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 $A(h,k,l)$.

& \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{aligned}$$ So, it is clear that the image of point (5,2, -7) with respect to XY-plane is (5,2,7). Now we should find the image of point (-4,0,7) with respect to line y=0. Now we should compare y=0 with $$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{aligned} & \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{aligned}$$ So, it is clear that the image of point (-4,0, 7) with respect to XZ-plane is (-4,0,7). Note: This sum can be solved in an alternative method also. We know that the image of a point $$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).