Q.For any vector $\vec{a}$, the value of $(\vec{a}\times\hat{i})^2+(\vec{a}\times\hat{j})^2+(\vec{a}\times\hat{k})^2$ is equal to
(A) $\vec{a}^{\,2}$
(B) $3\vec{a}^{\,2}$
(C) $4\vec{a}^{\,2}$
(D) $2\vec{a}^{\,2}$

Concept: Vector Triple Product identity and expansion of $(\vec{a} \times \hat{i})^2$ as $|\vec{a} \times \hat{i}|^2$.

Step 1: For any unit vector $\hat{u}$,

$|\vec{a} \times \hat{u}|^2 = a^2 - (\vec{a} \cdot \hat{u})^2$ (since $|\vec{a} \times \hat{u}| = a \sin\theta$).

Step 2: Apply to $\hat{i}, \hat{j}, \hat{k}$:

$|\vec{a} \times \hat{i}|^2 = a^2 - a_x^2$,

$|\vec{a} \times \hat{j}|^2 = a^2 - a_y^2$,

$|\vec{a} \times \hat{k}|^2 = a^2 - a_z^2$. …

This is a problem about the vector triple product — but not the triple product identity. Instead, it’s about the magnitude of cross products with the standard basis vectors. The key insight: when you cross a vector with a unit vector, the magnitude of the result depends only on the component of $\vec{a}$ perpendicular to that unit vector.

Let’s unpack that.

1. Write $\vec{a}$ in components. Let $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$. The cross product with $\hat{i}$ is:

Using $\hat{i}\times\hat{i}=0$, $\hat{j}\times\hat{i}=-\hat{k}$, $\hat{k}\times\hat{i}=\hat{j}$, we get:

So its squared magnitude is:

Notice: the $a_1$ component disappears because it’s parallel to $\hat{i}$.

2. Repeat for $\hat{j}$ and $\hat{k}$. Similarly:

3. Add them up. $(\vec{a}\times\hat{i})^2 + (\vec{a}\times\hat{j})^2 + (\vec{a}\times\hat{k})^2 = (a_2^2+a_3^2) + (a_1^2+a_3^2) + (a_1^2+a_2^2)$ …

Method: Simplifying sums of squared cross-products with the basis vectors

Use this for any expression of the form $(\vec{a}\times\hat{i})^2+(\vec{a}\times\hat{j})^2+(\vec{a}\times\hat{k})^2$, where the "square" of a vector means its squared magnitude.

Steps

Step 1: Write $\vec{a}$ in components

Let $\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$. This makes each cross product concrete.

Step 2: Use the "perpendicular part survives" identity

Crossing $\vec{a}$ with a unit vector $\hat{u}$ kills the part of $\vec{a}$ along $\hat{u}$ and keeps the rest:

So $|\vec{a}\times\hat{i}|^2=a_2^2+a_3^2$ (the $a_1$ term drops out), and similarly for $\hat{j}$ and $\hat{k}$. …

Common Mistakes

Mistake 1: Answering $3\vec{a}^{\,2}$

Why it's wrong: students assume each term contributes a full $|\vec{a}|^2$ and multiply by the three axes. But each cross product drops the component along its own axis, so each component of $\vec a$ is counted only twice, not three times. Correct approach: the total is $2|\vec{a}|^2$.

Mistake 2: Treating $(\vec{a}\times\hat{i})^2$ as a genuine vector, or confusing it with the dot product …