MHT CET · Maths · Three Dimensional Geometry
The foot of the perpendicular drawn from the origin to the plane \(x+y+3 z-4=0\)
is
- A \(\left(\frac{2}{11}, \frac{2}{11}, \frac{9}{11}\right)\)
- B \(\left(\frac{4}{11}, \frac{4}{11}, \frac{12}{11}\right)\)
- C \(\left(\frac{1}{7}, \frac{1}{7}, \frac{6}{7}\right)\)
- D \(\left(\frac{1}{5}, \frac{1}{5}, \frac{3}{5}\right)\)
Answer & Solution
Correct Answer
(B) \(\left(\frac{4}{11}, \frac{4}{11}, \frac{12}{11}\right)\)
Step-by-step Solution
Detailed explanation
d.r. of \(\perp\) er drawn from origin to the given plane are \(1,1,3\).
Hence equation of \(\perp e r\) line to the plane and passing through origin is
\(\frac{x}{1}=\frac{y}{1}=\frac{z}{3}=K\) ... say
Now let foot of the \(\perp\) er be \(P(K, K, 3 K)\)
Also this point P lies on given plane
\(\therefore \mathrm{K}+\mathrm{K}+9 \mathrm{~K}-4=0 \Rightarrow 11 \mathrm{~K}=4 \Rightarrow \mathrm{K}=\frac{4}{11}\)
Hence \(P \equiv\left(\frac{4}{11}, \frac{4}{11}, \frac{12}{11}\right)\)
Hence equation of \(\perp e r\) line to the plane and passing through origin is
\(\frac{x}{1}=\frac{y}{1}=\frac{z}{3}=K\) ... say
Now let foot of the \(\perp\) er be \(P(K, K, 3 K)\)
Also this point P lies on given plane
\(\therefore \mathrm{K}+\mathrm{K}+9 \mathrm{~K}-4=0 \Rightarrow 11 \mathrm{~K}=4 \Rightarrow \mathrm{K}=\frac{4}{11}\)
Hence \(P \equiv\left(\frac{4}{11}, \frac{4}{11}, \frac{12}{11}\right)\)
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