TS EAMCET · Maths · Vector Algebra
If \(\bar{a}=2 \bar{i}+\bar{j}-\bar{k}, \bar{b}=\bar{i}-\bar{j}+3 \bar{k}, \bar{x}=\left(\frac{\bar{a} \cdot \bar{b}}{|\bar{b}|^2}\right) \bar{b}, \bar{y}=\left(\frac{\bar{a} \cdot \bar{b}}{|\bar{a}|^2}\right) \bar{a}\) and \(\theta\) is angle between \(\bar{a}\) and \(\bar{b}\), then \(\mathrm{x}^2+\mathrm{y}^2=\)
- A \(17 \cos ^2 \theta\)
- B \((\sqrt{6}+\sqrt{11}) \cos ^2 \theta\)
- C \(17 \cos 2 \theta\)
- D \(17 \sin ^2 \theta\)
Answer & Solution
Correct Answer
(A) \(17 \cos ^2 \theta\)
Step-by-step Solution
Detailed explanation
Given \(\vec{a}=2 \hat{i}+\hat{j}-\hat{k}, \vec{b}=\hat{i}-\hat{j}+3 \hat{k}\) Then, \(\vec{x}=\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right) \cdot \vec{b}\)…
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