Question

Use the formula in Exercise 79 to find the (least) distance between the given point \(Q\) and line \(\mathbf{r}\). $$Q(6,6,7), \mathbf{r}(t)=\langle 3 t,-3 t, 4\rangle$$

Short answer

Answer: The least distance between point Q and line r is 18.

Step-by-step solution

Find a point on line r

Choose any value of \(t\) and find the corresponding coordinates on line r. Let's choose \(t=0\): $$\mathbf{r}(0) = \langle 3(0), -3(0), 4 \rangle = \langle 0, 0, 4 \rangle$$ Then, \(P(0,0,4)\) is a point on line r.

Find the vector PQ

Find the vector between points P and Q: $$\overrightarrow{PQ}= \overrightarrow{Q} - \overrightarrow{P} = \langle 6, 6, 7 \rangle - \langle 0, 0, 4 \rangle = \langle 6, 6, 3 \rangle$$

Find the direction vector of line r

From the equation of line r, we can extract the direction vector: $$\mathbf{d} = \langle 3, -3, 0 \rangle$$

Find a vector perpendicular to the direction vector of line r

Let's call the vector we seek as \(\mathbf{n}\). Calculate the cross product of \(\overrightarrow{PQ}\) and \(\mathbf{d}\): $$\mathbf{n} = \overrightarrow{PQ} \times \mathbf{d} = \begin{vmatrix} \mathbf{i} &\mathbf{j} & \mathbf{k} \\ 6 &6 &3 \\ 3 &-3 &0 \\ \end{vmatrix} = \langle 9, 9, -36 \rangle$$

Calculate the distance between Q and line r

Now, we can find the magnitude of the vector \(\mathbf{n}\). The magnitude of \(\mathbf{n}\) will give us the least distance between point Q and line r: $$d = \frac{|\mathbf{n}|}{|\mathbf{d}|} = \frac{\sqrt{9^2 + 9^2 + (-36)^2}}{\sqrt{3^2 + (-3)^2 + 0^2}} = \frac{54}{3}$$ Thus, the least distance between point Q and line r is 18.

Key concepts

Vectors

A vector is a mathematical entity that has both a magnitude and a direction, making it quite handy for describing physical quantities. Think of it like an arrow pointing in a particular direction with a certain length. In mathematics, vectors are often written in terms of their components, for example, \( \langle x, y, z \rangle \) in three dimensions. These components represent how far the vector moves along each axis.

In the context of geometry and spatial interpretations, you might use vectors to indicate positions or displacements. In our example, we consider the vector \( \overrightarrow{PQ} \), which represents the difference in position between points \( P \) and \( Q \). This is calculated as the coordinates of \( Q \) minus the coordinates of \( P \), giving us the vector \( \langle 6, 6, 3 \rangle \).

• **Direction**: Indicates the way in which the vector points.
• **Magnitude**: Tells us about how long the vector is, which we often calculate using its components.

Cross product

The cross product is a way to multiply two vectors in three-dimensional space, resulting in another vector that is perpendicular to both of the original vectors. This operation is only defined in three dimensions and is helpful for finding a vector normal to a plane given two vectors lying on that plane.

The cross product is represented by the symbol \( \times \), and you can think of it as creating a new vector often denoted by \( \mathbf{n} \) that is orthogonal, or perpendicular, to the two vectors being multiplied. We often use the determinant of a 3x3 matrix to compute the cross product, as shown in our example for \( \overrightarrow{PQ} \) and \( \mathbf{d} \) giving \( \mathbf{n} = \langle 9, 9, -36 \rangle \).

• **Orthogonal Vector**: The result of the cross product is perpendicular to both original vectors.
• **Determinant Method**: The cross product involves arranging the vectors in a matrix form and using determinants to find components.

Magnitude of a vector

The magnitude of a vector indicates its "length" and can be thought of as the distance from the vector's start point to its end point. It's a scalar quantity—which means it only has size, not direction—and is always non-negative.
The formula for finding the magnitude of a vector \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) is given by \( |\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \). In the problem solution, the magnitude of a vector \( \mathbf{n} \) was calculated as part of finding the distance between a point and a line.

The calculated magnitude helps us understand how much "stretch" or "reach" the vector has in space. To solve the distance problem, we divided the magnitude of the perpendicular vector by the magnitude of the direction vector \( \mathbf{d} \), yielding the shortest path or distance from the point to the line. This gives us the least distance formula:

• **Numerical Result**: The magnitude reflects a real number without any direction component.
• **Distance Formula Application**: Used to determine minimum distances or lengths in spatial problems.