NIMCET 2020 — Mathematics PYQ

1. Expand the Cross Product:

First, calculate $(\vec{a}+\vec{b})\times (\vec{a}+\vec{c})$ using the distributive property:

2. Use Cross Product Properties:

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

3. Perform the Dot Product:

Now, take the dot product of $(\vec{a}+\vec{b}+\vec{c})$ with the result from step 2:

4. Distribute the Dot Product:

Recall that the Scalar Triple Product $[\vec{x}\vec{y}\vec{z}]=\vec{x}\cdot (\vec{y}\times \vec{z})$ is zero if any two vectors are the same.

• With $\vec{a}$:

  $$\vec{a}\cdot (\vec{a}\times \vec{c})-\vec{a}\cdot (\vec{a}\times \vec{b})+\vec{a}\cdot (\vec{b}\times \vec{c})=0-0+[\vec{a}\vec{b}\vec{c}]$$

• With $\vec{b}$:

  $$\vec{b}\cdot (\vec{a}\times \vec{c})-\vec{b}\cdot (\vec{a}\times \vec{b})+\vec{b}\cdot (\vec{b}\times \vec{c})=[\vec{b}\vec{a}\vec{c}]-0+0$$

  Note: $[\vec{b}\vec{a}\vec{c}]=-[\vec{a}\vec{b}\vec{c}]$ (due to anti-commutative property of swapping two vectors).

• With $\vec{c}$:

  $$\vec{c}\cdot (\vec{a}\times \vec{c})-\vec{c}\cdot (\vec{a}\times \vec{b})+\vec{c}\cdot (\vec{b}\times \vec{c})=0-[\vec{c}\vec{a}\vec{b}]+0$$

  Note: $[\vec{c}\vec{a}\vec{b}]=[\vec{a}\vec{b}\vec{c}]$ (cyclic permutation).

5. Sum the terms:

Final Answer:

The value of the expression is: