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Question: The image of the point \(P(1,3,4)\) in the plane \(2x - y + z + 3 = 0\),is A - (3, - 5,2) B - ...

The image of the point P(1,3,4)P(1,3,4) in the plane 2xy+z+3=02x - y + z + 3 = 0,is
A - (3, - 5,2)
B - (3,5, - 2)
C - (3,5,2)
D - ( - 3,5,2)

Explanation

Solution

In this particular sum the student is expected to first find the coordinates of the point on the plane. Then assume the Image of point P. The point obtained on the plane would be mid-point for the Line PQ, where Q is the coordinates of the image of Point P. Using the midpoint formula the student can compute the coordinates of the image of point P.

Complete answer:
Let P(1,3,4)P(1,3,4) be the point.
Let OO be a point on the plane.
It is given that the equation of the plane is 2xy+z+3=02x - y + z + 3 = 0.
Now , since we have the equation of the plane and the P(1,3,4)P(1,3,4), the equation of the plane in another form can be written as
x12=y31=z41=k\dfrac{{x - 1}}{2} = \dfrac{{y - 3}}{{ - 1}} = \dfrac{{z - 4}}{1} = k
Any point on the above line, PO is of the form
x=2k+1.........(1) y=k+3.........(2) z=k+4.........(3) \begin{gathered} x = 2k + 1.........(1) \\\ y = - k + 3.........(2) \\\ z = k + 4.........(3) \\\ \end{gathered}
Substituting the above values of x,y,zx,y,z in the equation of the plane we get following equation
2(2k+1)(k+3)+(k+4)+3=02(2k + 1) - ( - k + 3) + (k + 4) + 3 = 0
Simplifying the above equation, we will get the value of kk
4k+2+k3+k+4+3=04k + 2 + k - 3 + k + 4 + 3 = 0
6k+6=0\therefore 6k + 6 = 0
k=1\therefore k = - 1
Substituting the value of kk in equation 1,2,31,2,3 we get the coordinates of point OO
x=2(1)+1.........(4) y=(1)+3.........(5) z=1+4.........(6) \begin{gathered} x = 2( - 1) + 1.........(4) \\\ y = - ( - 1) + 3.........(5) \\\ z = - 1 + 4.........(6) \\\ \end{gathered}
Coordinates of Point OO are (1,4,3)( - 1,4,3).
Let Q(x1,y1,z1)Q({x_1},{y_1},{z_1}) be the image of point P. Since it is an image we can say that Point O is the midpoint of line PQ.
Using the Midpoint formula we get the coordinates of the Q(x1,y1,z1)Q({x_1},{y_1},{z_1}) be the image of point P.
\-1=x1+12 4=y1+32 3=4+z12 \begin{gathered} \- 1 = \dfrac{{{x_1} + 1}}{2} \\\ 4 = \dfrac{{{y_1} + 3}}{2} \\\ 3 = \dfrac{{4 + {z_1}}}{2} \\\ \end{gathered}
x1=3 y1=5 z1=2 \begin{gathered} {x_1} = - 3 \\\ {y_1} = 5 \\\ {z_1} = 2 \\\ \end{gathered}
The image of the Point P has coordinates (3,5,2)( - 3,5,2)

Answer is Option D(3,5,2)D - ( - 3,5,2)

Note: The student should have a strong imaginative power in order to solve this sums quickly. Another way to improve on these types of sums is by remembering the type and then following the same steps as and when the sums of these types are asked. Also the student should make sure that he doesn’t confuse between the equation of a plane and equation of a line.