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Question: Find the image of (5, 2,-7) in the xy-plane. (a) (5,-2,-7) (b) (-5, 2, 7) (c) (5, 2, 7) (d) ...

Find the image of (5, 2,-7) in the xy-plane.
(a) (5,-2,-7)
(b) (-5, 2, 7)
(c) (5, 2, 7)
(d) (5, 2,-7)

Explanation

Solution

e know that the image (g,h,k) of a point (p,q,r) to a general plane ax+by+cz+d=0 is given by:
g−pa=h−qb=k−rc=−2(ap+bq+cr+d)a2+b2+c2\dfrac{g-p}{a}=\dfrac{h-q}{b}=\dfrac{k-r}{c}=\dfrac{-2\left( ap+bq+cr+d \right)}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}} . We also know that the equation of xy-plane is z=0. So, use the formula to get the answer to the above question.

Complete step-by-step answer :
We know that the image (g,h,k) of a point (p,q,r) to a general plane ax+by+cz+d=0 is given by:
g−pa=h−qb=k−rc=−2(ap+bq+cr+d)a2+b2+c2\dfrac{g-p}{a}=\dfrac{h-q}{b}=\dfrac{k-r}{c}=\dfrac{-2\left( ap+bq+cr+d \right)}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}} .
We also know that the equation of the xy-plane is z=0. The point of which we need to find the image is (5,2,-7), so p=5, q=2 and r=-7. As the equation of the plane is z=0, we can say that a=0,b=0,c=1 and d=0. If we put all this data in the formula, we get
g−50=h−20=k−(−7)1=−2(−7)02+02+12\dfrac{g-5}{0}=\dfrac{h-2}{0}=\dfrac{k-\left( -7 \right)}{1}=\dfrac{-2\left( -7 \right)}{{{0}^{2}}+{{0}^{2}}+{{1}^{2}}}
⇒g−50=h−20=k+71=141\Rightarrow \dfrac{g-5}{0}=\dfrac{h-2}{0}=\dfrac{k+7}{1}=\dfrac{14}{1}
So, let us see the equation we can get out of the above expression.
g−50=141\dfrac{g-5}{0}=\dfrac{14}{1}
If we cross-multiply, we get
g−5=14×0g-5=14\times 0
⇒g=5\Rightarrow g=5
The other equation we get is:
h−20=141\dfrac{h-2}{0}=\dfrac{14}{1}
If we cross-multiply, we get
h−2=0h-2=0
⇒h=2\Rightarrow h=2
The third equation we get is:
k+7=14k+7=14
⇒k=7\Rightarrow k=7
So, the image (g,h,k) is (5,2,7). Hence, the answer to the above question is option (c).

Note :In questions related to lines and planes the key thing is to remember the important formulas, like the foot of perpendicular to the plane, image of a point etc. Generally, it is seen that people miss the negative sign or the 2 in the formula of the image of a point. If you want you can solve the above question directly if you know that if the image of any point (a,b,c) is asked about the xy-plane, the image comes out to be (a,b,-c), i.e., only the sign of the z-coordinate is reversed.