TS EAMCET 11 May 2024 Shift 1 Solved Paper

To solve for the expression $2 a^2+3 b^2+4 c^2$, follow these steps:Given that $\hat{e}=a \hat{i}+b \hat{j}+c \hat{k}$ is a unit vector, we have:$a^2+b^2+c^2=1$The vector $\hat{e}$ is coplanar with the vectors $\hat{i}-3 \hat{j}+5 \hat{k}$ and $3 \hat{i}+\hat{j}-5 \hat{k}$. The condition for coplanarity implies the determinant formed by these vectors should be zero:$\begin{vmatrix} a & b & c \\ 1 & -3 & 5 \\ 3 & 1 & -5 \end{vmatrix}=0$Calculating the determinant gives:$a(10)-b(-20)+c(10)=0 \;\; \Rightarrow \;\; 10 a+20 b+10 c=0$Additionally, since $\hat{e}$ is perpendicular to $\hat{i}+\hat{j}+\hat{k}$, we have:$a+b+c=0$Solving the linear equations, we use the relation derived from coplanarity:$10 a+20 b+10 c=0 \;\; \Rightarrow \;\; a+2 b+c=0$Together with $a+b+c=0$, it follows:$\;\frac{a}{1}=\;\frac{-b}{0}=\;\frac{c}{-1}=\lambda$Thus: $a=\lambda, \;\; b=0, \;\; c=-\lambda$Substituting into the unit vector condition:$a^2+b^2+c^2=1 \;\; \Rightarrow \;\; \lambda^2+0+(-\lambda)^2=1 \;\; \Rightarrow \;\; 2 \lambda^2=1$Hence:$\lambda^2=\;\frac{1}{2} \;\; \Rightarrow \;\; \lambda= \pm \;\frac{1}{\sqrt{2}}$Therefore, the values for $a, b$, and $c$ are:$a=\;\frac{1}{\sqrt{2}}, \;\; b=0, \;\; c=-\;\frac{1}{\sqrt{2}}$Calculating $2 a^2+3 b^2+4 c^2$ :$2 \left(\frac{1}{2}\right) +3(0)+4 \left(\frac{1}{2}\right) =1+2=3$Thus, $2 a^2+3 b^2+4 c^2=3$.