Q72 / 160 · TS EAMCET 2024 (11 May Shift 1) · Mathematics

A unit vector $e=a \hat{i}+b \hat{j}+c k$ is coplanar with the vectors $\hat{i}-3 \hat{j}+5 \hat{k}$ and $3 \hat{i}+\hat{j}-5 \hat{k}$. If $\vec{e}$ is perpendicular to the vector $\hat{i}+\hat{j}+\hat{k}$ then $2 a^2+3 b^2+4 c^2=$

  • A
    $1$
  • Correct answer
    B
    $3$
  • C
    $-1$
  • D
    $\sqrt{2}$

Solution

$\vec{e}$ is coplanar with $\hat{i}-3 \hat{j}+5 \hat{k}$ and $3 \hat{i}+\hat{j}-5 \hat{k}$ $\left|\begin{array}{ccc}a & b & c \\ 1 & -3 & 5 \\ 3 & 1 & -5\end{array}\right|=0 \Rightarrow a+2 b+c=0$ ...(i) $(a \hat{i}+b \hat{j}+c \hat{k}) \cdot(\hat{i}+\hat{j}+\hat{k})=0 \Rightarrow a+b+c=0$ ...(ii) Solving (i) and (ii), $b=0, c=-a$ $\begin{aligned} & |\vec{e}|=1 \Rightarrow a^2+b^2+c^2=1 \Rightarrow 2 a^2=1 \Rightarrow a^2=\frac{1}{2} ; c^2=\frac{1}{2} \\ & 2 a^2+3 b^2+4 c^2=2 \times \frac{1}{2}+3 \times 0 \div 4 \times \frac{1}{2}=3 \end{aligned}$

Hence, the correct answer is option: 'B'.

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