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Exercise 10.4 · Q4

Q.Show that (a⃗−b⃗)×(a⃗+b⃗)=2(a⃗×b⃗)(\vec{a}-\vec{b})\times(\vec{a}+\vec{b})=2(\vec{a}\times\vec{b}).

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Using the distributive property of the cross product and the fact that a⃗×a⃗=b⃗×b⃗=0\vec{a}\times\vec{a}=\vec{b}\times\vec{b}=0, the expression simplifies directly to 2(a⃗×b⃗)2(\vec{a}\times\vec{b}).

This is a classic vector identity that looks intimidating at first but collapses beautifully once you apply two simple rules: cross product distributes over addition, and any vector crossed with itself gives zero.

The cross product is distributive over vector addition, meaning:

u⃗×(v⃗+w⃗)=u⃗×v⃗+u⃗×w⃗\vec{u} \times (\vec{v} + \vec{w}) = \vec{u} \times \vec{v} + \vec{u} \times \vec{w}

and similarly from the other side. This is the only tool we need — no geometry, no components, no determinants.

Watch out

The cross product is not commutative. In fact, a⃗×b⃗=−(b⃗×a⃗)\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a}). Keep track of order carefully — a sign slip here is the most common mistake.

Let’s work through it step by step.

  1. Expand the left-hand side using distribution. Treat (a⃗−b⃗)(\vec{a} - \vec{b}) as one vector and (a⃗+b⃗)(\vec{a} + \vec{b}) as the other. Distribute from left to right:

(a⃗−b⃗)×(a⃗+b⃗)=(a⃗−b⃗)×a⃗+(a⃗−b⃗)×b⃗(\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) = (\vec{a} - \vec{b}) \times \vec{a} + (\vec{a} - \vec{b}) \times \vec{b}

  1. Distribute again inside each term. For the first term:

(a⃗−b⃗)×a⃗=a⃗×a⃗−b⃗×a⃗(\vec{a} - \vec{b}) \times \vec{a} = \vec{a} \times \vec{a} - \vec{b} \times \vec{a}

For the second term:

(a⃗−b⃗)×b⃗=a⃗×b⃗−b⃗×b⃗(\vec{a} - \vec{b}) \times \vec{b} = \vec{a} \times \vec{b} - \vec{b} \times \vec{b}

  1. Apply the zero self-cross property. Any vector crossed with itself is the zero vector: a⃗×a⃗=0⃗\vec{a} \times \vec{a} = \vec{0} and b⃗×b⃗=0⃗\vec{b} \times \vec{b} = \vec{0}. So the expression becomes:

0⃗−b⃗×a⃗+a⃗×b⃗−0⃗\vec{0} - \vec{b} \times \vec{a} + \vec{a} \times \vec{b} - \vec{0}

which simplifies to: …

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