AP EAMCET · Maths · Vector Algebra
If the points \(A, B, C, D\) with position vectors \(\overline{\mathrm{i}}+\overline{\mathrm{j}}-\overline{\mathrm{k}}, \overline{\mathrm{i}}-\overline{\mathrm{j}}+2 \overline{\mathrm{k}}, \overline{\mathrm{i}}-2 \overline{\mathrm{j}}+\overline{\mathrm{k}}\), \(2 \overline{\mathrm{i}}+\overline{\mathrm{j}}+\overline{\mathrm{k}}\) respectively form a tetrahedron, then the angle between the faces ABC and ABD of the tetrahedron is
- A \(\operatorname{Cos}^{-1}\left(\frac{-4}{\sqrt{29}}\right)\)
- B \(\operatorname{Cos}^{-1}\left(\frac{-4}{5}\right)\)
- C \(\operatorname{Cos}^{-1}\left(\frac{3}{5}\right)\)
- D \(\operatorname{Cos}^{-1}\left(\frac{\sqrt{29}}{\sqrt{33}}\right)\)
Answer & Solution
Correct Answer
(A) \(\operatorname{Cos}^{-1}\left(\frac{-4}{\sqrt{29}}\right)\)
Step-by-step Solution
Detailed explanation
\( \vec{AB} = (\overline{i} - \overline{j} + 2 \overline{k}) - (\overline{i} + \overline{j} - \overline{k}) = -2\overline{j} + 3\overline{k} \)…
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