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Carry out the following steps to determine the (least) 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 \(\mathbf{u}\) corresponding to \(P\). c. Find \(\operatorname{proj}_{\mathbf{v}} \mathbf{u}\) 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 the line \(\ell\). e. Find \(\mathbf{w}\) and \(|\mathbf{w}| .\) Explain why \(|\mathbf{w}|\) is the least distance between \(P\) and \(\ell\). $$P(0,2,6) ; \ell \text { is parallel to }\langle 3,0,-4\rangle$$

Short Answer

Expert verified
Answer: The least distance between point P and line 饾搧 is 3.6 units.

Step by step solution

01

Find any vector in direction of 饾搧

From the given line 饾搧, we know that the direction vector is \(\langle 3, 0, -4 \rangle\). Therefore, any vector in the direction of 饾搧 is a scalar multiple of this vector. Let's use the \(\mathbf{v}\) as this vector. Thus, \(\mathbf{v} = \langle 3, 0, -4 \rangle\).
02

Find position vector of P

The coordinates of point P are given as (0, 2, 6). To find the position vector \(\mathbf{u}\) corresponding to P, we simply write the coordinates as components of the vector. So, \(\mathbf{u} = \langle 0, 2, 6 \rangle\).
03

Find projection of \(\mathbf{u}\) on \(\mathbf{v}\)

To find the projection of \(\mathbf{u}\) on \(\mathbf{v}\), we use the formula \(\operatorname{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v}\). First, compute the dot product of \(\mathbf{u}\) and \(\mathbf{v}\) and \(\mathbf{v} \cdot \mathbf{v}\). \(\mathbf{u} \cdot \mathbf{v} = (0 \times 3) + (2 \times 0) + (6 \times -4) = -24\) \(\mathbf{v} \cdot \mathbf{v} = (3 \times 3) + (0 \times 0) + (-4 \times -4) = 9 + 16 = 25\) Now, compute the projection: \(\operatorname{proj}_{\mathbf{v}} \mathbf{u} = \frac{-24}{25} \mathbf{v} = \left\langle -\frac{72}{25}, 0, \frac{96}{25} \right\rangle\)
04

Show that \(\mathbf{w}=\mathbf{u}-\operatorname{proj}_{\mathbf{v}} \mathbf{u}\) is orthogonal to \(\mathbf{v}\) and the length of \(\mathbf{w}\) is the distance between P and 饾搧

Compute the vector \(\mathbf{w}\): \(\mathbf{w} = \mathbf{u} - \operatorname{proj}_{\mathbf{v}} \mathbf{u} = \langle 0, 2, 6 \rangle - \left\langle -\frac{72}{25}, 0, \frac{96}{25} \right\rangle = \left\langle \frac{72}{25}, 2, \frac{54}{25} \right\rangle\) To check if \(\mathbf{w}\) is orthogonal to \(\mathbf{v}\), we'll compute their dot product. If the dot product is zero, the vectors are orthogonal: \(\mathbf{w} \cdot \mathbf{v} = \left(\frac{72}{25} \times 3\right) + (2 \times 0) + \left(\frac{54}{25} \times -4\right) = \frac{216}{25} - \frac{216}{25} = 0\) Since the dot product is zero, \(\mathbf{w}\) is indeed orthogonal to \(\mathbf{v}\). Now, the length of \(\mathbf{w}\) represents the distance between P and 饾搧, as it is the shortest distance between the point P and the line 饾搧 along the direction perpendicular to the line.
05

Find \(\mathbf{w}\) and \(|\mathbf{w}|\) and explain why it's the least distance between P and 饾搧

We already found the vector \(\mathbf{w}\) to be \(\left\langle \frac{72}{25}, 2, \frac{54}{25} \right\rangle\). Now let's compute the length (magnitude) of \(\mathbf{w}\): \(|\mathbf{w}| = \sqrt{\left(\frac{72}{25}\right)^2 + 2^2 + \left(\frac{54}{25}\right)^2} = \sqrt{\frac{5184}{625} + 4 + \frac{2916}{625}} = \sqrt{\frac{8100}{625}} = \frac{90}{25} = 3.6\) So, the least distance between point P and the line 饾搧 is 3.6 units. Finally, the reason why the length of \(\mathbf{w}\) represents the least distance between P and 饾搧 is due to the fact that \(\mathbf{w}\) represents the projection of the position vector of P perpendicular to the direction vector of 饾搧. By projecting \(\mathbf{u}\) onto the orthogonal direction of 饾搧, we guarantee that the resulting vector \(\mathbf{w}\) is the shortest possible distance between point P and the line 饾搧.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Distance Between Point and Line
Understanding the concept of calculating the distance between a point and a line is fundamental in geometry and vector analysis. In a three-dimensional space, the distance between a point and a line is the length of the shortest line segment (perpendicular to the line) that connects the point to the line.

To find this distance using vectors, we first need a position vector for the point and a direction vector for the line. We then project the position vector onto the direction vector to find the closest point on the line to our point. The vector connecting the point to this closest point on the line is perpendicular to the line, and its magnitude represents the shortest distance. Importantly, ensuring that the connecting vector is orthogonal to the direction vector of the line guarantees that we are calculating the minimal distance.
Position Vector
A position vector defines the position of a point in space relative to an origin point. In general terms, the position vector \textbf{u} for point P in 3D space with coordinates \textbf{(x, y, z)} is represented as \textbf{u} = \( \textbf{} \). The tail of the position vector is at the origin (0,0,0), and its head is at the given point P. This vector is crucial when determining distances and performing other vector operations, as it serves as the starting point for many calculations.
Orthogonal Vectors
Orthogonal vectors are two vectors that are perpendicular to each other in space. In simpler terms, if you were to draw two orthogonal vectors, they would form a 90-degree angle where they meet.

Mathematically, orthogonal vectors \textbf{a} and \textbf{b} have a dot product equal to zero; that is, \( \textbf{a} \cdot \textbf{b} = 0 \). Orthogonality is a pivotal concept in vector projection because when we subtract the projection of one vector onto another from the original vector, the result is a new vector that is orthogonal to the second vector. This principle is often used to find the distance between points and lines, as well as in many areas of mathematics, physics, and engineering.
Dot Product
The dot product or scalar product is a way of multiplying two vectors that gives us a scalar (a single number). For two vectors \textbf{a} = \( \textbf{} \) and \textbf{b} = \( \textbf{} \), the dot product is: \( \textbf{a} \cdot \textbf{b} = a1 \times b1 + a2 \times b2 + a3 \times b3 \).

The result can tell us about the relationship between the two vectors. If the dot product is zero, the vectors are orthogonal. The dot product is not only vital for determining orthogonality but is also used in calculating the vector projection, which is the fundamental operation used to find the distance from a point to a line.
Vector Magnitude
The magnitude of a vector is a measure of its length. For a vector \textbf{v} with components \( \textbf{} \), its magnitude is calculated using the square root of the sum of the squared components: \( |\textbf{v}| = \sqrt{v1^2 + v2^2 + v3^2} \).

The concept of magnitude is essential when discussing distances as the magnitude represents the distance from the origin to the point represented by the position vector. In the context of the distance between a point and a line, it corresponds to the shortest distance between them when the connecting vector's magnitude (which is perpendicular to the line) is taken.

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