NIMCET 2020 — Mathematics PYQ

Step 1: Simplify the cross product term

First, let's expand the cross product inside the square brackets:

$$(\vec{a}+\vec{b})\times (\vec{a}+\vec{c})$$

Using the distributive property of the cross product:

$$=\vec{a}\times \vec{a}+\vec{a}\times \vec{c}+\vec{b}\times \vec{a}+\vec{b}\times \vec{c}$$

Since the cross product of any vector with itself is zero ($\vec{a}\times \vec{a}=0$), the expression becomes:

$$=0+\vec{a}\times \vec{c}+\vec{b}\times \vec{a}+\vec{b}\times \vec{c}$$

$$=\vec{a}\times \vec{c}+\vec{b}\times \vec{a}+\vec{b}\times \vec{c}$$

Step 2: Take the dot product with $(\vec{a}+\vec{b}+\vec{c})$

Now, substitute this simplified cross product back into the original expression:

$$(\vec{a}+\vec{b}+\vec{c})\cdot (\vec{a}\times \vec{c}+\vec{b}\times \vec{a}+\vec{b}\times \vec{c})$$

Distributing each vector using the dot product rules:

1. For $\vec{a}$:

   $$\vec{a}\cdot (\vec{a}\times \vec{c})+\vec{a}\cdot (\vec{b}\times \vec{a})+\vec{a}\cdot (\vec{b}\times \vec{c})$$

   • Since a scalar triple product is zero if any two vectors are identical:

     • $\vec{a}\cdot (\vec{a}\times \vec{c})=0$

     • $\vec{a}\cdot (\vec{b}\times \vec{a})=0$

   • This leaves only: $\vec{a}\cdot (\vec{b}\times \vec{c})=[\vec{a}\vec{b}\vec{c}]$

2. For $\vec{b}$:

   $$\vec{b}\cdot (\vec{a}\times \vec{c})+\vec{b}\cdot (\vec{b}\times \vec{a})+\vec{b}\cdot (\vec{b}\times \vec{c})$$

   • Similarly, terms with repeated vectors become zero:

     • $\vec{b}\cdot (\vec{b}\times \vec{a})=0$

     • $\vec{b}\cdot (\vec{b}\times \vec{c})=0$

   • This leaves only: $\vec{b}\cdot (\vec{a}\times \vec{c})=[\vec{b}\vec{a}\vec{c}]$

3. For $\vec{c}$:

   $$\vec{c}\cdot (\vec{a}\times \vec{c})+\vec{c}\cdot (\vec{b}\times \vec{a})+\vec{c}\cdot (\vec{b}\times \vec{c})$$

   • Again, terms with repeated vectors become zero:

     • $\vec{c}\cdot (\vec{a}\times \vec{c})=0$

     • $\vec{c}\cdot (\vec{b}\times \vec{c})=0$

   • This leaves only: $\vec{c}\cdot (\vec{b}\times \vec{a})=[\vec{c}\vec{b}\vec{a}]$

Step 3: Combine and simplify the remaining terms

Adding the non-zero terms together:

$$=[\vec{a}\vec{b}\vec{c}]+[\vec{b}\vec{a}\vec{c}]+[\vec{c}\vec{b}\vec{a}]$$

Using the properties of the scalar triple product (interchanging two positions changes the sign):

• $[\vec{b}\vec{a}\vec{c}]=-[\vec{a}\vec{b}\vec{c}]$

• $[\vec{c}\vec{b}\vec{a}]=-[\vec{c}\vec{a}\vec{b}]=[\vec{a}\vec{c}\vec{b}]=-[\vec{a}\vec{b}\vec{c}]$

Substitute these values back:

$$=[\vec{a}\vec{b}\vec{c}]-[\vec{a}\vec{b}\vec{c}]-[\vec{a}\vec{b}\vec{c}]$$

$$=-[\vec{a}\vec{b}\vec{c}]$$