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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.(b) x + y + z = 1

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

  (b)     x + y + z = 1 

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Given the equation of the plane is x+y+z=1 or we can write 1x+1y+1z=1

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

Therefore \sqrt{1^2+1^2+1^2} =\sqrt{3}

Then dividing both sides of the plane equation by \sqrt{3}, we get

\frac{x}{\sqrt3}+\frac{y}{\sqrt3}+\frac{z}{\sqrt3}=\frac{1}{\sqrt3}

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{1}{\sqrt3},\ \frac{1}{\sqrt3},\ \frac{1}{\sqrt3} and the distance of the plane from the origin is \frac{1}{\sqrt3} units.

Posted by

Divya Prakash Singh

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