Write the distances of the point (7, −2, 3) from XY, YZ and XZ-planes.

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Solution

$The distance of a general point P (x, y, z) from X Y - plane is z . Thus, distance of (7, - 2, 3) from X Y - plane is 3 . Similarly, the distance of P (x, y, z) from Y Z - plane is x . Thus, distance of (7, - 2, 3) from Y Z - plane is 7 . The distance of P (x, y, z) from X Z - plane is y . Thus, distance of (7, - 2, 3) from X Z - plane is 2 .$

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Perpendicular distance of P(x, y, z ) from XY, YZ and XZ planes respectively are 1, 2 and 3. Which of the following could be the coordinates of P?

Find the image of :

(i) (-2, 3, 4) in the yz-plane.

(ii) (-5, 4, -3) in the xz-plane.

(iii) (5, 2, -7) in the xy-plane.

(iv)(-5, 0, 3) in the xz-plane.

(v)(-4, 0, 0) in the xy-plane.

x y -

y z -

z x -

3 - x y + y z - x z .

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