Explanation Approach: A vector from the initial point A ( x 1 , x 2 ) and the terminal point B ( y 1 , y 2 ) is A B → = 〈 y 1 − x 1 , y 2 − x 2 〉 . The dot product of two vectors, u = 〈 u 1 , u 2 〉 = u 1 i + u 2 j and v = 〈 v 1 , v 2 〉 = v 1 i + v 2 j , is defined as u ⋅ v = u 1 v 1 + u 2 v 2 . The length or magnitude of a vector u is denoted as ‖ u ‖ and is defined as ‖ u ‖ = u ⋅ u . The magnitude of a vector is always positive. The angle θ between two non-zero vectors u and v can be found by using the formula, cos θ = u ⋅ v ‖ u ‖ ‖ v ‖ . Given Information: The vertices of a triangle are A ( − 3 , 0 ) , B (2, 2) and C (0, 6). Calculation: A vector from A ( − 3 , 0 ) to B (2, 2) is A B = 〈 2 − ( − 3 ) , 2 − 0 〉 = 〈 5 , 2 〉 . A vector from B (2, 2) to C (0, 6) is B C = 〈 0 − 2 , 6 − 2 〉 = 〈 − 2 , 4 〉 . A vector from A ( − 3 , 0 ) to C (0, 6) is A C = 〈 0 − ( − 3 ) , 6 − 0 〉 = 〈 3 , 6 〉 . A B ⋅ A C = 〈 5 , 2 〉 ⋅ 〈 3 , 6 〉 = ( 5 ) ( 3 ) + ( 2 ) ( 6 ) = 15 + 12 = 27 A B ⋅ B C = 〈 5 , 2 〉 ⋅ 〈 − 2 , 4 〉 = ( 5 ) ( − 2 ) + ( 2 ) ( 4 ) = ( − 10 ) + 8 = − 2 ‖ A B ‖ 2 = A B ⋅ A B = 〈 5 , 2 〉 ⋅ 〈 5 , 2 〉 = ( 5 ) ( 5 ) + ( 2 ) ( 2 ) = 25 + 4 = 29 Therefore, ‖ A B ‖ = 29 . ‖ B C ‖ 2 = B C ⋅ B C = 〈 − 2 , 4 〉 ⋅ 〈 − 2 , 4 〉 = ( − 2 ) ( − 2 ) + ( 4 ) ( 4 ) = 4 + 16 = 20 Therefore, ‖ B C ‖ = 20 = 2 5

Bundle: Precalculus, Loose-leaf Version, 10th + WebAssign Printed Access Card for Larson's Precalculus, 10th Edition, Single-Term

10th Edition

ISBN: 9781337605090

Author: Ron Larson

Publisher: Cengage Learning

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Chapter 6.4, Problem 45E

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The interior angles of the triangle whose vertices are (−3,0), (2, 2) and (0, 6) using the concept of vector.

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Chapter 6.4, Problem 44E

Chapter 6.4, Problem 46E

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Chapter 6 Solutions

Bundle: Precalculus, Loose-leaf Version, 10th + WebAssign Printed Access Card for Larson's Precalculus, 10th Edition, Single-Term

Ch. 6.1 - Checkpoint For the triangle shown, A=30, B=45, and...Ch. 6.1 - Prob. 2CP Ch. 6.1 - Prob. 3CP Ch. 6.1 - Prob. 4CP Ch. 6.1 - Prob. 5CP Ch. 6.1 - Prob. 6CP Ch. 6.1 - Prob. 7CP Ch. 6.1 - Prob. 1E Ch. 6.1 - Prob. 2E Ch. 6.1 - Prob. 3E

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The interior angles of the triangle whose vertices are ( − 3 , 0 ) , (2, 2) and (0, 6) using the concept of vector .

Solution Summary: The author explains how to find the interior angles of the triangle whose vertices are (-3,0), (2, 2), and (0, 6) using the concept of vector.

To find:

The interior angles of the triangle whose vertices are (−3,0), (2, 2) and (0, 6) using the concept of vector.

Want to see the full answer?

Chapter 6 Solutions

The interior angles of the triangle whose vertices are ( − 3 , 0 ) , (2, 2) and (0, 6) using the concept of vector .

Solution Summary: The author explains how to find the interior angles of the triangle whose vertices are (-3,0), (2, 2), and (0, 6) using the concept of vector.

To find: The interior angles of the triangle whose vertices are (−3,0), (2, 2) and (0, 6) using the concept of vector.

Want to see the full answer?

Chapter 6 Solutions

To find:

The interior angles of the triangle whose vertices are (−3,0), (2, 2) and (0, 6) using the concept of vector.