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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

  1. A \(\operatorname{Cos}^{-1}\left(\frac{-4}{\sqrt{29}}\right)\)
  2. B \(\operatorname{Cos}^{-1}\left(\frac{-4}{5}\right)\)
  3. C \(\operatorname{Cos}^{-1}\left(\frac{3}{5}\right)\)
  4. D \(\operatorname{Cos}^{-1}\left(\frac{\sqrt{29}}{\sqrt{33}}\right)\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\operatorname{Cos}^{-1}\left(\frac{-4}{\sqrt{29}}\right)\)

Step-by-step Solution

Detailed explanation

\( \vec{AB} = (\overline{i} - \overline{j} + 2 \overline{k}) - (\overline{i} + \overline{j} - \overline{k}) = -2\overline{j} + 3\overline{k} \)…
From AP EAMCET
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