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3D Geometry question

2024 · 6 Apr · Shift 2 · Q40
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  5. /2024 · 6 Apr · Shift 2 · Q40

3D Geometry question

2024 · 6 Apr · Shift 2 · Q40

JEE MainMathematics3D GeometryMCQ+4 / −1
Let P(α,β,γ)\mathrm{P}(\alpha, \beta, \gamma)P(α,β,γ) be the image of the point Q(3,−3,1)\mathrm{Q}(3,-3,1)Q(3,−3,1) in the line x−01=y−31=z−1−1\frac{x-0}{1}=\frac{y-3}{1}=\frac{z-1}{-1}1x−0​=1y−3​=−1z−1​ and R\mathrm{R}R be the point (2,5,−1)(2,5,-1)(2,5,−1). If the area of the triangle PQR\mathrm{PQR}PQR is λ\lambdaλ and λ2=14 K\lambda^2=14 \mathrm{~K}λ2=14 K, then K\mathrm{K}K is equal to :
  1. A
    18
  2. B
    81
  3. C
    72
  4. D
    36
View written solutionFree

Correct answer: B

  1. Interpret the line

The line is

x1=y−31=z−1−1=t\frac{x}{1}=\frac{y-3}{1}=\frac{z-1}{-1}=t1x​=1y−3​=−1z−1​=t

So its parametric form is

x=t,y=3+t,z=1−t.x=t,\quad y=3+t,\quad z=1-t.x=t,y=3+t,z=1−t.

Hence, a point on the line is A=(0,3,1)A=(0,3,1)A=(0,3,1) and a direction vector is d⃗=(1,1,−1).\vec d=(1,1,-1).d=(1,1,−1).

  1. Find the image of Q(3,−3,1)Q(3,-3,1)Q(3,−3,1) in the line

“Image of a point in a line” means reflection of the point about the line.

Let MMM be the foot of the perpendicular from QQQ to the line. Then MMM lies on the line, so

M=(t,3+t,1−t).M=(t,3+t,1-t).M=(t,3+t,1−t).

Also,

QM→=M−Q=(t−3, t+6, −t).\overrightarrow{QM}=M-Q=(t-3,\, t+6,\, -t).QM​=M−Q=(t−3,t+6,−t).

Since QM⊥QM \perpQM⊥ line, we must have

QM→⋅d⃗=0.\overrightarrow{QM}\cdot \vec d=0.QM​⋅d=0.

Thus,

(t−3)(1)+(t+6)(1)+(−t)(−1)=0(t-3)(1)+(t+6)(1)+(-t)(-1)=0(t−3)(1)+(t+6)(1)+(−t)(−1)=0 t−3+t+6+t=0t-3+t+6+t=0t−3+t+6+t=0 3t+3=03t+3=03t+3=0 t=−1.t=-1.t=−1.

So,

M=(−1,2,2).M=(-1,2,2).M=(−1,2,2).

Now MMM is the midpoint of QPQPQP, because PPP is the reflection of QQQ in the line. Hence

P=2M−Q.P=2M-Q.P=2M−Q.

So,

P=2(−1,2,2)−(3,−3,1)=(−2,4,4)−(3,−3,1)=(−5,7,3).P=2(-1,2,2)-(3,-3,1)=(-2,4,4)-(3,-3,1)=(-5,7,3).P=2(−1,2,2)−(3,−3,1)=(−2,4,4)−(3,−3,1)=(−5,7,3).

Thus,

P(α,β,γ)=(−5,7,3).P(\alpha,\beta,\gamma)=(-5,7,3).P(α,β,γ)=(−5,7,3).
  1. Find the area of triangle PQRPQRPQR

Given

Q=(3,−3,1),R=(2,5,−1).Q=(3,-3,1),\qquad R=(2,5,-1).Q=(3,−3,1),R=(2,5,−1).

Take vectors from QQQ:

QP→=P−Q=(−5−3, 7−(−3), 3−1)=(−8,10,2),\overrightarrow{QP}=P-Q=(-5-3,\,7-(-3),\,3-1)=(-8,10,2),QP​=P−Q=(−5−3,7−(−3),3−1)=(−8,10,2), QR→=R−Q=(2−3, 5−(−3), −1−1)=(−1,8,−2).\overrightarrow{QR}=R-Q=(2-3,\,5-(-3),\,-1-1)=(-1,8,-2).QR​=R−Q=(2−3,5−(−3),−1−1)=(−1,8,−2).

Area of triangle PQRPQRPQR is

λ=12∣QP→×QR→∣.\lambda=\frac12\left|\overrightarrow{QP}\times \overrightarrow{QR}\right|.λ=21​​QP​×QR​​.

Compute the cross product:

QP→×QR→=∣i^j^k^−8102−18−2∣\overrightarrow{QP}\times \overrightarrow{QR} =\begin{vmatrix} \hat i & \hat j & \hat k\\ -8 & 10 & 2\\ -1 & 8 & -2 \end{vmatrix}QP​×QR​=​i^−8−1​j^​108​k^2−2​​ =i^(10⋅(−2)−2⋅8)−j^((−8)(−2)−2(−1))+k^((−8)(8)−10(−1))=\hat i(10\cdot(-2)-2\cdot 8)-\hat j((-8)(-2)-2(-1))+\hat k((-8)(8)-10(-1))=i^(10⋅(−2)−2⋅8)−j^​((−8)(−2)−2(−1))+k^((−8)(8)−10(−1)) =i^(−20−16)−j^(16+2)+k^(−64+10)=\hat i(-20-16)-\hat j(16+2)+\hat k(-64+10)=i^(−20−16)−j^​(16+2)+k^(−64+10) =(−36,−18,−54).=(-36,-18,-54).=(−36,−18,−54).

So,

∣QP→×QR→∣=(−36)2+(−18)2+(−54)2\left|\overrightarrow{QP}\times \overrightarrow{QR}\right| =\sqrt{(-36)^2+(-18)^2+(-54)^2}​QP​×QR​​=(−36)2+(−18)2+(−54)2​ =1296+324+2916=4536.=\sqrt{1296+324+2916}= \sqrt{4536}.=1296+324+2916​=4536​.

Now,

4536=324⋅14,4536=324\cdot 14,4536=324⋅14,

so

∣QP→×QR→∣=1814.\left|\overrightarrow{QP}\times \overrightarrow{QR}\right|=18\sqrt{14}.​QP​×QR​​=1814​.

Therefore,

λ=12(1814)=914.\lambda=\frac12(18\sqrt{14})=9\sqrt{14}.λ=21​(1814​)=914​.

Hence,

λ2=(914)2=81⋅14.\lambda^2=(9\sqrt{14})^2=81\cdot 14.λ2=(914​)2=81⋅14.

Given

λ2=14K,\lambda^2=14K,λ2=14K,

therefore

K=81.K=81.K=81.
  1. Check with options

The correct option is

81\boxed{81}81​

which is Option B.

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