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Question: The equation of the plane containing the line <img src="https://cdn.pureessence.tech/canvas_651.png?...

The equation of the plane containing the line and perpendicular to the plane r.n=q\mathrm { r } . \mathrm { n } = \mathrm { q } is

A

(rb)(n×a)=0( \mathrm { r } - \mathrm { b } ) \cdot ( \mathrm { n } \times \mathrm { a } ) = 0

B

(ra)(n×(a×b))=0( \mathrm { r } - \mathrm { a } ) \cdot ( \mathrm { n } \times ( \mathrm { a } \times \mathrm { b } ) ) = 0

C

(ra)(n×b)=0( \mathrm { r } - \mathrm { a } ) \cdot ( \mathrm { n } \times \mathrm { b } ) = 0

D
Answer

(ra)(n×b)=0( \mathrm { r } - \mathrm { a } ) \cdot ( \mathrm { n } \times \mathrm { b } ) = 0

Explanation

Solution

Since the required plane contains the line and is perpendicular to the plane .

∴ It passes through the point a and parallel to vectors b and n. Hence, it is perpendicular to the vector .

∴ Equation of the required plane is

.