JEE Mains · Maths · STD 12 - 11. three dimension geometry
The foot of the perpendicular from a point on the circle \(x ^{2}+ y ^{2}=1, z =0\) to the plane \(2 x +3 y + z =6\) lies on which one of the following curves?
- A \((6 x+5 y-12)^{2}+4(3 x+7 y-8)^{2}=1\), \(z=6-2 x-3 y\)
- B \((5 x+6 y-12)^{2}+4(3 x+5 y-9)^{2}=1\), \(z=6-2 x-3 y\)
- C \((6 x+5 y-14)^{2}+9(3 x+5 y-7)^{2}=1\), \(z=6-2 x-3 y\)
- D \((5 x+6 y-14)^{2}+9(3 x+7 y-8)^{2}=1\), \(z=6-2 x-3 y\)
Answer & Solution
Correct Answer
(B) \((5 x+6 y-12)^{2}+4(3 x+5 y-9)^{2}=1\), \(z=6-2 x-3 y\)
Step-by-step Solution
Detailed explanation
\(\frac{ h -\cos \theta}{2}=\frac{ k -\sin \theta}{3}=\frac{ w -0}{1}\) \(=\frac{-1(2 \cos \theta+3 \sin \theta-6)}{14}\) \(h =\cos \frac{-2(2 \cos \theta+3 \sin \theta-6)}{14}\) \(=\frac{10 \cos \theta-6 \sin \theta+12}{14}\)…
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