MHT CET · Maths · Three Dimensional Geometry
The magnitude of the projection of the vector \(2 \hat{i}+\hat{j}+\widehat{k}\), on the vector perpendicular to the plane containing the vectors \(\hat{i}+\hat{j}+\hat{k}\) and \(\hat{i}+2 \hat{j}+3 \hat{k}\), is
- A \(\frac{5}{\sqrt{6}}\) units
- B \(\frac{1}{\sqrt{6}}\) units
- C \(\sqrt{6}\) units
- D \(\frac{2}{\sqrt{6}}\) units
Answer & Solution
Correct Answer
(B) \(\frac{1}{\sqrt{6}}\) units
Step-by-step Solution
Detailed explanation
The perpendicular vector is \(\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 2 & 3\end{array}\right|=\hat{i}-2 \hat{j}+\hat{k}\)
the required projection
\(=\frac{(2 \hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}-2 \hat{j}+\hat{k})}{|\hat{i}-2 \hat{j}+\hat{k}|}=\frac{2-2+1}{\sqrt{1^2+(-2)^2+1^2}}=\frac{1}{\sqrt{6}}\)
the required projection
\(=\frac{(2 \hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}-2 \hat{j}+\hat{k})}{|\hat{i}-2 \hat{j}+\hat{k}|}=\frac{2-2+1}{\sqrt{1^2+(-2)^2+1^2}}=\frac{1}{\sqrt{6}}\)
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