JEE Advanced 2022 · Paper 2, Mathematics question 13

Read the matrix as a cross product. The left side is $\vec c\times\vec b$, and the right side is $\vec a-\vec c$:
$$\vec c\times\vec b=\vec a-\vec c.$$

Dot products.

• Dot with $\vec b$: $0=\vec a\cdot\vec b-\vec b\cdot\vec c=-\vec b\cdot\vec c$. So $\vec b\cdot\vec c=0$, and (B) is true.
• Dot with $\vec c$: $0=\vec a\cdot\vec c-|\vec c|^2$.

Solve for $\vec c$. Cross both sides with $\vec b$. Since $\vec b\cdot\vec c=0$:
$$(\vec c\times\vec b)\times\vec b=-|\vec b|^2\vec c=\vec a\times\vec b-(\vec a-\vec c),$$
so $\vec c=\dfrac{\vec a-\vec a\times\vec b}{1+|\vec b|^2}$.

Since $\vec a\perp\vec b$ and $|\vec a|^2=11$:
$$\vec a\cdot\vec c=|\vec c|^2=\frac{11}{1+|\vec b|^2}>0.$$
So (A) is false, and $|\vec c|\le\sqrt{11}$, so (D) is true.

$|\vec b|$. From $\vec a\cdot\vec b=3+b_2-b_3=0$, $b_3=b_2+3$. Then $b_2b_3>0$ means $b_2>0$ or $b_2<-3$, and
$$|\vec b|^2=1+b_2^2+(b_2+3)^2=2b_2^2+6b_2+10>10$$
on both ranges. So (C) is true. (B), (C), (D).