WBJEE · Maths · Three Dimensional Geometry
The plane \(2 x-y+3 z+5=0\) is rotated through \(90^{\circ}\) about its line of intersection with the plane \(x+y+z=1\). The equation of the plane in new position is
- A \(3 x+9 y+z+17=0\)
- B \(3 x+9 y+z=17\)
- C \(3 x-9 y-z=17\)
- D \(3 x+9 y-z=17\)
Answer & Solution
Correct Answer
(B) \(3 x+9 y+z=17\)
Step-by-step Solution
Detailed explanation
Hint : Equation of a plane passing through the line of intersection of the given planes is \(2 x-y+3 z+5+\lambda(x+y+z-1)=0\) \(\Rightarrow(2+\lambda) x-(1-\lambda) y+(3+\lambda) z+5-\lambda=0\) this will be perpendicular to the plane \(2 x-y+3 z+5=0\)…
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