A vector n is inclined to x -axis at 4 5 ∘ , to y -axis at 6 0 ∘ and at an acute angle to z -axis. If n is a normal to a plane passing through the point ( 2 , − 1 , 1 ) then the equation of the plan

Question

A vector n is inclined to x -axis at 4 5 ∘ , to y -axis at 6 0 ∘ and at an acute angle to z -axis. If n is a normal to a plane passing through the point ( 2 ​ , − 1 , 1 ) then the equation of the plane is :

TestHub · Accepted Answer

Direction cosines of n are 2 1 ​ , 4 1 ​ , 2 1 ​ . Equation of the plane, ⇒ ⇒ ⇒ ​ 2 1 ​ ( x − 2 ​ ) + 4 1 ​ ( y + 1 ) + 2 1 ​ ( z − 1 ) = 0 2 ( x − 2 ​ ) + ( y + 1 ) + 2 ( z − 1 ) = 0 2 x + y + 2 z = 2 2 ​ − 1 + 2 2 x + y + 2 z = 2 2 ​ + 1 ​

Mathematics - Vector Question with Solution | TestHub

Options:

Doubts & Discussion