MHT CET · Maths · Three Dimensional Geometry
The d.r.s. of the normal to the plane passing through the origin and the line of intersection of the planes \(x+2 y+3 z=4\) and \(4 x+3 y+2 z=1\) are
- A 3,2,1
- B 2,3,1
- C 1,2,1
- D 3,1,2
Answer & Solution
Correct Answer
(A) 3,2,1
Step-by-step Solution
Detailed explanation
Equation of the plane passing through the line of intersection of the given planes, is \((x z+2 y+3 z-4)+\lambda(4 x+3 y+2 z-1)=0\)
Since the plane (1) passes through origin, We get \(-4-\lambda=0 \quad \Rightarrow \lambda=-4\) Substituting value of \(\lambda\) in equation (1), we get
\(\begin{aligned}& -15 x-10 y-5 z=0 \quad \Rightarrow 3 x+2 y+z=0 \\& \therefore \text { d.r.s. are }(3,2,1)\end{aligned}\)
Since the plane (1) passes through origin, We get \(-4-\lambda=0 \quad \Rightarrow \lambda=-4\) Substituting value of \(\lambda\) in equation (1), we get
\(\begin{aligned}& -15 x-10 y-5 z=0 \quad \Rightarrow 3 x+2 y+z=0 \\& \therefore \text { d.r.s. are }(3,2,1)\end{aligned}\)
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