Three Dimensional Geometry: Mathematics | JEE Main
Let the image of the point P(1,2,3) in the plane 2x−y+z=9 be Q. If the coordinates of the point R are (6,10,7), then the square of the area of the triangle PQR is_
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Hint 1 of 3
Using the reflection formula ax−x1=by−y1=cz−z1=−2a2+b2+c2ax1+by1+cz1+d, what are the coordinates of the image point Q of P(1,2,3) in the plane 2x−y+z−9=0?
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Correct answer
The square of the area of triangle PQR is 594.
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Solution
StepWorking
01given
Point P(1,2,3), plane 2x−y+z=9, and point R(6,10,7). Q is the image of P in the plane.
02goal
Find the square of the area of ΔPQR, i.e., (Area(ΔPQR))2.
03approach
Use the standard formula for the image of a point (x1,y1,z1) in a plane ax+by+cz+d=0:
ax−x1=by−y1=cz−z1=a2+b2+c2−2(ax1+by1+cz1+d)
Then, calculate PQ and PR, find their cross product, and calculate Area2=41∣PQ×PR∣2.
04execute
Let Q=(α,β,γ).
2α−1=−1β−2=1γ−3=22+(−1)2+12−2(2(1)−(2)+(3)−9)=4+1+1−2(3−9)=6−2(−6)=2
Thus:
α=1+2(2)=5β=2−1(2)=0γ=3+1(2)=5
So, Q=(5,0,5).
05execute
Find vectors PQ and PR:
PQ=(5−1)i^+(0−2)j^+(5−3)k^=4i^−2j^+2k^PR=(6−1)i^+(10−2)j^+(7−3)k^=5i^+8j^+4k^
06execute
Compute the cross product PQ×PR:
PQ×PR=i^45j^−28k^24=i^((−2)(4)−(2)(8))−j^((4)(4)−(2)(5))+k^((4)(8)−(−2)(5))=i^(−8−16)−j^(16−10)+k^(32+10)=−24i^−6j^+42k^
07execute
Area of ΔPQR=21∣PQ×PR∣=21∣−24i^−6j^+42k^∣=∣−12i^−3j^+21k^∣.
(Area)2=(−12)2+(−3)2+212=144+9+441=594
✓verify
Check the midpoint of PQ: M=(21+5,22+0,23+5)=(3,1,4).
Substitute M into plane equation: 2(3)−(1)+(4)=6−1+4=9. Midpoint lies on the plane.
Vector PQ=(4,−2,2)=2(2,−1,1), which is parallel to the normal vector n=(2,−1,1). Thus, Q is correct.
Hints that build this answer step by step
Using the reflection formula ax−x1=by−y1=cz−z1=−2a2+b2+c2ax1+by1+cz1+d, what are the coordinates of the image point Q of P(1,2,3) in the plane 2x−y+z−9=0?
Q(5,0,5)
What is the cross product of the vectors PQ and PR, where P(1,2,3), Q(5,0,5), and R(6,10,7)?
−24i^−6j^+42k^
Using the relation Area2=41∣PQ×PR∣2, what is the square of the area of △PQR?
The foot of perpendicular from a point to a plane is halfway along the normal vector, so it uses −1. The reflection (image) is twice as far along the normal vector, so it uses −2.