TS EAMCET · Maths · Three Dimensional Geometry
Let \(\pi_1\) be the plane determined by the vectors \(\bar{i}+\bar{j}, \bar{i}+\bar{k}\) and \(\pi_2\) be the plane determined by the vectors \(\bar{j}-\bar{k}, \bar{k}-\bar{i}\). Let \(\bar{a}\) be a non-zero vector parallel to the line of intersection of the planes \(\pi_1\) and \(\pi_2\). If \(\bar{b}=\bar{i}+\bar{j}-\bar{k}\) then the angle between the vectors \(\bar{a}\) and \(\bar{b}\) is
- A \(\operatorname{Cos}^{-1}\left(\sqrt{\frac{2}{3}}\right)\)
- B \(\frac{\pi}{2}\)
- C \(\operatorname{Cos}^{-1}\left(\frac{1}{\sqrt{3}}\right)\)
- D \(\operatorname{Cos}^{-1}\left(\frac{\sqrt{2}}{3}\right)\)
Answer & Solution
Correct Answer
(A) \(\operatorname{Cos}^{-1}\left(\sqrt{\frac{2}{3}}\right)\)
Step-by-step Solution
Detailed explanation
\(\bar{n}_1 = (\bar{i}+\bar{j}) \times (\bar{i}+\bar{k}) = \bar{i} - \bar{j} - \bar{k}\) \(\bar{n}_2 = (\bar{j}-\bar{k}) \times (\bar{k}-\bar{i}) = \bar{i} + \bar{j} + \bar{k}\)…
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