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Chapter 9: Problem 38

The vertices of a triangle in space are \(\left(x_{1}, y_{1}, z_{1}\right),\left(x_{2}, y_{2}, z_{2}\right)\) and \(\left(x_{3}, y_{3}, z_{3}\right) .\) Explain how to find a vector perpendicular to the triangle.

Short Answer

Expert verified
To find the vector perpendicular to the triangle, calculate the cross product of two adjacent vectors (sides) of the triangle. This will give a vector that is perpendicular to the triangle's plane.

Step by step solution

01

Identify the Vectors of the Triangle

We can represent two sides of the triangle as vectors. Let's call them AB and AC, where A is the first vertex \((x_{1}, y_{1}, z_{1})\), B is the second vertex \((x_{2}, y_{2}, z_{2})\), and C is the third vertex \((x_{3}, y_{3}, z_{3})\). So, AB = B - A = \((x_{2} - x_{1}, y_{2} - y_{1}, z_{2} - z_{1})\) and AC = C - A = \((x_{3} - x_{1}, y_{3} - y_{1}, z_{3} - z_{1})\).
02

Calculate the Cross Product

To find a vector perpendicular to the plane of the triangle, we need to calculate the cross product of AB and AC. Using the formula of the cross product of two vectors, we have AB x AC = \((y_{2} - y_{1})*(z_{3} - z_{1}) - (z_{2} - z_{1})*(y_{3} - y_{1}), (z_{2} - z_{1})*(x_{3} - x_{1}) - (x_{2} - x_{1})*(z_{3} - z_{1}), (x_{2} - x_{1})*(y_{3} - y_{1}) - (y_{2} - y_{1})*(x_{3} - x_{1}))\). This vector is perpendicular to the plane of the triangle.
03

Finishing Up

The vector obtained from the cross product AB x AC is the vector perpendicular to the triangle defined by vertices A, B and C. It can also be normalized if a unit vector is necessary.

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Most popular questions from this chapter

In Exercises 71 and \(72,\) determine the values of \(c\) that satisfy the equation. Let \(\mathbf{u}=\mathbf{i}+2 \mathbf{j}+\mathbf{3 k}\) and \(\mathbf{v}=\mathbf{2} \mathbf{i}+\mathbf{2} \mathbf{j}-\mathbf{k}\) \(\|c \mathbf{v}\|=5\)

Find \((\mathbf{a}) \mathbf{u} \cdot \mathbf{v},(\mathbf{b}) \mathbf{u} \cdot \mathbf{u},(\mathbf{c})\|\mathbf{u}\|^{2},(\mathbf{d})(\mathbf{u} \cdot \mathbf{v}) \mathbf{v}\) and \((e) u \cdot(2 v)\). $$ \mathbf{u}=\langle-4,8\rangle, \quad \mathbf{v}=\langle 6,3\rangle $$

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Find the vector \(z,\) given that \(\mathbf{u}=\langle 1,2,3\rangle\) \(\mathbf{v}=\langle 2,2,-1\rangle,\) and \(\mathbf{w}=\langle 4,0,-4\rangle\) \(2 \mathbf{z}-3 \mathbf{u}=\mathbf{w}\)

In Exercises 57-60, determine which of the vectors is (are) parallel to \(\mathrm{z}\). Use a graphing utility to confirm your results. \(\mathrm{z}=\langle 3,2,-5\rangle\) (a) \langle-6,-4,10\rangle (b) \(\left\langle 2, \frac{4}{3},-\frac{10}{3}\right\rangle\) (c) \langle 6,4,10\rangle (d) \langle 1,-4,2\rangle
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