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3D Geometry question

2019 · Shift 1 · Q35
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3D Geometry question

2019 · Shift 1 · Q35

JEE AdvancedMathematics3D GeometryNumerical+3 / −1
Three lines are given by r=λi^, λ∈Rr = \lambda \widehat i,\,\lambda \in Rr=λi,λ∈R, r=μ(i^+j^), μ∈Rr = \mu (\widehat i + \widehat j),\,\mu \in Rr=μ(i+j​),μ∈R and r=v(i^+j^+k^), v ∈Rr = v(\widehat i + \widehat j + \widehat k),\,v\, \in Rr=v(i+j​+k),v∈R Let the lines cut the plane x + y + z = 1 at the points A, B and C respectively. If the area of the triangle ABC is Δ\DeltaΔ then the value of (6 Δ\DeltaΔ)2 equals ..............
Numerical answer
View written solutionFree

Correct answer: 0.75

  1. Find the points where the given lines meet the plane

The plane is x+y+z=1.x+y+z=1.x+y+z=1.

The three lines are:

  • r=λi^⇒(x,y,z)=(λ,0,0)r=\lambda \hat i \Rightarrow (x,y,z)=(\lambda,0,0)r=λi^⇒(x,y,z)=(λ,0,0)
  • r=μ(i^+j^)⇒(x,y,z)=(μ,μ,0)r=\mu(\hat i+\hat j) \Rightarrow (x,y,z)=(\mu,\mu,0)r=μ(i^+j^​)⇒(x,y,z)=(μ,μ,0)
  • r=v(i^+j^+k^)⇒(x,y,z)=(v,v,v)r=v(\hat i+\hat j+\hat k) \Rightarrow (x,y,z)=(v,v,v)r=v(i^+j^​+k^)⇒(x,y,z)=(v,v,v)

Now intersect each with the plane.

Point AAA: (λ,0,0)(\lambda,0,0)(λ,0,0) Substitute into x+y+z=1x+y+z=1x+y+z=1: λ=1\lambda=1λ=1 So, A=(1,0,0).A=(1,0,0).A=(1,0,0).

Point BBB: (μ,μ,0)(\mu,\mu,0)(μ,μ,0) Substitute into x+y+z=1x+y+z=1x+y+z=1: μ+μ=1⇒2μ=1⇒μ=12\mu+\mu=1 \Rightarrow 2\mu=1 \Rightarrow \mu=\frac12μ+μ=1⇒2μ=1⇒μ=21​ So, B=(12,12,0).B=\left(\frac12,\frac12,0\right).B=(21​,21​,0).

Point CCC: (v,v,v)(v,v,v)(v,v,v) Substitute into x+y+z=1x+y+z=1x+y+z=1: 3v=1⇒v=133v=1 \Rightarrow v=\frac133v=1⇒v=31​ So, C=(13,13,13).C=\left(\frac13,\frac13,\frac13\right).C=(31​,31​,31​).

  1. Form the side vectors of triangle ABCABCABC

Take AB→=B−A=(−12,12,0),\overrightarrow{AB}=B-A=\left(-\frac12,\frac12,0\right),AB=B−A=(−21​,21​,0), AC→=C−A=(−23,13,13).\overrightarrow{AC}=C-A=\left(-\frac23,\frac13,\frac13\right).AC=C−A=(−32​,31​,31​).

  1. Use cross product to find area

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

Compute the cross product:

\begin{vmatrix} \hat i & \hat j & \hat k\\ -\frac12 & \frac12 & 0\\ -\frac23 & \frac13 & \frac13 \end{vmatrix}$$ $$=\hat i\left(\frac12\cdot\frac13-0\cdot\frac13\right) -\hat j\left(-\frac12\cdot\frac13-0\cdot\left(-\frac23\right)\right) +\hat k\left(-\frac12\cdot\frac13-\frac12\cdot\left(-\frac23\right)\right)$$ $$=\left(\frac16,\frac16,\frac16\right).$$ Hence, $$\left|\overrightarrow{AB}\times\overrightarrow{AC}\right| =\sqrt{\left(\frac16\right)^2+\left(\frac16\right)^2+\left(\frac16\right)^2} =\frac{\sqrt3}{6}.$$ Therefore, $$\Delta=\frac12\cdot\frac{\sqrt3}{6}=\frac{\sqrt3}{12}.$$ --- 4. **Compute $(6\Delta)^2$** $$6\Delta=6\cdot\frac{\sqrt3}{12}=\frac{\sqrt3}{2}$$ So, $$(6\Delta)^2=\left(\frac{\sqrt3}{2}\right)^2=\frac34.$$ --- 5. **Compare with stored answer** Derived answer: $$\frac34=0.75$$ This matches the stored correct answer.
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