Question

Find the distance from the point to the line. \(Q=(2,5,6), \quad \vec{\ell}(t)=\langle-1,1,1\rangle+t\langle 1,0,1\rangle\)

Short answer

The distance from the point to the line is \(3\sqrt{2}\).

Step-by-step solution

Identify Key Components

To find the distance from a point to a line in three-dimensional space \(\mathbb{R}^3\), we need the point \(Q = (2, 5, 6)\) and the parametric equation of the line \(\vec{\ell}(t) = \langle -1, 1, 1 \rangle + t \langle 1, 0, 1 \rangle\). The direction vector of the line is \(\vec{d} = \langle 1, 0, 1 \rangle\), and a point on the line \(P\) when \(t=0\) is \((-1, 1, 1)\).

Find Vector from Line Point to Q

Calculate the vector from a point on the line \(P = (-1, 1, 1)\) to the point \(Q = (2, 5, 6)\). This is done by subtracting coordinates: \(\vec{PQ} = \langle 2 - (-1), 5 - 1, 6 - 1 \rangle = \langle 3, 4, 5 \rangle\).

Calculate Cross Product

Compute the cross product \(\vec{PQ} \times \vec{d}\) to find a vector perpendicular to both \(\vec{PQ}\) and the line's direction vector \(\vec{d}\):\[\vec{PQ} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 4 & 5 \ 1 & 0 & 1 \end{vmatrix} = \hat{i}(4\cdot1 - 0\cdot5) - \hat{j}(3\cdot1 - 1\cdot5) + \hat{k}(3\cdot0 - 4\cdot1)\]\[= \langle 4, -2, -4 \rangle.\]

Compute Magnitudes

Calculate the magnitude of the cross product vector, \(\|\vec{PQ} \times \vec{d}\|\), and the direction vector, \(\|\vec{d}\| = \sqrt{1^2 + 0^2 + 1^2} = \sqrt{2}\). For the cross product: \[\|\vec{PQ} \times \vec{d}\| = \sqrt{4^2 + (-2)^2 + (-4)^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6.\]

Calculate Distance

The distance from point \(Q\) to the line is given by the formula:\[\text{Distance} = \frac{\|\vec{PQ} \times \vec{d}\|}{\|\vec{d}\|} = \frac{6}{\sqrt{2}} = \frac{6\sqrt{2}}{2} = 3\sqrt{2}.\]

Key concepts

Vector Calculations

When calculating the distance from a point to a line in three-dimensional space, vector calculations are an essential tool. Vectors are mathematical objects that have both magnitude and direction. They are used to represent physical quantities like force and velocity.
For our problem, we need to compute various vectors. First, understand how to calculate a vector from one point to another. You achieve this by subtracting the coordinates of the initial point from the coordinates of the endpoint. In our exercise, we find the vector \( \vec{PQ} \) by subtracting point \( P = (-1, 1, 1) \) from point \( Q = (2, 5, 6) \).
Additionally, vectors are used to describe directions, like the direction vector of a line, which is a crucial part of finding distances in space.

Parametric Equations

A parametric equation is a way of defining a relationship using parameters. In three-dimensional space, lines can be expressed using parametric equations, which represent every point on the line.
In our exercise, the line described by \( \vec{\ell}(t) = \langle -1, 1, 1 \rangle + t \langle 1, 0, 1 \rangle \) uses \( t \) as a parameter. The vector \( \langle -1, 1, 1 \rangle \) indicates a specific point on the line, and adding the direction vector \( \langle 1, 0, 1 \rangle \) multiplied by \( t \) allows us to find any point on the line.
Understanding parametric equations helps in determining the position of points on a line, which is helpful for calculating distances and solving geometric problems.

Cross Product

The cross product is a mathematical operation on two vectors in three-dimensional space. It results in a vector that is perpendicular to both of the original vectors. This property is particularly useful in finding distances.
In our task, we calculate the cross product \( \vec{PQ} \times \vec{d} \). This involves using the components of the vectors to set up a determinant, which helps you find the perpendicular vector easily. The formula from the solution is:

• \[ \vec{PQ} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 4 & 5 \ 1 & 0 & 1 \end{vmatrix} = \langle 4, -2, -4 \rangle \]

The magnitude of the cross product vector gives the area of the parallelogram formed by the original two vectors. This is then used to determine the distance from the point to the line when divided by the magnitude of the direction vector.

Three-Dimensional Space

Three-dimensional space, often noted as \( \mathbb{R}^3 \), provides a setting for finding distances, orientations, and relationships between objects like points and lines. Here, concepts like vectors and parametric equations come alive.
In 3D space, every point is defined by three coordinates \( (x, y, z) \), allowing for a more complex analysis compared to two-dimensional planes. This additional dimension lets us evaluate how objects relate directly in space, which we use to determine distances between them.
The whole process, from defining vectors to using parametric equations and cross products, relies on the capabilities of three-dimensional space. It allows students to solve practical problems and understand the spatial arrangement of objects better, giving a real-world perspective on geometry.