Q73 / 160 · TS EAMCET 2024 (11 May Shift 1) · Mathematics

$\vec{a}=\hat{i}+\hat{j}-2 \hat{k}, \vec{b}=\hat{i}-2 \hat{j}+\hat{k}$ and $\vec{c}=2 \hat{i}+\hat{j}-\hat{k}$ are three vectors. If $\vec{d}$ is a normal to the plane of $\vec{a}$ and $\vec{b}$ and $\vec{d} \cdot \vec{c}=2$, then $|\vec{d}|=$

  • A
    $\sqrt{6}$
  • B
    $2 \sqrt{3}$
  • Correct answer
    C
    $\sqrt{3}$
  • D
    $2$

Solution

$\begin{aligned} & \text { } \vec{d}=\lambda(\vec{a} \times \vec{b}) \\ & \vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -2 \\ 1 & -2 & 1\end{array}\right|=-3 \hat{i}-3 \hat{j}-3 \hat{k} \\ & \vec{d}=\lambda(\hat{i}+\hat{j}+\hat{k}) \Rightarrow \vec{d} \cdot \vec{c}=2 \Rightarrow \lambda(2+1-1)=2 \Rightarrow \lambda=1 \\ & \vec{d}=\hat{i}+\hat{j}+\hat{k} \Rightarrow|\vec{d}|=\sqrt{3} .\end{aligned}$

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

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