JEE Challenger
More from Three Dimensional Geometry
MathMedAdv 2019 • Paper 1

Determine Squared Scaled Area of Triangle Formed by Intersections in 3D Space

Three straight lines are defined as follows: r⃗=λi^,λ∈R\vec{r} = \lambda \hat{i}, \quad \lambda \in \mathbb{R} r⃗=μ(i^+j^),μ∈R\vec{r} = \mu (\hat{i} + \hat{j}), \quad \mu \in \mathbb{R} r⃗=ν(i^+j^+k^),ν∈R\vec{r} = \nu (\hat{i} + \hat{j} + \hat{k}), \quad \nu \in \mathbb{R}

Suppose these lines intersect the plane x+y+z=1x + y + z = 1 at the points AA, BB, and CC respectively. If Δ\Delta represents the area of triangle ABCABC, find the value of (6Δ)2(6\Delta)^2.

Official Numerical Answer0.75

Topics & Concepts

Three Dimensional GeometryLine in 3DPlane in 3D

Step-by-Step Solution

To find the value of (6Δ)2(6\Delta)^2, we first determine the position vectors of the points of intersection AA, BB, and CC of the three given lines with the plane x+y+z=1x + y + z = 1.

1. Point AA: The first line is given by: r⃗=(λ,0,0)\vec{r} = (\lambda, 0, 0) Substituting the coordinates (λ,0,0)(\lambda, 0, 0) into the equation of the plane x+y+z=1x + y + z = 1: λ+0+0=1  ⟹  λ=1\lambda + 0 + 0 = 1 \implies \lambda = 1 Thus, point AA is: A=(1,0,0)A = (1, 0, 0)

2. Point BB: The second line is given by: r⃗=(μ,μ,0)\vec{r} = (\mu, \mu, 0) Substituting the coordinates (μ,μ,0)(\mu, \mu, 0) into the plane equation: μ+μ+0=1  ⟹  2μ=1  ⟹  μ=12\mu + \mu + 0 = 1 \implies 2\mu = 1 \implies \mu = \frac{1}{2} Thus, point BB is: B=(12,12,0)B = \left(\frac{1}{2}, \frac{1}{2}, 0\right)

3. Point CC: The third line is given by: r⃗=(ν,ν,ν)\vec{r} = (\nu, \nu, \nu) Substituting the coordinates (ν,ν,ν)(\nu, \nu, \nu) into the plane equation: ν+ν+ν=1  ⟹  3ν=1  ⟹  ν=13\nu + \nu + \nu = 1 \implies 3\nu = 1 \implies \nu = \frac{1}{3} Thus, point CC is: C=(13,13,13)C = \left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)

4. Vectors AB⃗\vec{AB} and AC⃗\vec{AC}: AB⃗=(12−1)i^+(12−0)j^+(0−0)k^=−12i^+12j^\vec{AB} = \left(\frac{1}{2} - 1\right)\hat{i} + \left(\frac{1}{2} - 0\right)\hat{j} + (0 - 0)\hat{k} = -\frac{1}{2}\hat{i} + \frac{1}{2}\hat{j} AC⃗=(13−1)i^+(13−0)j^+(13−0)k^=−23i^+13j^+13k^\vec{AC} = \left(\frac{1}{3} - 1\right)\hat{i} + \left(\frac{1}{3} - 0\right)\hat{j} + \left(\frac{1}{3} - 0\right)\hat{k} = -\frac{2}{3}\hat{i} + \frac{1}{3}\hat{j} + \frac{1}{3}\hat{k}

5. Area of Triangle ABCABC (Δ\Delta): The vector area is given by the cross product AB⃗×AC⃗\vec{AB} \times \vec{AC}:

AB⃗×AC⃗=∣i^j^k^−12120−231313∣\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -\frac{1}{2} & \frac{1}{2} & 0 \\ -\frac{2}{3} & \frac{1}{3} & \frac{1}{3} \end{vmatrix}

Evaluating the determinant: AB⃗×AC⃗=i^(16−0)−j^(−16−0)+k^(−16−(−26))\vec{AB} \times \vec{AC} = \hat{i}\left(\frac{1}{6} - 0\right) - \hat{j}\left(-\frac{1}{6} - 0\right) + \hat{k}\left(-\frac{1}{6} - \left(-\frac{2}{6}\right)\right) AB⃗×AC⃗=16i^+16j^+16k^=16(i^+j^+k^)\vec{AB} \times \vec{AC} = \frac{1}{6}\hat{i} + \frac{1}{6}\hat{j} + \frac{1}{6}\hat{k} = \frac{1}{6}(\hat{i} + \hat{j} + \hat{k})

The area Δ\Delta is half the magnitude of this cross product: Δ=12∣AB⃗×AC⃗∣=12⋅1612+12+12=312\Delta = \frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{1}{2} \cdot \frac{1}{6} \sqrt{1^2 + 1^2 + 1^2} = \frac{\sqrt{3}}{12}

6. Calculation of (6Δ)2(6\Delta)^2: 6Δ=6⋅312=326\Delta = 6 \cdot \frac{\sqrt{3}}{12} = \frac{\sqrt{3}}{2} (6Δ)2=(32)2=34=0.75(6\Delta)^2 = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} = 0.75

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