Chapter 12: Problem 3
Why is \(\mathbf{r}(t)=\langle f(t), g(t), h(t)\rangle\) called a vector-valued function?
/*! 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
Why is \(\mathbf{r}(t)=\langle f(t), g(t), h(t)\rangle\) called a vector-valued function?
All the tools & learning materials you need for study success - in one app.
Get started for free
The points \(O(0,0,0), P(1,4,6),\) and \(Q(2,4,3)\) lie at three vertices of a parallelogram. Find all possible locations of the fourth vertex.
Assume that \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) are vectors in \(\mathrm{R}^{3}\) that form the sides of a triangle (see figure). Use the following steps to prove that the medians intersect at a point that divides each median in a 2: 1 ratio. The proof does not use a coordinate system. a. Show that \(\mathbf{u}+\mathbf{v}+\mathbf{w}=\mathbf{0}\) b. Let \(\mathbf{M}_{1}\) be the median vector from the midpoint of \(\mathbf{u}\) to the opposite vertex. Define \(\mathbf{M}_{2}\) and \(\mathbf{M}_{3}\) similarly. Using the geometry of vector addition show that \(\mathbf{M}_{1}=\mathbf{u} / 2+\mathbf{v} .\) Find analogous expressions for \(\mathbf{M}_{2}\) and \(\mathbf{M}_{3}\) c. Let \(a, b,\) and \(c\) be the vectors from \(O\) to the points one-third of the way along \(\mathbf{M}_{1}, \mathbf{M}_{2},\) and \(\mathbf{M}_{3},\) respectively. Show that \(\mathbf{a}=\mathbf{b}=\mathbf{c}=(\mathbf{u}-\mathbf{w}) / 3\) d. Conclude that the medians intersect at a point that divides each median in a 2: 1 ratio.
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. Use the vectors \(\mathbf{u}=\langle\sqrt{a}, \sqrt{b}\rangle\) and \(\mathbf{v}=\langle\sqrt{b}, \sqrt{a}\rangle\) to show that \(\sqrt{a b} \leq(a+b) / 2,\) where \(a \geq 0\) and \(b \geq 0\).
The points \(P, Q, R,\) and \(S,\) joined by the vectors \(\mathbf{u}, \mathbf{v}, \mathbf{w},\) and \(\mathbf{x},\) are the vertices of a quadrilateral in \(\mathrm{R}^{3}\). The four points needn't lie in \(a\) plane (see figure). Use the following steps to prove that the line segments joining the midpoints of the sides of the quadrilateral form a parallelogram. The proof does not use a coordinate system. a. Use vector addition to show that \(\mathbf{u}+\mathbf{v}=\mathbf{w}+\mathbf{x}\) b. Let \(m\) be the vector that joins the midpoints of \(P Q\) and \(Q R\) Show that \(\mathbf{m}=(\mathbf{u}+\mathbf{v}) / 2\) c. Let n be the vector that joins the midpoints of \(P S\) and \(S R\). Show that \(\mathbf{n}=(\mathbf{x}+\mathbf{w}) / 2\) d. Combine parts (a), (b), and (c) to conclude that \(\mathbf{m}=\mathbf{n}\) e. Explain why part (d) implies that the line segments joining the midpoints of the sides of the quadrilateral form a parallelogram.
Determine the equation of the line that is perpendicular to the lines \(\mathbf{r}(t)=\langle-2+3 t, 2 t, 3 t\rangle\) and \(\mathbf{R}(s)=\langle-6+s,-8+2 s,-12+3 s\rangle\) and passes through the point of intersection of the lines \(\mathbf{r}\) and \(\mathbf{R}\).
What do you think about this solution?
We value your feedback to improve our textbook solutions.