Vector triple product — prove a × (b × c) = b(a·c) - c(a·b)

Question

Prove the vector triple product identity:

\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b}(\vec{a} \cdot \vec{c}) - \vec{c}(\vec{a} \cdot \vec{b})

(JEE Advanced 2023, similar pattern)

Solution — Step by Step

Let $\vec{a} = (a_1, a_2, a_3)$ , $\vec{b} = (b_1, b_2, b_3)$ , $\vec{c} = (c_1, c_2, c_3)$ .

We can simplify by choosing $\vec{b}$ along the x-axis and $\vec{c}$ in the xy-plane (since the identity is coordinate-independent):

$\vec{b} = (b_1, 0, 0)$ and $\vec{c} = (c_1, c_2, 0)$ .

\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ b_1 & 0 & 0 \\ c_1 & c_2 & 0 \end{vmatrix} = (0, 0, b_1 c_2)

\vec{a} \times (0, 0, b_1 c_2) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ 0 & 0 & b_1 c_2 \end{vmatrix}

= (a_2 b_1 c_2, -a_1 b_1 c_2, 0) \quad \cdots (*)

$\vec{a} \cdot \vec{c} = a_1 c_1 + a_2 c_2$

$\vec{a} \cdot \vec{b} = a_1 b_1$

\vec{b}(\vec{a} \cdot \vec{c}) - \vec{c}(\vec{a} \cdot \vec{b})

= (b_1, 0, 0)(a_1 c_1 + a_2 c_2) - (c_1, c_2, 0)(a_1 b_1)

= (b_1 a_1 c_1 + b_1 a_2 c_2, 0, 0) - (a_1 b_1 c_1, a_1 b_1 c_2, 0)

= (b_1 a_2 c_2, -a_1 b_1 c_2, 0)

This matches $(*)$ exactly. $\blacksquare$

Why This Works

The key insight: $\vec{a} \times (\vec{b} \times \vec{c})$ must lie in the plane of $\vec{b}$ and $\vec{c}$ (because $\vec{b} \times \vec{c}$ is perpendicular to both $\vec{b}$ and $\vec{c}$ , and crossing with $\vec{a}$ brings us back to the plane of $\vec{b}$ and $\vec{c}$ ). So the result must be of the form $\alpha\vec{b} + \beta\vec{c}$ .

The proof shows that $\alpha = \vec{a} \cdot \vec{c}$ and $\beta = -\vec{a} \cdot \vec{b}$ . We used a special coordinate system for simplicity, but since the identity is a vector equation, it holds in all coordinate systems.

The mnemonic BAC-CAB helps remember the formula: $\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b}(\vec{a} \cdot \vec{c}) - \vec{c}(\vec{a} \cdot \vec{b})$ .

Alternative Method — Direct component computation

Without choosing special coordinates, write out all 9 components of $\vec{b} \times \vec{c}$ , then compute the cross product with $\vec{a}$ , and verify equality with the RHS. It works but involves significantly more algebra.

The BAC-CAB rule is one of the most powerful vector identities in JEE. It converts a messy triple cross product into dot products, which are much easier to compute. Remember: the “middle” vector ( $\vec{b}$ ) goes with the dot product of the “outer” vectors ( $\vec{a} \cdot \vec{c}$ ), and the “last” vector ( $\vec{c}$ ) goes with the other dot product ( $\vec{a} \cdot \vec{b}$ ), with a minus sign.

Caution: $\vec{a} \times (\vec{b} \times \vec{c}) \neq (\vec{a} \times \vec{b}) \times \vec{c}$ in general. The cross product is NOT associative.

Common Mistake

Students frequently apply the BAC-CAB rule with the wrong vector as the “outer” one. In $\vec{a} \times (\vec{b} \times \vec{c})$ , the outer vector is $\vec{a}$ (the one outside the brackets). The formula is $\vec{b}(\vec{a} \cdot \vec{c}) - \vec{c}(\vec{a} \cdot \vec{b})$ . If you accidentally use $(\vec{a} \times \vec{b}) \times \vec{c}$ , the outer vector is $\vec{c}$ , and the result is $\vec{b}(\vec{a} \cdot \vec{c}) - \vec{a}(\vec{b} \cdot \vec{c})$ — a different expression entirely.

Question

Solution — Step by Step

Why This Works

Alternative Method — Direct component computation

Common Mistake

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