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Chapter 12: Problem 67

Find vectors parallel to \(\mathbf{v}\) of the given length. $$\mathbf{v}=\langle 6,-8,0\rangle ; \text { length }=20$$

Short Answer

Expert verified
Question: Given the vector, \(\mathbf{v}=\langle 6,-8,0\rangle \), find the vector parallel to \(\mathbf{v}\) with length 20. Answer: The vector parallel to \(\mathbf{v}\) with length 20 is \(\mathbf{u} = 20\cdot\frac{\langle 6,-8,0 \rangle}{\sqrt{6^2 + (-8)^2 + 0^2}}\).

Step by step solution

01

Calculate the magnitude of the given vector.

To find a unit vector parallel to \(\mathbf{v}\), we first need to find the magnitude of \(\mathbf{v}\). The magnitude of a vector \(\mathbf{v}=\langle a,b,c\rangle\) is given by: $$ ||\mathbf{v}||=\sqrt{a^2 + b^2 + c^2} $$ For the given vector \(\mathbf{v}=\langle 6,-8,0\rangle \), the magnitude is: $$ ||\mathbf{v}||=\sqrt{6^2 + (-8)^2 + 0^2} $$
02

Determine the unit vector of \(\mathbf{v}\).

Once we have the magnitude of \(\mathbf{v}\), we can find its unit vector by dividing each component of the vector by its magnitude. The unit vector is given by: $$ \mathbf{\hat{v}}=\frac{\mathbf{v}}{||\mathbf{v}||} $$ So in our case, the unit vector \(\mathbf{\hat{v}}\) is: $$ \mathbf{\hat{v}}=\frac{\langle 6,-8,0\rangle}{||\mathbf{v}||}$$
03

Scale the unit vector to the desired length.

Now that we have found the unit vector \(\mathbf{\hat{v}}\), we can scale it to the desired length by simply multiplying it by the given length. The new vector with length 20 parallel to \(\mathbf{v}\) is given by: $$ \mathbf{u}=20\cdot\mathbf{\hat{v}} $$ So the new vector parallel to \(\mathbf{v}\) with length 20 is: $$ \mathbf{u}=20\cdot\frac{\langle 6,-8,0 \rangle}{||\mathbf{v}||} $$
04

Calculate the final vector.

Now, we just need to compute the result to find the final vector. Plugging in the values calculated in the previous steps, we get: $$ \mathbf{u} = 20\cdot\frac{\langle 6,-8,0 \rangle}{\sqrt{6^2 + (-8)^2 + 0^2}} $$ Compute the final vector \(\mathbf{u}\) to complete the solution for this problem.

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Most popular questions from this chapter

An object moves on the helix \(\langle\cos t, \sin t, t\rangle,\) for \(t \geq 0\) a. Find a position function \(\mathbf{r}\) that describes the motion if it occurs with a constant speed of \(10 .\) b. Find a position function \(\mathbf{r}\) that describes the motion if it occurs with speed \(t\)

Find the points (if they exist) at which the following planes and curves intersect. $$y=1 ; \mathbf{r}(t)=\langle 10 \cos t, 2 \sin t, 1\rangle, \text { for } 0 \leq t \leq 2 \pi$$

Let \(\mathbf{u}=\left\langle u_{1}, u_{2}, u_{3}\right\rangle\) \(\mathbf{v}=\left\langle v_{1}, v_{2}, v_{3}\right\rangle\), and \(\mathbf{w}=\) \(\left\langle w_{1}, w_{2}, w_{3}\right\rangle\). Let \(c\) be a scalar. Prove the following vector properties. \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|\)

Consider the curve \(\mathbf{r}(t)=(\cos t, \sin t, c \sin t),\) for \(0 \leq t \leq 2 \pi,\) where \(c\) is a real number. It can be shown that the curve lies in a plane. Prove that the curve is an ellipse in that plane.

For the following vectors u and \(\mathbf{v}\) express u as the sum \(\mathbf{u}=\mathbf{p}+\mathbf{n},\) where \(\mathbf{p}\) is parallel to \(\mathbf{v}\) and \(\mathbf{n}\) is orthogonal to \(\mathbf{v}\). \(\mathbf{u}=\langle-2,2\rangle, \mathbf{v}=\langle 2,1\rangle\)
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