Question
Question: The image of the point having the position vector \[\hat i + 3\hat j + 4\hat k\] in the plane \[\ove...
The image of the point having the position vector i^+3j^+4k^ in the plane r×(2i^−j^+k^)+3=0
A) 3i^+2j^+k^
B) 3i^+5j^+2k^
C) −3i^+5j^+2k^
D) 3i^+2j^−5k^
Solution
We will use the given equation of the plane and write the Cartesian equation of the plane. Then we will assume that there is a point lying on the line. So, we will write the equation for that point. Then we will find the midpoint formula and on simplification, we will get the image of the given point.
Complete step by step solution:
The position vector i^+3j^+4k^ is given
So, we can see that the point is P(1,3,4)
The plane given is r×(2i^−j^+k^)+3=0
The cartesian equation of the plane is 2x−y+z+3=0
Let I(a,b,c) be the image of the point P in the plane.
Consider a point Q lying on the line IP, such that Q is the midpoint of IP.
Hence, PQ is also the line.
The direction ratios of the line PQ is given by
⇒2x−1=−1y−3=1z−4=u
The point on the line PQ is given by
⇒x=2u+1;y=3−u;z=u+4 …. (1)
Substituting the values of x, y and z in the equation
⇒2(2u+1)−(−u+3)+u+4+3=0
On simplification, we will get the value of u
⇒u=−1
We put the value of u in equation
⇒x=2×(−1)+1;y=3−(−1);z=(−1)+4
On simplification we get
⇒x=−1;y=4;z=−3
Hence the point Q is
Q(−1,4,3)
Q is the midpoint of PI. Then we use midpoint formula and we get
⇒2x+1=−1;2y+3=4;2z+4=3
By cross multiplication, we get
⇒x=−3,y=5,z=2
The image of the point P(1,3,4) in the plane r×(2i^−j^+k^)+3=0 is (−3,5,2).
Note:
You can make mistakes in writing the equation of the line and plane. You can get confused when you write the equation for direction ratios, so avoid those small mistakes when solving this type of problem.