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Recall that any line in space may be represented parametrically by a vector- valued function. a. Find a vector-valued function \(\mathbf{r}\) that parameterizes the line through (-2,1,4) in the direction of the vector \(\mathbf{v}=\langle 3,2,-5\rangle\) b. Find a vector-valued function \(\mathbf{r}\) that parameterizes the line of intersection of the planes \(x+2 y-z=4\) and \(3 x+y-2 z=1\) c. Determine the point of intersection of the lines given by $$ \begin{array}{l} x=2+3 t, y=1-2 t, z=4 t \\ x=3+1 s, y=3-2 s, z=2 s \end{array} $$ Then, find a vector-valued function \(\mathbf{r}\) that parameterizes the line that passes through the point of intersection you just found and is perpendicular to both of the given lines.

Short Answer

Expert verified
The solutions for parts a, b, and c of the given question are as follows: a. The vector-valued function that parameterizes the line through (-2,1,4) in the direction of the vector \(\mathbf{v}=\langle 3,2,-5\rangle\) is: \(\mathbf{r}(t) = \begin{pmatrix}-2+3t\\1+2t\\4-5t\end{pmatrix}\) b. The vector-valued function that parameterizes the line of intersection of the planes \(x+2y-z=4\) and \(3x+y-2z=1\) is: \(\mathbf{r}(t) = \begin{pmatrix}3t\\2+t\\5t\end{pmatrix}\) c. The point of intersection of the given lines is (5,-3,2), and the vector-valued function that parameterizes the line that passes through the point of intersection and is perpendicular to both of the given lines is: \(\mathbf{r}(t) = \begin{pmatrix}5+16t\\-3+2t\\2+4t\end{pmatrix}\)

Step by step solution

01

Identify a Point and a Direction Vector

Given a point A(-2, 1, 4) and a direction vector \(\mathbf{v}=\langle 3,2,-5\rangle\).
02

Write the Vector-Valued Function

The vector-valued function for the line can be written as follows: \(\mathbf{r}(t) = \mathbf{A} + t\mathbf{v}.\)
03

Substitute Point A and the Direction Vector into the Formula

Plug A and \(\mathbf{v}\) into the formula: \(\mathbf{r}(t) = \begin{pmatrix}-2\\1\\4\end{pmatrix} + t\begin{pmatrix}3\\2\\-5\end{pmatrix}.\)
04

Simplify the Result

Simplify the vector-valued function: \(\mathbf{r}(t) = \begin{pmatrix}-2+3t\\1+2t\\4-5t\end{pmatrix}\). The solution for part a is: $$\mathbf{r}(t) = \begin{pmatrix}-2+3t\\1+2t\\4-5t\end{pmatrix}$$ ## Part b: ##
05

Compute the Cross Product of the Normal Vectors of the Planes

The cross product \(\mathbf{n}_{1} \times \mathbf{n}_{2}\) gives the direction vector of the intersection line. The normal vectors are \(\mathbf{n}_{1}=\begin{pmatrix}1\\2\\-1\end{pmatrix}\) and \(\mathbf{n}_{2}=\begin{pmatrix}3\\1\\-2\end{pmatrix}\). Their cross product is \(\mathbf{v}=\begin{pmatrix}3\\1\\5\end{pmatrix}\).
06

Find a Point on the Intersection Line

To find a point on the intersection of two planes, we can use the equation of the plane to eliminate one variable and solve one of the remaining variables and use back-substitution. For example, set z = 0: we get the simultaneous equations \(x + 2y = 4\) and \(3x + y=1\). Solving these equations, we get point P(0, 2, 0).
07

Write the Vector-Valued Function for the Intersection Line

With the direction vector \(\mathbf{v}\) and point P, the vector-valued function for the intersection line \(\mathbf{r}(t)=\mathbf{P}+t\mathbf{v}\).
08

Substitute Point P and the Direction Vector into the Formula

Plug P and \(\mathbf{v}\) into the formula: \(\mathbf{r}(t) = \begin{pmatrix}0\\2\\0\end{pmatrix} + t\begin{pmatrix}3\\1\\5\end{pmatrix}\). The solution for part b is: $$\mathbf{r}(t) = \begin{pmatrix}3t\\2+t\\5t\end{pmatrix}$$ ## Part c: ##
09

Find the Point of Intersection

To find the intersection, set the parametric equations equal: \[ \begin{array}{l} 2+3 t = 3+1 s \\ 1-2t = 3-2s \\ 4t = 2s \end{array} \] Solving the simultaneous equations, we get point Q(5, -3, 2).
10

