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Chapter 12: Problem 25

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

Short Answer

Expert verified
Answer: The area of the triangle with vertices A, B, and C is \(2\sqrt{10}\).

Step by step solution

01

Calculate the distance between the points A, B and C

To calculate the distance AB, BC, and AC, we can use the distance formula for 3D space: $$AB=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$$ Apply the formula for each pair of vertices: $$AB=\sqrt{(3-0)^2 + (0-0)^2 + (1-0)^2} = \sqrt{9 + 0 + 1}=\sqrt{10}$$ $$BC=\sqrt{(1-3)^2 + (1-0)^2 + (0-1)^2} = \sqrt{4 + 1 + 1}=\sqrt{6}$$ $$AC=\sqrt{(1-0)^2 + (1-0)^2 + (0-0)^2} = \sqrt{1 + 1 + 0}=\sqrt{2}$$
02

Calculate the semiperimeter of the triangle

The semiperimeter (s) of a triangle is defined as the half of its perimeter. To find the semiperimeter, add the distances AB, BC, and AC and divide the result by 2: $$s=\frac{AB+BC+AC}{2}$$ Plug in the values we found in Step 1: $$s=\frac{\sqrt{10}+\sqrt{6}+\sqrt{2}}{2}$$
03

Apply Heron's formula to find the area of the triangle

Heron's formula states that the area of a triangle with sides a, b, and c and semiperimeter s is: $$Area=\sqrt{s(s-a)(s-b)(s-c)}$$ Apply the formula for the area using the values from the previous steps: $$Area=\sqrt{\frac{\sqrt{10}+\sqrt{6}+\sqrt{2}}{2}\left(\frac{\sqrt{10}+\sqrt{6}+\sqrt{2}}{2}-\sqrt{10}\right)\left(\frac{\sqrt{10}+\sqrt{6}+\sqrt{2}}{2}-\sqrt{6}\right)\left(\frac{\sqrt{10}+\sqrt{6}+\sqrt{2}}{2}-\sqrt{2}\right)}$$ After evaluating the expression, we get: $$Area=\frac{\sqrt{40}}{2} = 2\sqrt{10}$$ Hence, the area of the triangle with vertices A, B, and C is \(2\sqrt{10}\).

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Most popular questions from this chapter

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.

\(\mathbb{R}^{2}\) Consider the vectors \(\mathbf{I}=\langle 1 / \sqrt{2}, 1 / \sqrt{2}\rangle\) and \(\mathbf{J}=\langle-1 / \sqrt{2}, 1 / \sqrt{2}\rangle\). Write the vector \langle 2,-6\rangle in terms of \(\mathbf{I}\) and \(\mathbf{J}\).

Alternative derivation of the curvature Derive the computational formula for curvature using the following steps. a. Use the tangential and normal components of the acceleration to show that \(\left.\mathbf{v} \times \mathbf{a}=\kappa|\mathbf{v}|^{3} \mathbf{B} . \text { (Note that } \mathbf{T} \times \mathbf{T}=\mathbf{0} .\right)\) b. Solve the equation in part (a) for \(\kappa\) and conclude that \(\kappa=\frac{|\mathbf{v} \times \mathbf{a}|}{\left|\mathbf{v}^{3}\right|},\) as shown in the text.

An object moves along a path given by \(\mathbf{r}(t)=\langle a \cos t+b \sin t, c \cos t+d \sin t\rangle, \quad\) for \(0 \leq t \leq 2 \pi\) a. What conditions on \(a, b, c,\) and \(d\) guarantee that the path is a circle? b. What conditions on \(a, b, c,\) and \(d\) guarantee that the path is an ellipse?

Find the domains of the following vector-valued functions. $$\mathbf{r}(t)=\frac{2}{t-1} \mathbf{i}+\frac{3}{t+2} \mathbf{j}$$
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