UPSC NDA I 2023 Mathematics Solved Paper

Concept:Given $\vec{a}=3\hat{i}+3\hat{j}+3\hat{k}$ and $\vec{c}=\hat{j}-\hat{k}$, with $\vec{a}\cdot\vec{b}=27$ and $\vec{a}\times\vec{b}=9\vec{c}$. The dot product gives a linear equation in components; the cross product gives three component equations.Explanation:Let $\vec{b}=x\hat{i}+y\hat{j}+z\hat{k}$. From $\vec{a}\cdot\vec{b}=27$: $3x+3y+3z=27 \Rightarrow x+y+z=9\quad$ (1). Compute $\vec{a}\times\vec{b}$: $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 3 & 3 \\ x & y & z \end{vmatrix} = 3(z-y)\hat{i} - 3(z-x)\hat{j} + 3(y-x)\hat{k}.$$ Given $\vec{a}\times\vec{b}=9\vec{c}=9\hat{j}-9\hat{k}$, equate components: $3(z-y)=0 \Rightarrow y=z\quad$ (2). $-3(z-x)=9 \Rightarrow z-x=-3\quad$ (3). $3(y-x)=-9 \Rightarrow y-x=-3\quad$ (4). From (2), replace $y$ by $z$ in (4): $z-x=-3$ (same as (3)). Solve (1), (2), (3): Using (2), write $y=z$ in (1): $x+z+z=9 \Rightarrow x+2z=9$. From (3): $x=z+3$. Substitute: $(z+3)+2z=9 \Rightarrow 3z=6 \Rightarrow z=2$. Thus $y=2$, $x=5$. Therefore $\vec{b}=5\hat{i}+2\hat{j}+2\hat{k}$.Answer:$\vec{b}=5\hat{i}+2\hat{j}+2\hat{k}$, which corresponds to option B.