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MHT CET · Maths · Three Dimensional Geometry

The distance of the point \((7,5,2)\) from the plane \(3 x+4 y+z-8=0\) measured parallel to the line \(\frac{x-1}{3}=\frac{y-2}{6}=\frac{z+1}{2}\)

  1. A \(\sqrt{74}\) units
  2. B \(\sqrt{47}\) units
  3. C 6 units
  4. D 7 units
Verified Solution

Answer & Solution

Correct Answer

(D) 7 units

Step-by-step Solution

Detailed explanation

Let \(\quad P=(7,5,2)\)
Eq. of line passing through \(P\) and parallel to given line is
\(
\frac{x-7}{3}=\frac{y-5}{6}=\frac{z-2}{2}=r \text { (say) }
\)
Hence coordinates of any point on this line are \((3 r+7,6 r+5,2 r+2) \equiv Q(\) say \()\)
We have \(3 x+4 y+z-8=0\)
\(\therefore 3(3 r+7)+4(6 r+5)+(2 r+2)-8=0\)
\(\therefore 9 r+24 r+2 r+21+20+2-8=0 \Rightarrow\) \(35 r=-35 \Rightarrow r=-1\)
\(\therefore Q \equiv(-3+7,-6+5,-2+2)\) i.e. \((4,-1,0)\)
Distance between \(\mathrm{PQ} \equiv \sqrt{(7-4)^{2}+(5+1)^{2}+(2-0)^{2}}=\) \(\sqrt{9+36+4}=7\)