KCET 2026 Mathematics Solved Paper

Concept:Use cross product and dot product to set up equations for the unknown components of $\vec{c}$.Explanation:Let $\vec{c}=x\hat{i}+y\hat{j}+z\hat{k}$.Given $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{a}\times\vec{c}=\vec{b}$, where $\vec{b}=\hat{j}-\hat{k}$.Compute the cross product:$$\vec{a}\times\vec{c}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\1 & 1 & 1 \\x & y & z\end{vmatrix}$$Expanding:$$\vec{a}\times\vec{c}=(z-y)\hat{i}+(x-z)\hat{j}+(y-x)\hat{k}$$Since $\vec{a}\times\vec{c}=0\hat{i}+1\hat{j}-1\hat{k}$, compare coefficients:$$z-y=0,\quad x-z=1,\quad y-x=-1$$From $z-y=0$, we get $z=y$.From $x-z=1$, using $z=y$, we get $x-y=1$.Now use the dot product condition:$$\vec{a}\cdot\vec{c}=3$$$$(\hat{i}+\hat{j}+\hat{k})\cdot(x\hat{i}+y\hat{j}+z\hat{k})=3$$So $x+y+z=3$. Since $z=y$, this becomes $x+2y=3$.From $x-y=1$, we have $x=y+1$.Substitute into $x+2y=3$:$$y+1+2y=3 \Rightarrow 3y=2 \Rightarrow y=\frac{2}{3}$$Thus $z=\frac{2}{3}$ and $x=y+1=\frac{5}{3}$.Therefore:$$\vec{c}=\frac{5}{3}\hat{i}+\frac{2}{3}\hat{j}+\frac{2}{3}\hat{k}$$Answer:$\vec{c}=\frac{5}{3}\hat{i}+\frac{2}{3}\hat{j}+\frac{2}{3}\hat{k}$, which is Option C.