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Chapter 12: Problem 28

For the given points \(A, B,\) and \(C,\) find the area of the triangle with vertices \(A, B,\) and \(C .\) $$A(-1,-5,-3), B(-3,-2,-1), C(0,-5,-1)$$

Short Answer

Expert verified
Answer: The area of the triangle is $\frac{7}{2}$.

Step by step solution

01

Find vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\)

To find the vectors that connect points \(A\) and \(B\), and points \(A\) and \(C\), we have to subtract the coordinates of point \(A\) from points \(B\) and \(C\) respectively: $$\overrightarrow{AB}=B-A=(-3,-2,-1)-(-1,-5,-3)=(2,3,2)$$ $$\overrightarrow{AC}=C-A=(0,-5,-1)-(-1,-5,-3)=(1,0,2)$$
02

Compute the cross product of \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\)

The cross product of two vectors, also known as the vector product, is given by: $$\overrightarrow{AB}\times\overrightarrow{AC} = \begin{vmatrix} \bold{i} & \bold{j} & \bold{k} \\ 2 & 3 & 2 \\ 1 & 0 & 2 \end{vmatrix}$$ To find the cross product, we perform the following calculation: $$\overrightarrow{AB}\times\overrightarrow{AC} = \bold{i}(3\cdot2 - 2\cdot0)- \bold{j}(2\cdot2 - 2\cdot1)+ \bold{k}(2\cdot0 - 1\cdot3)$$ $$\overrightarrow{AB}\times\overrightarrow{AC}=(6\bold{i} - 2\bold{j} - 3\bold{k})$$
03

Compute the magnitude of the cross product

The magnitude of a vector \(\vec{v}=(x, y, z)\) is given by: $$\lVert\vec{v}\rVert=\sqrt{x^2+y^2+z^2}$$ Applying this to the cross product: $$\lVert\overrightarrow{AB}\times\overrightarrow{AC}\rVert=\sqrt{6^2+(-2)^2+(-3)^2}=\sqrt{49}=7$$
04

Calculate the area of the triangle

Now that we have the magnitude of the cross product, we can find the area of the triangle \(ABC\). The area of a triangle formed by two vectors \(\vec{u}\) and \(\vec{v}\) is given by half of the magnitude of the cross product of these two vectors: $$Area(\triangle ABC)=\frac{1}{2}\lVert\overrightarrow{AB}\times\overrightarrow{AC}\rVert$$ Substituting the value from step 3, we have: $$Area(\triangle ABC)=\frac{1}{2}(7)=\frac{7}{2}$$ Thus, the area of the triangle with vertices \(A, B\) , and \(C\) is \(\boxed{\frac{7}{2}}\).

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Most popular questions from this chapter

Determine whether the following statements are true and give an explanation or counterexample. a. The line \(\mathbf{r}(t)=\langle 3,-1,4\rangle+t\langle 6,-2,8\rangle\) passes through the origin. b. Any two nonparallel lines in \(\mathbb{R}^{3}\) intersect. c. The curve \(\mathbf{r}(t)=\left\langle e^{-t}, \sin t,-\cos t\right\rangle\) approaches a circle as \(t \rightarrow \infty\). d. If \(\mathbf{r}(t)=e^{-t^{2}}\langle 1,1,1\rangle\) then \(\lim _{t \rightarrow \infty} \mathbf{r}(t)=\lim _{t \rightarrow-\infty} \mathbf{r}(t)\).

A pair of lines in \(\mathbb{R}^{3}\) are said to be skew if they are neither parallel nor intersecting. Determine whether the following pairs of lines are parallel, intersecting, or skew. If the lines intersect. determine the point(s) of intersection. $$\begin{aligned} &\mathbf{r}(t)=\langle 1+2 t, 7-3 t, 6+t\rangle;\\\ &\mathbf{R}(s)=\langle-9+6 s, 22-9 s, 1+3 s\rangle \end{aligned}$$

An object moves along an ellipse given by the function \(\mathbf{r}(t)=\langle a \cos t, b \sin t\rangle,\) for \(0 \leq t \leq 2 \pi,\) where \(a > 0\) and \(b > 0\) a. Find the velocity and speed of the object in terms of \(a\) and \(b\) for \(0 \leq t \leq 2 \pi\) b. With \(a=1\) and \(b=6,\) graph the speed function, for \(0 \leq t \leq 2 \pi .\) Mark the points on the trajectory at which the speed is a minimum and a maximum. c. Is it true that the object speeds up along the flattest (straightest) parts of the trajectory and slows down where the curves are sharpest? d. For general \(a\) and \(b\), find the ratio of the maximum speed to the minimum speed on the ellipse (in terms of \(a\) and \(b\) ).

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 \(\mathbf{v}\). \(\mathbf{u}=\langle 4,3\rangle, \mathbf{v}=\langle 1,1\rangle\)

Suppose \(\mathbf{u}\) and \(\mathbf{v}\) are nonzero vectors in \(\mathbb{R}^{3}\). a. Prove that the equation \(\mathbf{u} \times \mathbf{z}=\mathbf{v}\) has a nonzero solution \(\mathbf{z}\) if and only if \(\mathbf{u} \cdot \mathbf{v}=0 .\) (Hint: Take the dot product of both sides with v.) b. Explain this result geometrically.
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