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Chapter 5: Problem 49

Find the area of the triangle with the given vertices. Use the fact that the area \(A\) of the triangle having \(u\) and \(v\) as adjacent sides is \(A=\frac{1}{2}\|\mathbf{u} \times \mathbf{v}\|\). $$(3,5,7),(5,5,0),(-4,0,4)$$

Short Answer

Expert verified
The area of the triangle is \( A = \frac{1}{2} * \sqrt{1226} \)

Step by step solution

01

Convert the vertices to vectors

Find two vectors u and v by subtracting the coordinates of the vertices. Let vertices be: A = (3,5,7), B = (5,5,0), C = (-4,0,4). Let \( \mathbf{u} = \mathbf{B} - \mathbf{A} \) and \( \mathbf{v} = \mathbf{C} - \mathbf{A} \). This gives us vectors \( \mathbf{u} = (2,0,-7) \) and \( \mathbf{v} = (-7,-5,-3) \).
02

Find the cross-product

Calculate the cross product of vectors u and v: \( \mathbf{u} \times \mathbf{v} \). The cross product of two vectors can be calculated as follows: \( \mathbf{u} \times \mathbf{v} = \begin{bmatrix} i & j & k \ 2 & 0 & -7 \ -7 & -5 & -3 \end{bmatrix} \), where i, j, k are unit vectors. The cross product is calculated as follows: \( \mathbf{u} \times \mathbf{v} = 35i - 49j -10k \).
03

Compute the magnitude

Compute the magnitude (||u x v||) of the cross-product, which is: \( \sqrt{(35)^2 + (-49)^2 + (-10)^2} = \sqrt{1226} \)
04

Compute the area of the triangle

The Area of the triangle is given by \( A = \frac{1}{2} \|\mathbf{u} \times \mathbf{v}\| \). Using the magnitude calculated in the previous step, we can find that the area is: \( A = \frac{1}{2} * \sqrt{1226} \)

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Most popular questions from this chapter

Prove that the angle \(\theta\) between \(\mathbf{u}\) and \(v\) is found using \(\|\mathbf{u} \times \mathbf{v}\|=\|\mathbf{u}\|\|\mathbf{v}\| \sin \theta\).

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