Determine the Direction Vector of Intersection Line

The direction vector of the intersection line is the cross product of direction vectors of the given lines \((\mathbf{a} \times \mathbf{b})\). The given direction vectors for the lines are \(\mathbf{a} = \begin{pmatrix}3\\-2\\4\end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix}1\\-2\\2\end{pmatrix}\). Their cross product is \(\mathbf{u} = \begin{pmatrix}16\\2\\4\end{pmatrix}\).
11

Write the Vector-Valued Function for the Line of Intersection

With the direction vector \(\mathbf{u}\) and point Q, the vector-valued function for the intersection line is \(\mathbf{r}(t)=\mathbf{Q}+t\mathbf{u}\).
12

Substitute Point Q and the Direction Vector into the Formula

Plug Q and \(\mathbf{u}\) into the formula: \(\mathbf{r}(t) = \begin{pmatrix}5\\-3\\2\end{pmatrix} + t\begin{pmatrix}16\\2\\4\end{pmatrix}\). The solution for part c is: $$\mathbf{r}(t) = \begin{pmatrix}5+16t\\-3+2t\\2+4t\end{pmatrix}$$

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Key Concepts

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

Vector-Valued Functions and Parametric Equations
Vector-valued functions are powerful tools in mathematics for representing lines and curves in a multi-dimensional space. They encapsulate both a direction and a magnitude, allowing us to describe the motion of objects or the position of points relative to some starting point.

Parametric equations are the underlying component of vector-valued functions. Instead of expressing a graph or line as a function of x and y alone, parametric equations introduce an independent parameter, typically t, to represent each variable – x(t), y(t), z(t) – independently.

For instance, a vector-valued function \( \mathbf{r}(t) = \mathbf{A} + t\mathbf{v} \) parameterizes a line through a point A with position vector \( \mathbf{A} \) in the direction of the vector \( \mathbf{v} \). By varying t, we can map out the entire line in space. This approach is particularly useful as it easily adapts to various dimensions, making it widely applicable in fields such as physics and engineering.
Line Intersection in Space
Finding the intersection of lines in space involves solving systems of equations that represent each line. But when two lines are not in the same plane or when we deal with planes intersecting, the solution is not as straightforward.

When dealing with the intersection of planes to find a line, we can employ the concept of vector-valued functions and parametric equations. Given two planes, each plane's equation can help us find a point that lies on both planes. We can then use the cross product of the planes' normal vectors to find a direction vector representing the line of their intersection.

For instance, in our exercise, after computing the cross product to find the direction vector and determining the intersecting point, we formed a vector-valued function \( \mathbf{r}(t) \) that fully describes the line of intersection of the two planes.
Cross Product
The cross product is an operation on two vectors in three-dimensional space that results in a third vector which is perpendicular to both of the original vectors. It is denoted by \( \times \) and plays a vital role in many areas of mathematics and physics.

For two vectors \( \mathbf{a} \) and \( \mathbf{b} \), their cross product \( \mathbf{a} \times \mathbf{b} \) is a vector with a magnitude that represents the area of the parallelogram formed by \( \mathbf{a} \) and \( \mathbf{b} \). The direction is given by the right-hand rule. To calculate the cross product, we use the determinants of a matrix formed by the unit vectors \( \hat{i}, \hat{j}, \hat{k} \) and the components of \( \mathbf{a} \) and \( \mathbf{b} \).

In the context of our exercise, the cross product is used to find the direction vector of the line that lies at the intersection of two planes or that is perpendicular to two given lines.
Direction Vector
A direction vector provides information about the orientation of a line in space. For lines, it can be thought of as the 'slope' in three-dimensional space and determines the trajectory of the line. When we have a vector-valued function for a line, the coefficient of the parameter t in \( \mathbf{r}(t) \) is the direction vector.

In practical terms, to draw a line, we need a starting point and a direction to continue indefinitely. The direction vector gives us this path, allowing us to extend from our initial point in exactly one direction, creating a straight path. It is important to note that any scalar multiple of a direction vector also describes the same direction. Therefore, when provided in parametric form, it is the proportional relationship of its components that matters, rather than their specific values.

This concept was applied in our exercise where the vector \( \mathbf{v} \) or \( \mathbf{u} \) represents the direction vector for the line. It tells us in which direction to 'move' from our known point to trace out the line.

