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High School Math Calculus Precalculus with Limits: A Graphing Approach

The process of finding a vector perpendicular to the triangle.

The process of finding a vector perpendicular to the triangle.

Precalculus with Limits: A Graphing Approach

Precalculus with Limits: A Graphing Approach

7th Edition

ISBN: 9781305071711

Author: Ron Larson

Publisher: Brooks/Cole

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Question

Chapter 10, Problem 78RE

To determine

To explain: The process of finding a vector perpendicular to the triangle.

Expert Solution & Answer

Answer to Problem 78RE

The vector

{(y2−y1)(z3−z1)−(y3−y1)(z2−z1)}i−{(x2−x1)(z3−z1)−(x3−x1)(z2−z1)}j+{(x2−x1)(y3−y1)−(x3−x1)(y2−y1)}k

is perpendicular to a triangle having vertices (x1,y1,z1),(x2,y2,z2) and (x3,y3,z3) .

Explanation of Solution

Given information:

The vertices of the triangle are,

A=(x1,y1,z1)B=(x2,y2,z2)C=(x3,y3,z3)

Calculation:

Calculate the vector of two sides.

AB=(x2−x1,y2−y1,z2−z1)AB=(x2−x1)i+(y2−y1)j+(z2−z1)kAC=(x3−x1,y3−y1,z3−z1)AC=(x3−x1)i+(y3−y1)j+(z3−z1)k .

So, the two side vectors of the triangle are

AB=(x2−x1)i+(y2−y1)j+(z2−z1)kAC=(x3−x1)i+(y3−y1)j+(z3−z1)k .

Find the cross product of these two vectors.

AB×AC=|ijk(x2−x1)(y2−y1)(z2−z1)(x3−x1)(y3−y1)(z3−z1)|=[{(y2−y1)(z3−z1)−(y3−y1)(z2−z1)}i−{(x2−x1)(z3−z1)−(x3−x1)(z2−z1)}j+{(x2−x1)(y3−y1)−(x3−x1)(y2−y1)}k]

The vectors obtain from the cross product of two side vectors of the triangle is said to be a vector perpendicular to a triangle.

Therefore, the vector

{(y2−y1)(z3−z1)−(y3−y1)(z2−z1)}i−{(x2−x1)(z3−z1)−(x3−x1)(z2−z1)}j+{(x2−x1)(y3−y1)−(x3−x1)(y2−y1)}k

is perpendicular to a triangle having vertices (x1,y1,z1),(x2,y2,z2) and (x3,y3,z3) .

Chapter 10, Problem 77RE

Chapter 10, Problem 1CT

Chapter 10 Solutions

Precalculus with Limits: A Graphing Approach

Ch. 10.1 - Prob. 1E Ch. 10.1 - Prob. 2E Ch. 10.1 - Prob. 3E Ch. 10.1 - Prob. 4E Ch. 10.1 - Prob. 5E Ch. 10.1 - Prob. 6E Ch. 10.1 - Prob. 7E Ch. 10.1 - Prob. 8E Ch. 10.1 - Prob. 9E Ch. 10.1 - Prob. 10E

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