TS EAMCET · Maths · Vector Algebra
\(\vec{a}=\hat{i}+\hat{j}-2 \hat{k}, \vec{b}=\hat{i}-2 \hat{j}+\hat{k}\) 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 \(2 \sqrt{3}\)
- C \(\sqrt{3}\)
- D \(2\)
Answer & Solution
Correct Answer
(C) \(\sqrt{3}\)
Step-by-step Solution
Detailed explanation
\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})…
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