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3D Geometry question

2023 · 6 Apr · Shift 2 · Q42
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3D Geometry question

2023 · 6 Apr · Shift 2 · Q42

JEE MainMathematics3D GeometryNumerical+4 / −1
If the lines x−12=2−y−3=z−3α\frac{x-1}{2}=\frac{2-y}{-3}=\frac{z-3}{\alpha}2x−1​=−32−y​=αz−3​ and x−45=y−12=zβ\frac{x-4}{5}=\frac{y-1}{2}=\frac{z}{\beta}5x−4​=2y−1​=βz​ intersect, then the magnitude of the minimum value of 8αβ8 \alpha \beta8αβ is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 18

  1. Write the lines in parametric form

For the first line,

x−12=2−y−3=z−3α=t\frac{x-1}{2}=\frac{2-y}{-3}=\frac{z-3}{\alpha}=t2x−1​=−32−y​=αz−3​=t

So,

Thus line 1 is

(x,y,z)=(1,2,3)+t(2,3,α).(x,y,z)=(1,2,3)+t(2,3,\alpha).(x,y,z)=(1,2,3)+t(2,3,α).

For the second line,

x−45=y−12=zβ=s\frac{x-4}{5}=\frac{y-1}{2}=\frac{z}{\beta}=s5x−4​=2y−1​=βz​=s

So,

Thus line 2 is

(x,y,z)=(4,1,0)+s(5,2,β).(x,y,z)=(4,1,0)+s(5,2,\beta).(x,y,z)=(4,1,0)+s(5,2,β).
  1. Use the condition that the lines intersect

If the lines intersect, then for some parameters t,st,st,s,

1+2t=4+5s...(1)1+2t=4+5s \quad ...(1)1+2t=4+5s...(1) 2+3t=1+2s...(2)2+3t=1+2s \quad ...(2)2+3t=1+2s...(2) 3+αt=βs...(3)3+\alpha t=\beta s \quad ...(3)3+αt=βs...(3)

From (1),

2t−5s=3.2t-5s=3.2t−5s=3.

From (2),

3t−2s=−1.3t-2s=-1.3t−2s=−1.

Solve these two equations:

Multiply the first by 333:

6t−15s=96t-15s=96t−15s=9

Multiply the second by 222:

6t−4s=−26t-4s=-26t−4s=−2

Subtract:

−11s=11⇒s=−1.-11s=11 \Rightarrow s=-1.−11s=11⇒s=−1.

Then from 3t−2s=−13t-2s=-13t−2s=−1,

3t−2(−1)=−13t-2(-1)=-13t−2(−1)=−1 3t+2=−13t+2=-13t+2=−1 3t=−3⇒t=−1.3t=-3 \Rightarrow t=-1.3t=−3⇒t=−1.

So the intersection occurs at t=s=−1.t=s=-1.t=s=−1.

  1. Relate α\alphaα and β\betaβ

Using equation (3):

3+α(−1)=β(−1)3+\alpha(-1)=\beta(-1)3+α(−1)=β(−1) 3−α=−β3-\alpha=-\beta3−α=−β α−β=3.\alpha-\beta=3.α−β=3.

So,

α=β+3.\alpha=\beta+3.α=β+3.
  1. Minimize 8αβ8\alpha\beta8αβ

We need the magnitude of the minimum value of 8αβ8\alpha\beta8αβ subject to

α−β=3.\alpha-\beta=3.α−β=3.

Let

α=β+3.\alpha=\beta+3.α=β+3.

Then

8αβ=8β(β+3)=8(β2+3β).8\alpha\beta=8\beta(\beta+3)=8(\beta^2+3\beta).8αβ=8β(β+3)=8(β2+3β).

Complete the square:

β2+3β=(β+32)2−94.\beta^2+3\beta=\left(\beta+\frac{3}{2}\right)^2-\frac{9}{4}.β2+3β=(β+23​)2−49​.

Hence,

8αβ=8[(β+32)2−94].8\alpha\beta=8\left[\left(\beta+\frac{3}{2}\right)^2-\frac{9}{4}\right].8αβ=8[(β+23​)2−49​].

The minimum occurs when

β=−32,α=32.\beta=-\frac{3}{2},\quad \alpha=\frac{3}{2}.β=−23​,α=23​.

Then

8αβ=8⋅32⋅(−32)=8⋅(−94)=−18.8\alpha\beta=8\cdot \frac{3}{2}\cdot\left(-\frac{3}{2}\right)=8\cdot\left(-\frac{9}{4}\right)=-18.8αβ=8⋅23​⋅(−23​)=8⋅(−49​)=−18.

So the minimum value is −18-18−18, and its magnitude is

∣−18∣=18.|-18|=18.∣−18∣=18.
  1. Final answer

The magnitude of the minimum value of 8αβ8\alpha\beta8αβ is

18.\boxed{18}.18​.

This matches the stored correct answer.

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