Let a line L passing through the point P(1, 1, 1) be perpendicular to the lines (x-4)/4 = (y-1)/1 =

Q: Why do we use the cross product to find the direction of line $L$?

The cross product $\vec{a} \times \vec{b}$ is inherently perpendicular to both $\vec{a}$ and $\vec{b}$. Since line $L$ must be perpendicular to both given lines, its direction vector must be parallel to the cross product of their direction vectors. It is the most direct way to find a common perpendicular.

Q: Wait, why is the $x$-coordinate zero on the $yz$-plane?

Think of the 3D coordinate system. The $yz$-plane is the flat surface where you move only up/down ($z$) and left/right ($y$). You never move forward or backward ($x$). Thus, $x$ is always zero on this plane. Setting the $x$-component of your parametric line equation to zero is the standard way to find where it pierces this plane.

Q: Can I use the area formula $\frac{1}{2} |\vec{PQ} \times \vec{PS}|$?

Careful! That formula is for a triangle. For a parallelogram, you do not divide by two. The area is simply the magnitude of the cross product of the two adjacent sides, $\vec{PQ}$ and $\vec{PS}$. Always double-check your geometric formulas!

Question

Let a line $L$ passing through the point $P(1, 1, 1)$ be perpendicular to the lines $\frac{x-4}{4} = \frac{y-1}{1} = \frac{z-1}{1}$ and $\frac{x-17}{1} = \frac{y-71}{1} = \frac{z}{0}$. Let the line $L$ intersect the $yz$-plane at the point $Q$. Another line parallel to $L$ and passing through the point $S(1, 0, -1)$ intersects the $yz$-plane at the point $R$. Then the square of the area of the parallelogram $PQRS$ is equal to ____.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at point $P(1, 1, 1)$ in a three-dimensional coordinate space. We are tasked with defining a line $L$ that passes through $P$ and is perpendicular to two given lines with direction vectors $\vec{d_1} = 4\hat{i} + \hat{j} + \hat{k}$ and $\vec{d_2} = \hat{i} + \hat{j} + 0\hat{k}$.

The direction vector of line $L$, denoted as $\vec{d_L}$, is found by calculating the cross product of $\vec{d_1}$ and $\vec{d_2}$:

$$
\vec{d_L} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 4 & 1 & 1 \ 1 & 1 & 0 \end{vmatrix}
$$

Expanding this determinant, we obtain:

$$
\vec{d_L} = \hat{i}(0 - 1) - \hat{j}(0 - 1) + \hat{k}(4 - 1) = -\hat{i} + \hat{j} + 3\hat{k}
$$

Thus, the direction vector is $\vec{d_L} = (-1, 1, 3)$.

### The Intersection with the $yz$-Plane

We define the line $L$ parametrically as $(1 - t, 1 + t, 1 + 3t)$. Since $L$ intersects the $yz$-plane at point $Q$, we set the $x$-coordinate to zero:

$$
1 - t = 0 \Rightarrow t = 1
$$

Substituting $t=1$ into the parametric equations, we find the coordinates of $Q$ to be $(0, 2, 4)$.

Next, we consider a second line parallel to $L$ passing through $S(1, 0, -1)$. Its parametric form is $(1 - s, s, -1 + 3s)$. Setting its $x$-coordinate to zero:

$$
1 - s = 0 \Rightarrow s = 1
$$

This yields the point $R(0, 1, 2)$. We have now identified the vertices of the parallelogram as $P(1, 1, 1)$, $Q(0, 2, 4)$, $R(0, 1, 2)$, and $S(1, 0, -1)$.

### The Final Calculation

To find the area of the parallelogram $PQRS$, we determine two adjacent vectors originating from point $P$:

$$
\vec{PQ} = Q - P = (0-1, 2-1, 4-1) = (-1, 1, 3)
$$

$$
\vec{PS} = S - P = (1-1, 0-1, -1-1) = (0, -1, -2)
$$

The area of the parallelogram is given by the magnitude of the cross product $\vec{A} = \vec{PQ} 	imes \vec{PS}$:

$$
\vec{A} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 1 & 3 \ 0 & -1 & -2 \end{vmatrix} = \hat{i}(-2 - (-3)) - \hat{j}(2 - 0) + \hat{k}(1 - 0) = (1, -2, 1)
$$

The area is the magnitude of vector $\vec{A}$:

$$
	ext{Area} = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}
$$

The problem asks for the square of the area. Therefore, the final result is:

**$(\sqrt{6})^2 = 6$**

Enter Numerical Value:

Visualized Solution

Understanding the Geometry

Direction Vectors of Given Lines

Setting up the Cross Product

Calculating Direction of Line $L$

Equation of Line $L$

Finding Point $Q$

Equation of the Second Line

Finding Point $R$

Defining the Parallelogram

Setting up the Area Vector

Calculating the Area Vector

Final Answer: Square of the Area

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Intersection with the $y z$ -Plane

The Final Calculation

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