+4 marks−1 if incorrectNumericalPrevious-year question
Intersection of a Line and Plane, Foot of a Perpendicular, and Triangle Area
Let the line L:(x-1)/2=(y+1)/(-1)=(z-3)/1 intersect the plane 2x+y+3z=16 at the point P. Let the point Q be the foot of perpendicular from the point R(1,-1,-3) on the line L. If α is the area of triangle PQR, then α^2 is equal to
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Hint 1 of 4
What is the strategic roadmap to compute the area squared, α2, of △PQR?
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Step-by-step solutionView
Correct answer
The value of α2 is 180.
Option analysis
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Reviewed route
Solution
StepWorking
01Given
Line L:2x−1=−1y+1=1z−3, plane Π:2x+y+3z=16, and point R(1,−1,−3). Point P is L∩Π, Q is the foot of perpendicular from R on L, and α is the area of △PQR.
02Goal
Find the squared area of △PQR, denoted by α2.
03Approach
Parameterize any point on L as (2λ+1,−λ−1,λ+3). Substitute into the plane equation to find P. Then find parameter λ corresponding to foot Q by setting QR⋅dL=0. Since Q is on L and RQ⊥L, △PQR is a right triangle at Q. The area is α=21∣QP∣∣QR∣, so α2=41∣QP∣2∣QR∣2.
04Execute
Find P: substitute (2λ+1,−λ−1,λ+3) into 2x+y+3z=16:
2(2λ+1)+(−λ−1)+3(λ+3)=16⟹6λ+10=16⟹λ=1.
Thus, P=(3,−2,4).
05Execute
Find Q: let Q=(2μ+1,−μ−1,μ+3).
Vector RQ=Q−R=(2μ,−μ,μ+6).
Since RQ⊥dL=(2,−1,1):
2(2μ)−1(−μ)+1(μ+6)=0⟹6μ+6=0⟹μ=−1.
Thus, Q=(−1,0,2).
06Execute
Since P,Q lie on line L and RQ⊥L, △PQR has a right angle at Q.
∣QP∣2=(3−(−1))2+(−2−0)2+(4−2)2=42+(−2)2+22=16+4+4=24.
∣QR∣2=(1−(−1))2+(−1−0)2+(−3−2)2=22+(−1)2+(−5)2=4+1+25=30.
Then α2=41∣QP∣2∣QR∣2=41×24×30=6×30=180.
✓Verify
Check cross product: QP=2dL=(4,−2,2) and QR=(−2,1,5) (or 2i^−j^−5k^). QP⋅QR=4(−2)+(−2)(1)+2(5)=−8−2+10=0 (verified perpendicular). Cross product magnitude squared ∣QP×QR∣2=∣(−12,−24,0)∣2=144+576=720. Then α2=4720=180.
Hints that build this answer step by step
What is the strategic roadmap to compute the area squared, α2, of △PQR?
Find point P by intersecting line L with the plane, find foot Q by projecting R onto L, and evaluate α2=41∣PQ∣2∣QR∣2 since ∠PQR=90∘.
A general point on line L has coordinates (2t+1,−t−1,t+3). What are the coordinates of the intersection point P with the plane 2x+y+3z=16?
P(3,−2,4)
What is the parameter λ corresponding to point Q(2λ+1,−λ−1,λ+3), the foot of perpendicular from R(1,−1,−3) onto L?
λ=−1
With P(3,−2,4), Q(−1,0,2), and R(1,−1,−3), what is the value of α2?
Why can we just use 1/2 * base * height instead of the cross product?
Because Q is explicitly the foot of perpendicular from R to line L, and P lies on line L. Therefore, RQ is perpendicular to QP, making △PQR a right-angled triangle at Q.
