TS EAMCET · Maths · Three Dimensional Geometry
Let \(P(1,-2,5)\) be the foot of the perpendicular drawn from the origin to the plane \(\pi_1\) and the same \(P\) be the foot of the perpendicular from \((1,2,-1)\) to the plane \(\pi_2\). Then the acute angle between the planes \(\pi_1\) and \(\pi_2\) is
- A \(\cos ^{-1}\left(\frac{19}{\sqrt{390}}\right)\)
- B \(\cos ^{-1}\left(\frac{19}{\sqrt{340}}\right)\)
- C \(\cos ^{-1}\left(\frac{19}{\sqrt{370}}\right)\)
- D \(\cos ^{-1}\left(\frac{19}{\sqrt{350}}\right)\)
Answer & Solution
Correct Answer
(A) \(\cos ^{-1}\left(\frac{19}{\sqrt{390}}\right)\)
Step-by-step Solution
Detailed explanation
According to given informations, the direction ratios of normal to the plane \(\pi_1\) are \(1-0,-2-0\), \(5-0\) i.e. \(1,-2,5\). And similarly direction ratios of normal to the plane \(\pi_2\) are \(1-1,2-(-2),-1-5\) i.e. \(0,4,-6\) So, the acute angle between planes \(\pi_1\)…
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