Q1 of 230 Page 30

Find the image of the point (0, 0, 0) in the plane 3x + 4y – 6z + 1 = 0.

Let point P = (0, 0, 0) and M be the image of P in the plane 3x + 4y – 6z + 1 = 0.


Direction ratios of PM are proportional to 3, 4, – 6 as PM is normal to the plane.


Recall the equation of the line passing through (x1, y1, z1) and having direction ratios proportional to l, m, n is given by



Here, (x1, y1, z1) = (0, 0, 0) and (l, m, n) = (3, 4, –6)


Hence, the equation of PM is




⇒ x = 3α, y = 4α, z = –6α


Let M = (3α, 4α, –6α).


As M is the image of P in the given plane, the midpoint of PM lies on the plane.


Using the midpoint formula, we have




This point lies on the given plane, which means this point satisfies the plane equation.







We have M = (3α, 4α, –6α)




Thus, the image of (0, 0, 0) in the plane 3x + 4y – 6z + 1 = 0 is.


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