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The equation of a plane through the line of intersection of the planes x+2y=3,   y-2z+1=0, and perpendicular to the first plane is :

Option: 1

2x-y-10z=9


Option: 2

2x-y+7z=11


Option: 3

2x-y+10z=11


Option: 4

2x-y-9z=10


Answers (1)

best_answer

Equation of a plane through the line of intersection of the planes x + 2y = 3 and y - 2 z + 1 = 0 is

\\ (x+2 y-3)+\lambda(y-2 z+1)=0 \\ \Rightarrow x+(2+\lambda) y-2 \lambda(z)-3+\lambda=0

 This plane is perpendicular to x+ 2y = 3

therefore, their dot product is zero

\\\text{i.e. }\quad 1+2(2+\lambda)=0 \Rightarrow \lambda=-\frac{5}{2}\\\text {Thus, required plane is }\\ \\ x+\left(2-\frac{5}{2}\right) y-2 \times \frac{-5}{2}(z)-3-\frac{5}{2}=0 \\ \Rightarrow \quad x-\frac{y}{2}+5 z-\frac{11}{2}=0 \\\Rightarrow \quad 2 x-y+10 z-11=0 

Posted by

Shailly goel

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