TS EAMCET · Maths · Vector Algebra
\(\vec{r}\) is a vector perpendicular to the piane determined by the vectors \(2 \hat{i}-\hat{j}\) and \(\hat{j}+2 \hat{k}\). If the magnitude of the projection of \(\vec{r}\) on the vector \(2 \hat{i}+\hat{j}+2 \hat{k}\) is 1 , then \(|\vec{r}|=\)
- A \(\sqrt{6}\)
- B \(3 \sqrt{6}\)
- C \(\frac{2 \sqrt{6}}{3}\)
- D \(\frac{3 \sqrt{6}}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{3 \sqrt{6}}{2}\)
Step-by-step Solution
Detailed explanation
Normal of plane \(\vec{r}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 0 \\ 0 & 1 & 2 \end{array}\right|=-2 \hat{i}-4 \hat{j}+2 \hat{k}\) Since, \(\vec{r}\) perpendicular to plane \(\therefore \vec{r}=\lambda(2 \hat{i}+4 \hat{j}-2 \hat{k})\) Projection of…
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