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Chapter 12: Problem 26

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

Short Answer

Expert verified
Answer: The area of the triangle with vertices A, B, and C is \(\frac{1}{2}\sqrt{17}\) square units.

Step by step solution

01

Calculate vectors from the given points

Create vectors AB and AC by subtracting the coordinates of points B and C from point A, respectively. $$AB = B - A = (5 - 1, 1 - 2, 5 - 3) = (4, -1, 2)$$ $$AC = C - A = (2 - 1, 3 - 2, 3 - 3) = (1, 1, 0)$$
02

Calculate the cross product of the vectors

The cross product of vectors AB and AC, noted as ABxAC, equals the following: $$AB \times AC = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -1 & 2 \\ 1 & 1 & 0 \end{vmatrix}$$ Calculate the determinant of the matrix to find the cross product. $$AB \times AC = \mathbf{i}((-1)(0)-(2)(1)) - \mathbf{j}((4)(0)-(2)(1)) + \mathbf{k}((4)(1)-(1)(1))$$ $$AB \times AC = -2\mathbf{i} + 2\mathbf{j} + 3\mathbf{k} = (-2, 2, 3)$$
03

Find the magnitude of the cross product vector

The magnitude of a vector (x, y, z) is calculated as: $$||AB \times AC|| = \sqrt{(-2)^2 + 2^2 + 3^2} = \sqrt{17}$$
04

Compute the area of the triangle

The area of the triangle can be found as half of the magnitude of the cross product of the vectors AB and AC. $$Area = \frac{1}{2} ||AB \times AC|| = \frac{1}{2} \sqrt{17}$$ Thus, the area of the triangle with vertices A, B, and C is \(\frac{1}{2}\sqrt{17}\) (square units).

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Most popular questions from this chapter

Consider the curve \(\mathbf{r}(t)=(a \cos t+b \sin t) \mathbf{i}+(c \cos t+d \sin t) \mathbf{j}+(e \cos t+f \sin t) \mathbf{k}\) where \(a, b, c, d, e,\) and fare real numbers. It can be shown that this curve lies in a plane. Assuming the curve lies in a plane, show that it is a circle centered at the origin with radius \(R\) provided \(a^{2}+c^{2}+e^{2}=b^{2}+d^{2}+f^{2}=R^{2}\) and \(a b+c d+e f=0\).

Consider the ellipse \(\mathbf{r}(t)=\langle a \cos t, b \sin t\rangle\) for \(0 \leq t \leq 2 \pi,\) where \(a\) and \(b\) are real numbers. Let \(\theta\) be the angle between the position vector and the \(x\) -axis. a. Show that \(\tan \theta=(b / a) \tan t\) b. Find \(\theta^{\prime}(t)\) c. Note that the area bounded by the polar curve \(r=f(\theta)\) on the interval \([0, \theta]\) is \(A(\theta)=\frac{1}{2} \int_{0}^{\theta}(f(u))^{2} d u\) Letting \(f(\theta(t))=|\mathbf{r}(\theta(t))|,\) show that \(A^{\prime}(t)=\frac{1}{2} a b\) d. Conclude that as an object moves around the ellipse, it sweeps out equal areas in equal times.

For the given points \(P, Q,\) and \(R,\) find the approximate measurements of the angles of \(\triangle P Q R\). $$P(0,-1,3), Q(2,2,1), R(-2,2,4)$$

Prove that for integers \(m\) and \(n\), the curve $$\mathbf{r}(t)=\langle a \sin m t \cos n t, b \sin m t \sin n t, c \cos m t\rangle$$ lies on the surface of a sphere provided \(a^{2}+b^{2}=c^{2}\).

Evaluate the following limits. $$\lim _{t \rightarrow \infty}\left(e^{-t} \mathbf{i}-\frac{2 t}{t+1} \mathbf{j}+\tan ^{-1} t \mathbf{k}\right)$$
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