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InterviewSolution
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Find the distance of the plane 3x+4y-5z-7=0.(a) \(\frac{7}{\sqrt{40}}\)(b) \(\frac{6}{\sqrt{34}}\)(c) \(\frac{8}{\sqrt{50}}\)(d) \(\frac{7}{\sqrt{50}}\)I have been asked this question during an interview for a job.Origin of the question is Three Dimensional Geometry topic in section Three Dimensional Geometry of Mathematics – Class 12 |
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Answer» CORRECT answer is (d) \(\frac{7}{\sqrt{50}}\) To explain: From the given equation, the direction ratios of the normal to the plane are 3, 4, -5; the direction cosines are \(\frac{3}{\sqrt{3^2+4^2+(-5)^2}},\frac{4}{\sqrt{3^2+4^2+(-5)^2}},\frac{-5}{\sqrt{3^2+4^2+(-5)^2}}\),i.e. \(\frac{3}{\sqrt{50}},\frac{4}{\sqrt{50}},\frac{-5}{\sqrt{50}}\) Dividing the equation throughout by √50, we get \(\frac{3}{\sqrt{50}} x+\frac{4}{\sqrt{50}} y-\frac{5}{\sqrt{50}} z=\frac{7}{\sqrt{50}}\) The above equation is in the form of lx+my+nz=d, where d is the DISTANCE of the plane from the origin. So, the distance of the plane from the origin is \(\frac{7}{\sqrt{50}}\). |
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