaring) before the square root.