JEE Mains · Maths · STD 12 - 11. three dimension geometry
The plane containing the line \(\frac{{x - 3}}{2} = \frac{{y + 2}}{{ - 1}} = \frac{{z - 1}}{3}\) and also containing its projection on the plane \(2x + 3y -z = 5,\) contains which one of the following points?
- A \((2, 2, 0)\)
- B \((-2, 2, 2)\)
- C \((0, -2, 2)\)
- D \( (2, 0, -2)\)
Answer & Solution
Correct Answer
(D) \( (2, 0, -2)\)
Step-by-step Solution
Detailed explanation
Equation of required plane is \(a(x-3)+b(y+2)+c(z-1)=0\) [as it contains the point \((3,-2,1)]\) \(2 a-b+3 c=0\) \(2 a+3 b-c=0\) \(\Rightarrow \frac{a}{-8}=\frac{b}{8}=\frac{c}{8}\) \(\Rightarrow \frac{a}{+1}=\frac{b}{-1}=\frac{c}{1}\) \(\therefore\) equation of plane is…
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