Let P be a plane passing through the points (2$$, $$1$$, $$0), (4$$, $$1$$, $$1) and (5$$, $$0$$,$$1) and R be any point (2$$, $$1$$,$$6). Then the image of R in the plane P is:

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Infinity Learn · Accepted Answer

The equation of the plane passing through three points (2$$, $$1$$, $$0), (4$$, $$1$$, $$1) and (5$$, $$0$$,$$1)  $$isx-2y-1z2013-11=0Expand$$ the determinant as below. (x-2)(0+1)-(y-1)(-1)+z(-2)=0$$⇒$$x-2+y-1-2z=0$$⇒$$x+y-2z=3Suppose$$ that the image of the point $$(2$$,$$1$$,$$6)  in the plane $$x+y-2z=3$$ is (h$$,k,$$l)If$$ the image of a point $$Px1$$,$$y1$$,$$z1$$ in the plane $$ax+by+cz+d=0$$ is h,k,l  then $$h-x1a=k-y1b=l-z1c=-2ax1+by1+ca2+b2Hence$$, $$h-21=k-11=l-6-2=-2(-12)6$$⇒$$(h$$,k,$$l)=(6$$,$$5$$,$$-2)

Let P be a plane passing through the points (2, 1, 0), (4, 1, 1) and (5, 0,1) and R be any point (2, 1,6). Then the image of R in the plane P is:

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