AP EAMCET · Maths · Three Dimensional Geometry
Let \(\mathrm{A}=(2,0,-1), \mathrm{B}=(1,-2,0), \mathrm{C}=(1,2,-1)\) and \(\mathrm{D}=(0,-1,-2)\) be four points. If \(\theta\) is the acute angle between the plane determined by \(A, B, C\) and the plane determined by \(\mathrm{A}, \mathrm{C}, \mathrm{D}\), then \(\tan \theta=\)
- A \(\sqrt{\frac{14}{5}}\)
- B \(\frac{3}{\sqrt{14}}\)
- C \(\frac{3}{\sqrt{5}}\)
- D \(\frac{\sqrt{5}}{3}\)
Answer & Solution
Correct Answer
(C) \(\frac{3}{\sqrt{5}}\)
Step-by-step Solution
Detailed explanation
\(\\vec{AB} = (1-2, -2-0, 0-(-1)) = (-1, -2, 1)\) \(\\vec{AC} = (1-2, 2-0, -1-(-1)) = (-1, 2, 0)\) \(\\vec{n_1} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & -2 & 1 \\ -1 & 2 & 0 \end{vmatrix} = (-2, -1, -4)\)…
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