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  • In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin. (c) 2 x+ 3 y-z = 5

1    In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

  (c)       2x + 3y - z = 5

Answers (1)

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Given the equation of plane is 2x+3y-z=5  

So, the direction ratios of normal from the above equation are, 2,\3,\ and\ -1.

Therefore \sqrt{2^2+3^2+(-1)^2} =\sqrt{14}

Then dividing both sides of the plane equation by \sqrt{14}, we get

\frac{2x}{\sqrt{14}}+\frac{3y}{\sqrt{14}}-\frac{z}{\sqrt{14}}=\frac{5}{\sqrt{14}}

So, this is the form of  lx+my+nz = d the plane, where l,\ m,\ n are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.

\therefore The direction cosines of the given line are \frac{2}{\sqrt{14}},\ \frac{3}{\sqrt{14}},\ \frac{-1}{\sqrt{14}} and the distance of the plane from the origin is \frac{5}{\sqrt{14}} units.

Posted by

Divya Prakash Singh

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