MHT CET · Maths · Three Dimensional Geometry
Let the vectors \(\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}\) and \(\overline{\mathrm{d}}\) be such that \((\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times(\overline{\mathrm{c}} \times \overline{\mathrm{d}})=\overline{0}\). Let \(\mathrm{P}_1\) and \(\mathrm{P}_2\) be the planes determined by the pair of vectors \(\stackrel{\rightharpoonup}{\mathrm{a}}, \overline{\mathrm{b}}\) and \(\overline{\mathrm{c}}, \overline{\mathrm{d}}\) respectively, then the angle between \(P_1\) and \(P_2\) is
- A 0
- B \(\frac{\pi}{4}\)
- C \(\frac{\pi}{3}\)
- D \(\frac{\pi}{2}\)
Answer & Solution
Correct Answer
(A) 0
Step-by-step Solution
Detailed explanation
According to the given condition, we get Normal to the plane \(P_1\) is parallel to \(\overline{\mathrm{a}} \times \overline{\mathrm{b}}\) and normal to the plane \(\mathrm{P}_2\) is parallel to \(\overline{\mathrm{c}} \times \overline{\mathrm{d}}\).
Given that \((\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times(\overline{\mathrm{c}} \times \overline{\mathrm{d}})=0\)
\(\therefore \quad(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \|(\overline{\mathrm{c}} \times \overline{\mathrm{d}})\)
\(\therefore \quad\) Angle between \(P_1\) and \(P_2\) is 0 .
Given that \((\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times(\overline{\mathrm{c}} \times \overline{\mathrm{d}})=0\)
\(\therefore \quad(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \|(\overline{\mathrm{c}} \times \overline{\mathrm{d}})\)
\(\therefore \quad\) Angle between \(P_1\) and \(P_2\) is 0 .
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