The distance of the point (1$$,$$0$$,−$$3) from the plane x−y−$$z=9$$ measured parallel to the line: x−$$22=y+23=z$$−$$6$$−$$6$$

Question

The distance of the point (1$$,$$0$$,−$$3) from the plane x−y−$$z=9$$ measured parallel to the line: x−$$22=y+23=z$$−$$6$$−$$6$$

Infinity Learn · Accepted Answer

The given plane is x−y−$$z=9-----(1)$$ The given line AB is x−$$22=y+23=z$$−$$6$$−$$6-------(2)$$ The equation of the line passing through (1$$,$$0$$,−$$3) and parallel  to x−$$22=y+23=z$$−$$6$$−$$6$$ is x−$$12=y$$−$$03=z+3$$−$$6=r-----(3)$$ Coordinate of any point on (3) may be given as $$P(2r+1$$ , $$3r$$,−$$6r$$−$$3)$$ If P is the point of the intersection of $$1$$ and,$$3$$ then it must  lie on (1). Therefore, (2r+1)−(3r)−$$($$−$$6r$$−$$3)=92r+1$$−$$3r+6r+3=9$$ or $$r=1$$ Therefore, the coordinates of P are $$3$$,$$3$$,−$$9$$   Distance between $$Q(1$$,$$0$$,−$$3)$$ and $$P(3$$,$$3$$,−$$9)=(3$$−$$1)2+(3$$−$$0)2+($$−$$9+3)2=4+9+36=7$$

$The distance of the point (1, 0, - 3) from the plane x - y - z = 9 measured parallel to the line: \frac{x - 2}{2} = \frac{y + 2}{3} = \frac{z - 6}{- 6}$

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The distance of the point (1,0,−3) from the plane x−y−z=9 measured parallel to the line: x−22=y+23=z−6−6

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$The distance of the point (1, 0, - 3) from the plane x - y - z = 9 measured parallel to the line: \frac{x - 2}{2} = \frac{y + 2}{3} = \frac{z - 6}{- 6}$