Question

Find the image of point (5,2, -7) in XY-plane and image of point (-4,0,7) in XZ-plane.

Answer 1

Hint: We should know that equation of XY-plane is z=0, equation of YZ-plane is x=0 and equation of XZ-plane is y=0. We should know that a point \[A(h,k,l)\] is image of \[B({{x}_{1}},{{y}_{1}},{{z}_{1}})\] with respect to the plane \[ax+by+cz=d\] if \[\dfrac{h-{{x}_{1}}}{a}=\dfrac{k-{{y}_{1}}}{b}=\dfrac{l-{{z}_{1}}}{c}=\dfrac{-2(a{{x}_{1}}+b{{y}_{1}}+c)}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\]. Now we should find the image of point (5,2, -7) in XY-plane and image of point (-4,0,7) in XZ-plane by using the above formula.

Complete step-by-step answer:
From the question, we were given to find the image of point (5,2, -7) in XY-plane.
Before solving the problem, we should know that the equation of XY-plane is z=0, equation of YZ-plane is y=0 and equation of XZ-plane is y=0.
We should know that a point \[A(h,k,l)\] is image of \[B({{x}_{1}},{{y}_{1}},{{z}_{1}})\] with respect to the plane \[ax+by+cz=d\] if \[\dfrac{h-{{x}_{1}}}{a}=\dfrac{k-{{y}_{1}}}{b}=\dfrac{l-{{z}_{1}}}{c}=\dfrac{-2(a{{x}_{1}}+b{{y}_{1}}+c)}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\].
Now we should find the image of point B (5,2, -7) with respect to line z=0.
Now we should compare z=0 with \[ax+by+cz=d\].
We get a=0, b=0, c=1 and d=0.
Let us compare \[B({{x}_{1}},{{y}_{1}},{{z}_{1}})\] with B (5,2, -7), then \[{{x}_{1}}=5,{{y}_{1}}=2,{{z}_{1}}=-7\].
We should know that a point \[A(h,k,l)\] is image of \[B({{x}_{1}},{{y}_{1}},{{z}_{1}})\] with respect to the plane \[ax+by+cz=d\] if \[\dfrac{h-{{x}_{1}}}{a}=\dfrac{k-{{y}_{1}}}{b}=\dfrac{l-{{z}_{1}}}{c}=\dfrac{-2(a{{x}_{1}}+b{{y}_{1}}+c)}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\].
Let us assume the image of B (5,2, -7) on the XY-plane is \[A(h,k,l)\].
\[\begin{align}
  & \Rightarrow \dfrac{h-5}{0}=\dfrac{k-2}{0}=\dfrac{z+7}{1}=\dfrac{-2(-7)}{1} \\
 & \Rightarrow \dfrac{h-5}{0}=\dfrac{k-2}{0}=\dfrac{z+7}{1}=14 \\
 & \Rightarrow h=5,k=2,z=7 \\
\end{align}\]
So, it is clear that the image of point (5,2, -7) with respect to XY-plane is (5,2,7).
Now we should find the image of point (-4,0,7) with respect to line y=0.
Now we should compare y=0 with \[ax+by+cz=d\].
We get a=0, b=1, c=0 and d=0.
Let us compare \[B({{x}_{1}},{{y}_{1}},{{z}_{1}})\] with C (-4,0, 7), then \[{{x}_{1}}=-4,{{y}_{1}}=0,{{z}_{1}}=7\].
We should know that a point \[A(h,k,l)\] is image of \[B({{x}_{1}},{{y}_{1}},{{z}_{1}})\] with respect to the plane \[ax+by+cz=d\] if \[\dfrac{h-{{x}_{1}}}{a}=\dfrac{k-{{y}_{1}}}{b}=\dfrac{l-{{z}_{1}}}{c}=\dfrac{-2(a{{x}_{1}}+b{{y}_{1}}+c)}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\].
Let us assume the image of C (-4,0,7) with respect to line y=0.
\[\begin{align}
  & \Rightarrow \dfrac{h+4}{0}=\dfrac{k-0}{1}=\dfrac{z-7}{0}=\dfrac{-2(0)}{1} \\
 & \Rightarrow \dfrac{h+4}{0}=\dfrac{k-0}{1}=\dfrac{z-7}{0}=0 \\
 & \Rightarrow h=-4,k=0,z=7 \\
\end{align}\]
So, it is clear that the image of point (-4,0, 7) with respect to XZ-plane is (-4,0,7).

Note: This sum can be solved in an alternative method also.
We know that the image of a point \[B({{x}_{1}},{{y}_{1}},{{z}_{1}})\] with respect to XY-plane is \[({{x}_{1}},{{y}_{1}},-{{z}_{1}})\], the image of point \[B({{x}_{1}},{{y}_{1}},{{z}_{1}})\] with respect to YZ-plane is \[(-{{x}_{1}},{{y}_{1}},{{z}_{1}})\] and \[B({{x}_{1}},{{y}_{1}},{{z}_{1}})\] with respect to XZ-plane is \[({{x}_{1}},-{{y}_{1}},{{z}_{1}})\].
So, the image of point (5,2, -7) with respect to XY-plane is (5,2,7). The image of point C (4,0, -7) with respect to XZ-plane is (4,0, -7).