JEE Advanced 2022 Paper 2 · Q13 · Vector Algebra

Question

Let $\hat{i}, \hat{j}$ and $\hat{k}$ be the unit vectors along the three positive coordinate axes. Let $\vec a = 3\hat i + \hat j - \hat k$, $\vec b = \hat i + b_{2}\hat j + b_{3}\hat k$, $b_{2},b_{3}\in\mathbb{R}$, and $\vec c = c_{1}\hat i + c_{2}\hat j + c_{3}\hat k$, $c_{1},c_{2},c_{3}\in\mathbb{R}$, be three vectors such that $b_{2}b_{3}>0$, $\vec a\cdot\vec b = 0$ and $\begin{pmatrix}0 & -c_{3} & c_{2}\ c_{3} & 0 & -c_{1}\ -c_{2} & c_{1} & 0\end{pmatrix}\begin{pmatrix}1\ b_{2}\ b_{3}\end{pmatrix} = \begin{pmatrix}3-c_{1}\ 1-c_{2}\ -1-c_{3}\end{pmatrix}$. Then, which of the following is/are TRUE?

SolveKar AI · Accepted Answer

The skew-symmetric matrix multiplication encodes $\vec c imes \vec b = \vec a - \vec c$ (matching the columns of $\vec a=(3,1,-1)$). So $\vec c imes\vec b = \vec a - \vec c$. Taking dot product with $\vec b$: $(\vec c imes\vec b)\cdot\vec b = 0 = \vec a\cdot\vec b - \vec c\cdot\vec b = 0 - \vec b\cdot\vec c$, hence $\vec b\cdot\vec c = 0$, so (B) true. Taking dot with $\vec c$: $0 = \vec a\cdot\vec c - |\vec c|^2$, so $\vec a\cdot\vec c = |\vec c|^2 e 0$ (generically), so (A) false. From $\vec a\cdot\vec b = 0$: $3+b_2-b_3=0$, so $b_3 = b_2+3$. Since $b_2 b_3 > 0$, either both positive or both negative; if both negative, $b_2<-3$. Compute $|\vec b|^2 = 1+b_2^2+(b_2+3)^2 = 2b_2^2+6b_2+10 = 2(b_2+3/2)^2 + 11/2$. For $b_2>0$: $|\vec b|^2 > 1+0+9=10$ at the limit, strictly $>10$. For $b_2<-3$: $|\vec b|^2 > 2(9/4)+10=14.5>10$. So $|\vec b|>\sqrt{10}$, (C) true. For (D), from $\vec c imes\vec b=\vec a-\vec c$ taking magnitude squared and using $\vec b\cdot\vec c=0$: $|\vec c|^2|\vec b|^2 = |\vec a|^2 - 2\vec a\cdot\vec c + |\vec c|^2 = 11 - 2|\vec c|^2 + |\vec c|^2 = 11 - |\vec c|^2$. So $|\vec c|^2(|\vec b|^2+1) = 11$, giving $|\vec c|^2 = \dfrac{11}{|\vec b|^ … [truncated]

JEE Advanced 2022 Paper 2 · Q13 · Vector Algebra

Solution