Question

Carry out the following steps to determine the (smallest) distance between the point \(P\) and the line \(\ell\) through the origin. a. Find any vector \(\mathbf{v}\) in the direction of \(\ell\) b. Find the position vector u corresponding to \(P\). c. Find \(\operatorname{proj}_{\mathbf{v}} \mathbf{u}\). d. Show that \(\mathbf{w}=\mathbf{u}-\) projy \(\mathbf{u}\) is a vector orthogonal to \(\mathbf{v}\) whose length is the distance between \(P\) and the line \(\ell\) e. Find \(\mathbf{w}\) and \(|\mathbf{w}| .\) Explain why \(|\mathbf{w}|\) is the distance between \(P\) and \(\ell\). \(P(-12,4) ; \ell: y=2 x\)

Short answer

Answer: The shortest distance between point P and line 𝑙 is \(\frac{28}{5}\sqrt{5}\).

Step-by-step solution

a. Find any vector 𝐯 in the direction of 𝑙

Since the line equation is given by \(y = 2x\), we can find its direction vector by setting \(x = 1\), then \(y = 2(1) = 2\). So the direction vector can be represented as \(\mathbf{v} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}\).

b. Find the position vector u corresponding to P

To find the position vector of the point P, we use the coordinates of the point. So the position vector can be represented as \(\mathbf{u} = \begin{pmatrix} -12 \\ 4 \end{pmatrix}\).

c. Find \(\operatorname{proj}_{\mathbf{v}} \mathbf{u}\)

To find the projection of \(\mathbf{u}\) onto \(\mathbf{v}\), we use the projection formula: $$\operatorname{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u}\cdot\mathbf{v}}{\mathbf{v}\cdot\mathbf{v}}\mathbf{v}$$ Calculating the dot products, we get: $$\mathbf{u}\cdot\mathbf{v} = (-12)(1) + (4)(2) = -4$$ $$\mathbf{v}\cdot\mathbf{v}= (1)(1) + (2)(2) = 5$$ Now, we can find the projection of \(\mathbf{u}\) onto \(\mathbf{v}\): $$\operatorname{proj}_{\mathbf{v}} \mathbf{u} = \frac{-4}{5}\begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} -\frac{4}{5} \\ -\frac{8}{5} \end{pmatrix}$$

d. Show that \(\mathbf{w}=\mathbf{u}-\operatorname{proj}_{\mathbf{v}} \mathbf{u}\) is a vector orthogonal to \(\mathbf{v}\) whose length is the distance between P and 𝑙

To find \(\mathbf{w}\), subtract \(\operatorname{proj}_{\mathbf{v}} \mathbf{u}\) from \(\mathbf{u}\): $$\mathbf{w} = \mathbf{u} - \operatorname{proj}_{\mathbf{v}} \mathbf{u} = \begin{pmatrix} -12 \\ 4 \end{pmatrix} - \begin{pmatrix} -\frac{4}{5} \\ -\frac{8}{5} \end{pmatrix} = \begin{pmatrix} -\frac{56}{5} \\ \frac{28}{5} \end{pmatrix}$$ To show that \(\mathbf{w}\) is orthogonal to \(\mathbf{v}\), we need to show that their dot product is 0: $$\mathbf{w}\cdot\mathbf{v} = \left(-\frac{56}{5}\right)(1) + \left(\frac{28}{5}\right)(2) = 0$$ Since their dot product is 0, \(\mathbf{w}\) is orthogonal to \(\mathbf{v}\). The length of \(\mathbf{w}\) will represent the distance between point P and line 𝑙.

e. Find \(\mathbf{w}\) and \(|\mathbf{w}|\)

We have already found \(\mathbf{w}\) as \(\begin{pmatrix} -\frac{56}{5} \\ \frac{28}{5} \end{pmatrix}\). Now we need to find the magnitude of this vector, which will give us the distance between point P and line 𝑙: $$|\mathbf{w}| = \sqrt{\left(-\frac{56}{5}\right)^2 + \left(\frac{28}{5}\right)^2} = \frac{28}{5}\sqrt{5}$$ So, the shortest distance between the point P and line 𝑙 is \(\frac{28}{5}\sqrt{5}\).

Key concepts

Projection of Vector

The concept of projecting one vector onto another is crucial for many calculations in vector mathematics. Imagine a torchlight shining onto an object. The shadow cast by the object on a flat surface is similar to a vector projection. In mathematics, when you want to find how much of one vector lies along the direction of another, you're looking for the projection.

Specifically, the projection of vector \( \mathbf{u} \) onto vector \( \mathbf{v} \) is calculated using the formula:

• \( \operatorname{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v} \)

Here, \( \mathbf{u} \cdot \mathbf{v} \) denotes the dot product of vectors \( \mathbf{u} \) and \( \mathbf{v} \). The dot product is a way of multiplying two vectors that give a scalar, providing insight into how aligned the vectors are. By dividing by \( \mathbf{v} \cdot \mathbf{v} \), we normalize the direction of \( \mathbf{v} \). This formula effectively "compresses" \( \mathbf{u} \) into the portion that lines up with \( \mathbf{v} \).

Understanding vector projection is key to solving many problems involving vectors, like finding distances or decomposing movements along specific directions.

Orthogonal Vectors

Two vectors are said to be orthogonal when they are perpendicular to each other. This is much like how the hands of a clock are at 90 degrees when one points to 12 and the other to 3.

Mathematically, this is confirmed when the dot product of the two vectors equals zero:

• \( \mathbf{a} \cdot \mathbf{b} = 0 \)

If two vectors have a dot product of zero, it indicates that they have no component or influence in each other's direction. This orthogonality is fundamental in determining distances between points and lines, especially when deriving formulas to calculate the shortest distance.

In the context of point-to-line distance problems, once the projection of a point's vector is subtracted from the original vector, the resulting vector \( \mathbf{w} \) is orthogonal to the line's direction vector. This earned orthogonality validates that \( \mathbf{w} \) points straight toward the closest point on the line, explaining why its magnitude reveals the shortest distance.

Magnitude of a Vector

The magnitude of a vector can be thought of as its length. It tells us how long a vector is in Euclidean space. So, when considering a vector in a plane, like \( \mathbf{w} \), we imagine it as an arrow with a certain length pointing from the start to the endpoint.

The mathematical formula to compute the magnitude \( |\mathbf{v}| \) of a vector \( \mathbf{v} = \begin{pmatrix} x \ y \end{pmatrix} \) is:

• \( |\mathbf{v}| = \sqrt{x^2 + y^2} \)

This formula derives from the Pythagorean theorem. It ensures that the length accounts for both the horizontal and vertical components, giving a true representation of the vector's span.

In distance-related exercises, the magnitude of vector \( \mathbf{w} \) not only represents its true length but also provides an immediate measure of how far a point is from a line. Thus, finding the magnitude in such problems isn't just a step for completion; it's the solution.