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3D Geometry question

2022 · 26 Jul · Shift 1 · Q45
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  5. /2022 · 26 Jul · Shift 1 · Q45

3D Geometry question

2022 · 26 Jul · Shift 1 · Q45

JEE MainMathematics3D GeometryNumerical+4 / −1
Let Q\mathrm{Q}Q and R\mathrm{R}R be two points on the line x+12=y+23=z−12\frac{x+1}{2}=\frac{y+2}{3}=\frac{z-1}{2}2x+1​=3y+2​=2z−1​ at a distance 26\sqrt{26}26​ from the point P(4,2,7)P(4,2,7)P(4,2,7). Then the square of the area of the triangle PQRP Q RPQR is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 153

  1. Write the line in parametric form

Given x+12=y+23=z−12=t\frac{x+1}{2}=\frac{y+2}{3}=\frac{z-1}{2}=t2x+1​=3y+2​=2z−1​=t so a general point on the line is Q(t)=(−1+2t, −2+3t, 1+2t).Q(t)=(-1+2t,\,-2+3t,\,1+2t).Q(t)=(−1+2t,−2+3t,1+2t).

Let a point on the line be A(t)A(t)A(t). We are told that its distance from P(4,2,7)P(4,2,7)P(4,2,7) is 26\sqrt{26}26​. Hence PA2=26.PA^2=26.PA2=26.

  1. Form the distance equation

Coordinates difference: A(t)−P=(−1+2t−4, −2+3t−2, 1+2t−7)=(2t−5, 3t−4, 2t−6).A(t)-P=(-1+2t-4,\,-2+3t-2,\,1+2t-7)=(2t-5,\,3t-4,\,2t-6).A(t)−P=(−1+2t−4,−2+3t−2,1+2t−7)=(2t−5,3t−4,2t−6).

Thus (2t−5)2+(3t−4)2+(2t−6)2=26.(2t-5)^2+(3t-4)^2+(2t-6)^2=26.(2t−5)2+(3t−4)2+(2t−6)2=26.

Expand: (4t2−20t+25)+(9t2−24t+16)+(4t2−24t+36)=26.(4t^2-20t+25)+(9t^2-24t+16)+(4t^2-24t+36)=26.(4t2−20t+25)+(9t2−24t+16)+(4t2−24t+36)=26.

So 17t2−68t+77=2617t^2-68t+77=2617t2−68t+77=26 17t2−68t+51=0.17t^2-68t+51=0. 17t2−68t+51=0.

Divide by 171717: t2−4t+3=0t^2-4t+3=0t2−4t+3=0 (t−1)(t−3)=0.(t-1)(t-3)=0.(t−1)(t−3)=0.

Hence the two points are obtained for t=1,t=3.t=1,\quad t=3.t=1,t=3.

So Q=(1,1,3),R=(5,7,7).Q=(1,1,3),\qquad R=(5,7,7).Q=(1,1,3),R=(5,7,7).

  1. Find the area of triangle PQRPQRPQR

Take vectors from PPP: PQ→=Q−P=(1−4,1−2,3−7)=(−3,−1,−4),\overrightarrow{PQ}=Q-P=(1-4,1-2,3-7)=(-3,-1,-4),PQ​=Q−P=(1−4,1−2,3−7)=(−3,−1,−4), PR→=R−P=(5−4,7−2,7−7)=(1,5,0).\overrightarrow{PR}=R-P=(5-4,7-2,7-7)=(1,5,0).PR=R−P=(5−4,7−2,7−7)=(1,5,0).

Area of triangle: Δ=12∣PQ→×PR→∣.\Delta = \frac12\left|\overrightarrow{PQ}\times\overrightarrow{PR}\right|.Δ=21​​PQ​×PR​.

Compute the cross product:

=\begin{vmatrix} \mathbf i & \mathbf j & \mathbf k\\ -3 & -1 & -4\\ 1 & 5 & 0 \end{vmatrix}.$$ $$=\mathbf i\big((-1)(0)-(-4)(5)\big)-\mathbf j\big((-3)(0)-(-4)(1)\big)+\mathbf k\big((-3)(5)-(-1)(1)\big)$$ $$=(20,-4,-14).$$ Its magnitude squared is $$20^2+(-4)^2+(-14)^2=400+16+196=612.$$ Therefore $$\Delta^2=\frac14\cdot 612=153.$$ 4. **Final answer** The square of the area of triangle $PQR$ is $$\boxed{153}.$$ 5. **Comparison with stored answer** Stored correct answer = $153$. Our derived answer matches it exactly.
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