JEE 2022 — Mathematics PYQ

1. Identify the given relations

We are given the following equations:

• $\vec{a}=\hat{i}-2\hat{j}+3\hat{k}$

• $\vec{b}=\hat{i}+\hat{j}+\hat{k}$

• $\vec{a}+(\vec{b}\times \vec{c})=\vec{0}⟹\vec{b}\times \vec{c}=-\vec{a}$

• $\vec{b}\cdot \vec{c}=5$

2. Take the cross product with $\vec{b}$

To find $\vec{c}$, we take the cross product of $\vec{b}$ with the equation $\vec{b}\times \vec{c}=-\vec{a}$:

Using the Vector Triple Product formula, $\vec{x}\times (\vec{y}\times \vec{z})=(\vec{x}\cdot \vec{z})\vec{y}-(\vec{x}\cdot \vec{y})\vec{z}$, we get:

3. Calculate the constants

• $\vec{b}\cdot \vec{c}=5$ (Given)

• $|\vec{b}{|}^2=\vec{b}\cdot \vec{b}=1^2+1^2+1^2=3$

Substitute these into the equation:

4. Find $3(\vec{c}\cdot \vec{a})$

Dot product both sides with $\vec{a}$:

Since the scalar triple product $(\vec{b}\times \vec{a})\cdot \vec{a}=0$ (because it involves two identical vectors), we are left with:

5. Calculate $\vec{b}\cdot \vec{a}$

6. Final Result

The value is 10.