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Most popular questions from this chapter

A force (like gravity) has both a magnitude and a direction. If two forces \(\mathbf{u}\) and \(\mathbf{v}\) are applied to an object at the same point, the resultant force on the object is the vector sum of the two forces. When a force is applied by a rope or a cable, we call that force tension. Vectors can be used to determine tension. As an example, suppose a painting weighing 50 pounds is to be hung from wires attached to the frame as illustrated in Figure \(9.2 .10 .\) We need to know how much tension will be on the wires to know what kind of wire to use to hang the picture. Assume the wires are attached to the frame at point \(O\). Let \(\mathbf{u}\) be the vector emanating from point \(O\) to the left and \(\mathbf{v}\) the vector emanating from point \(O\) to the right. Assume \(\mathbf{u}\) makes a \(60^{\circ}\) angle with the horizontal at point \(O\) and \(\mathbf{v}\) makes a \(45^{\circ}\) angle with the horizontal at point \(O\). Our goal is to determine the vectors \(\mathbf{u}\) and \(\mathbf{v}\) in order to calculate their magnitudes. a. Treat point \(O\) as the origin. Use trigonometry to find the components \(u_{1}\) and \(u_{2}\) so that \(\mathbf{u}=u_{1} \mathbf{i}+u_{2} \mathbf{j} .\) Since we don't know the magnitude of \(\mathbf{u},\) your components will be in terms of \(|\mathbf{u}|\) and the cosine and sine of some angle. Then find the components \(v_{1}\) and \(v_{2}\) so that \(\mathbf{v}=v_{1} \mathbf{i}+v_{2} \mathbf{j}\). Again, your components will be in terms of \(|\mathbf{v}|\) and the cosine and sine of some angle. b. The total force holding the picture up is given by \(\mathbf{u}+\mathbf{v}\). The force acting to pull the picture down is given by the weight of the picture. Find the force vector \(\mathbf{w}\) acting to pull the picture down. c. The picture will hang in equilibrium when the force acting to hold it up is equal in magnitude and opposite in direction to the force acting to pull it down. Equate these forces to find the components of the vectors \(\mathbf{u}\) and \(\mathbf{v}\).

Use the geometric definition of the cross product and the properties of the cross product to make the following calculations. (a) \(((\vec{i}+\vec{j}) \times \vec{i}) \times \vec{j}=\)_________ (b) \((\vec{j}+\vec{k}) \times(\vec{j} \times \vec{k})=\)_________ (c) \(5 \vec{i} \times(\vec{i}+\vec{j})=\)_________ (d) \((\vec{k}+\vec{j}) \times(\vec{k}-\vec{j})=\)_________

One of the properties of the cross product is that \((\mathbf{u}+\mathbf{v}) \times \mathbf{w}=(\mathbf{u} \times \mathbf{w})+\) \((\mathbf{v} \times \mathbf{w})\). That is, the cross product distributes over vector addition on the right. Here we investigate whether the cross product distributes over vector addition on the left. a. Let \(\mathbf{u}=\langle 1,2,-1\rangle, \mathbf{v}=\langle 4,-3,6\rangle,\) and \(\mathbf{v}=\langle 4,7,2\rangle .\) Calculate $$ \mathbf{u} \times(\mathbf{v}+\mathbf{w}) \quad \text { and } \quad(\mathbf{u} \times \mathbf{v})+(\mathbf{u} \times \mathbf{w}) $$ What do you notice? b. Use the properties of the cross product to show that in general $$ \mathbf{x} \times(\mathbf{y}+\mathbf{z})=(\mathbf{x} \times \mathbf{y})+(\mathbf{x} \times \mathbf{z}) $$ for any vectors \(\mathbf{x}, \mathbf{y},\) and \(\mathbf{z}\) in \(\mathbb{R}^{3}\).

Consider the function \(h\) defined by \(h(x, y)=8-\sqrt{4-x^{2}-y^{2}}\). a. What is the domain of \(h ?\) (Hint: describe a set of ordered pairs in the plane by explaining their relationship relative to a key circle.) b. The range of a function is the set of all outputs the function generates. Given that the range of the square root function \(g(t)=\sqrt{t}\) is the set of all nonnegative real numbers, what do you think is the range of \(h ?\) Why? c. Choose 4 different values from the range of \(h\) and plot the corresponding level curves in the plane. What is the shape of a typical level curve? d. Choose 5 different values of \(x\) (including at least one negative value and zero), and sketch the corresponding traces of the function \(h\). e. Choose 5 different values of \(y\) (including at least one negative value and zero), and sketch the corresponding traces of the function \(h\). f. Sketch an overall picture of the surface generated by \(h\) and write at least one sentence to describe how the surface appears visually. Does the surface remind you of a familiar physical structure in nature?

(a) Find a unit vector from the point \(P=(1,2)\) and toward the point \(Q=(6,14)\) \(\vec{u}=\)_________ (b) Find a vector of length 26 pointing in the same direction. \(\vec{v}=\)_________

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