If the projection of the line $$x2=y$$−$$12=z$$−$$11$$ on a plane P is $$x1=y$$−$$11=z$$−$$1$$−$$1$$. Then the distance of plane P from origin is

Question

If the projection of the line $$x2=y$$−$$12=z$$−$$11$$ on a plane P is $$x1=y$$−$$11=z$$−$$1$$−$$1$$. Then the distance of plane P from origin is

Infinity Learn · Accepted Answer

Projection of line $$x2=y$$−$$12=z$$−$$11$$ on a plane P is $$x1=y$$−$$11=z$$−$$1$$−$$1$$ Plane through these lines is perpendicular to the plane P Normal to the plane determined by the given lines $$isi^$$    $$j^$$    $$k^2$$    $$2$$    $$11$$    $$1$$    −$$1=$$−$$3i^+3j^Direction$$ ratios normal to the required plane is Equation of the plane is $$x+y+2z$$−$$3=0$$ as it passes through $$0$$,$$1$$,$$1$$ Its distance from origin is $$32$$.

$If the projection of the line \frac{x}{2} = \frac{y - 1}{2} = \frac{z - 1}{1} on a plane P is \frac{x}{1} = \frac{y - 1}{1} = \frac{z - 1}{- 1} . Then the distance of plane P from origin is$

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If the projection of the line x2=y−12=z−11 on a plane P is x1=y−11=z−1−1. Then the distance of plane P from origin is

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$If the projection of the line \frac{x}{2} = \frac{y - 1}{2} = \frac{z - 1}{1} on a plane P is \frac{x}{1} = \frac{y - 1}{1} = \frac{z - 1}{- 1} . Then the distance of plane P from origin is$