Three lines are given by $$\begin{aligned} & \vec{r}=\lambda \hat{i}, \lambda \in \mathbb{R} \\ & \vec{r}=\mu(\hat{i}+\hat{j}), \mu \in \mathbb{R} \text { and } \\ & \vec{r}=v(\hat{i}+\hat{j}+\hat{k}), v \in \mathbb{R} . \end{aligned}$$ Let the lines cut the plane $\mathrm{x}+\mathrm{y}+\mathrm{z}=1$ at the points $\mathrm{A}, \mathrm{B}$ and C respectively. If the area of the triangle ABC is $\Delta$ then the value of $(6 \Delta)^{2}$ equals ____

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Three Dimensional Geometry

Three lines are given by

r = λ \hat{i}, λ \in R r = μ (\hat{i} + \hat{j}), μ \in R and r = v (\hat{i} + \hat{j} + \hat{k}), v \in 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

then the value of

(6Δ)^{2}

equals ____

Answer: 0.75

Finding intersection points with the plane

x + y + z = 1

:

For line 1:

(λ, 0, 0)

gives

λ = 1

, so

A (1, 0, 0)

For line 2:

(μ, μ, 0)

gives

2 μ = 1

, so

B (\frac{1}{2}, \frac{1}{2}, 0)

For line 3:

(v, v, v)

gives

3 v = 1

, so

C (\frac{1}{3}, \frac{1}{3}, \frac{1}{3})

Vectors:

AB = - \frac{1}{2} \hat{i} + \frac{1}{2} \hat{j}, AC = - \frac{2}{3} \hat{i} + \frac{1}{3} \hat{j} + \frac{1}{3} \hat{k}

Cross product:

AB \times AC = \frac{1}{6} \hat{i} + \frac{1}{6} \hat{j} - \frac{1}{6} \hat{k}

Area:

Δ = \frac{1}{2} ∣ AB \times AC ∣ = \frac{1}{2} \cdot \frac{3}{6} = \frac{3}{12}

Therefore:

(6Δ)^{2} = (\frac{3}{2})^{2} = \frac{3}{4} = 0.75

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Numerical answer