Chapter 12: Problem 3
What is the magnitude of the cross product of two parallel vectors?
/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none}
Learning Materials
Features
Discover
Chapter 12: Problem 3
What is the magnitude of the cross product of two parallel vectors?
All the tools & learning materials you need for study success - in one app.
Get started for free
Show that the two-dimensional trajectory $$x(t)=u_{0} t+x_{0}\( and \)y(t)=-\frac{g t^{2}}{2}+v_{0} t+y_{0},\( for \)0 \leq t \leq T$$ of an object moving in a gravitational field is a segment of a parabola for some value of \(T>0 .\) Find \(T\) such that \(y(T)=0\)
Prove that the midpoint of the line segment joining \(P\left(x_{1}, y_{1}, z_{1}\right)\) and \(Q\left(x_{2}, y_{2}, z_{2}\right)\) is $$\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}, \frac{z_{1}+z_{2}}{2}\right)$$
The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|(\text {because}|\cos \theta| \leq 1) .\) This inequality, known as the Cauchy-Schwarz Inequality, holds in any number of dimensions and has many consequences. Verify that the Cauchy-Schwarz Inequality holds for \(\mathbf{u}=\langle 3,-5,6\rangle\) and \(\mathbf{v}=\langle-8,3,1\rangle\).
Imagine three unit spheres (radius equal to 1 ) with centers at \(O(0,0,0), P(\sqrt{3},-1,0)\) and \(Q(\sqrt{3}, 1,0) .\) Now place another unit sphere symmetrically on top of these spheres with its center at \(R\) (see figure). a. Find the coordinates of \(R\). (Hint: The distance between the centers of any two spheres is 2.) b. Let \(\mathbf{r}_{i j}\) be the vector from the center of sphere \(i\) to the center of sphere \(j .\) Find \(\mathbf{r}_{O P}, \mathbf{r}_{O Q}, \mathbf{r}_{P Q}, \mathbf{r}_{O R},\) and \(\mathbf{r}_{P R}\).
In contrast to the proof in Exercise \(81,\) we now use coordinates and position vectors to prove the same result. Without loss of generality, let \(P\left(x_{1}, y_{1}, 0\right)\) and \(Q\left(x_{2}, y_{2}, 0\right)\) be two points in the \(x y\) -plane and let \(R\left(x_{3}, y_{3}, z_{3}\right)\) be a third point, such that \(P, Q,\) and \(R\) do not lie on a line. Consider \(\triangle P Q R\). a. Let \(M_{1}\) be the midpoint of the side \(P Q\). Find the coordinates of \(M_{1}\) and the components of the vector \(\overrightarrow{R M}_{1}\) b. Find the vector \(\overrightarrow{O Z}_{1}\) from the origin to the point \(Z_{1}\) two-thirds of the way along \(\overrightarrow{R M}_{1}\). c. Repeat the calculation of part (b) with the midpoint \(M_{2}\) of \(R Q\) and the vector \(\overrightarrow{P M}_{2}\) to obtain the vector \(\overrightarrow{O Z}_{2}\) d. Repeat the calculation of part (b) with the midpoint \(M_{3}\) of \(P R\) and the vector \(\overline{Q M}_{3}\) to obtain the vector \(\overrightarrow{O Z}_{3}\) e. Conclude that the medians of \(\triangle P Q R\) intersect at a point. Give the coordinates of the point. f. With \(P(2,4,0), Q(4,1,0),\) and \(R(6,3,4),\) find the point at which the medians of \(\triangle P Q R\) intersect.
What do you think about this solution?
We value your feedback to improve our textbook solutions.