Question
Let the area of the triangle formed by the lines $x+2=y-1=z$, $\frac{x-3}{5} = \frac{y}{-1} = \frac{z-1}{1}$ and $\frac{x}{-3} = \frac{y-3}{3} = \frac{z-2}{1}$ be $A$. Then $A^2$ is equal to
Finding Triangle Area ($A^2$) from Three Lines
Line Equations and Parameters
The equations of the three lines are given. We first represent them in a consistent symmetric form to identify direction vectors and points.
- Line 1: $x+2=y-1=z$ can be written as $\frac{x-(-2)}{1} = \frac{y-1}{1} = \frac{z-0}{1}$.
- A point on the line: $P_1 = (-2, 1, 0)$.
- Direction vector: $\vec{d_1} = \langle 1, 1, 1 \rangle$.
- Line 2: $\frac{x-3}{5} = \frac{y}{-1} = \frac{z-1}{1}$.
- A point on the line: $P_2 = (3, 0, 1)$.
- Direction vector: $\vec{d_2} = \langle 5, -1, 1 \rangle$.
- Line 3: $\frac{x}{-3} = \frac{y-3}{3} = \frac{z-2}{1}$.
- A point on the line: $P_3 = (0, 3, 2)$.
- Direction vector: $\vec{d_3} = \langle -3, 3, 1 \rangle$.
Step 1: Find Vertices of the Triangle
The vertices of the triangle are the intersection points of the lines taken pairwise.
Intersection of Line 1 and Line 2 ($V_1$):
Using parametric forms $x = -2+t_1, y = 1+t_1, z = t_1$ and $x = 3+5t_2, y = -t_2, z = 1+t_2$. Equating coordinates yields $t_1=0$ and $t_2=-1$.
Vertex $V_1 = (-2, 1, 0)$.
Intersection of Line 1 and Line 3 ($V_2$):
Using parametric forms $x = -2+t_1, y = 1+t_1, z = t_1$ and $x = -3t_3, y = 3+3t_3, z = 2+t_3$. Equating coordinates yields $t_1=2$ and $t_3=0$.
Vertex $V_2 = (0, 3, 2)$.
Intersection of Line 2 and Line 3 ($V_3$):
Using parametric forms $x = 3+5t_2, y = -t_2, z = 1+t_2$ and $x = -3t_3, y = 3+3t_3, z = 2+t_3$. Equating coordinates yields $t_2=0$ and $t_3=-1$.
Vertex $V_3 = (3, 0, 1)$.
Step 2: Compute Area Vector
The area of the triangle can be found using the cross product of two vectors representing two sides of the triangle originating from the same vertex. Let's use vertex $V_1$.
Define vectors $\vec{a}$ and $\vec{b}$:
- $\vec{a} = \vec{V_1V_2} = V_2 - V_1 = \langle 0 - (-2), 3 - 1, 2 - 0 \rangle = \langle 2, 2, 2 \rangle$.
- $\vec{b} = \vec{V_1V_3} = V_3 - V_1 = \langle 3 - (-2), 0 - 1, 1 - 0 \rangle = \langle 5, -1, 1 \rangle$.
Calculate the cross product $\vec{a} \times \vec{b}$:
$ \vec{a} \times \vec{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 2 & 2 \\ 5 & -1 & 1 \end{vmatrix} $ $ = \mathbf{i}(2 \cdot 1 - 2 \cdot (-1)) - \mathbf{j}(2 \cdot 1 - 2 \cdot 5) + \mathbf{k}(2 \cdot (-1) - 2 \cdot 5) $ $ = \mathbf{i}(2 + 2) - \mathbf{j}(2 - 10) + \mathbf{k}(-2 - 10) $ $ = \langle 4, 8, -12 \rangle $Step 3: Calculate Area ($A$) and $A^2$
The area $A$ of the triangle is half the magnitude of the cross product vector.
Magnitude of $\vec{a} \times \vec{b}$:
$ |\vec{a} \times \vec{b}| = \sqrt{4^2 + 8^2 + (-12)^2} = \sqrt{16 + 64 + 144} = \sqrt{224} $Area $A$:
$ A = \frac{1}{2} |\vec{a} \times \vec{b}| = \frac{1}{2} \sqrt{224} = \frac{1}{2} \sqrt{16 \times 14} = \frac{1}{2} (4\sqrt{14}) = 2\sqrt{14} $Finally, calculate $A^2$:
$ A^2 = (2\sqrt{14})^2 = 4 \times 14 = 56 $The value of $A^2$ is 56.
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