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Find the angle between the vectors a = (4, 4, -2) and b = (5, -3, 3). Solution Since |al = V42 + 42 + (-2)2 and |b| = V52 + (-3)2 + 32 = and since a ·b = (4)(( + (4)(-3) + (-2)(O) we have from this corollary a·b cos(0) |a||b| So the angle between a and b is as follows. (Round your final answer to two decin 0 = cos rad

Find the angle between the vectors a = (4, 4, -2) and b = (5, -3, 3). Solution Since |al = V42 + 42 + (-2)2 and |b| = V52 + (-3)2 + 32 = and since a ·b = (4)(( + (4)(-3) + (-2)(O) we have from this corollary a·b cos(0) |a||b| So the angle between a and b is as follows. (Round your final answer to two decin 0 = cos rad

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Example Video Example Find the angle between the vectors a = (4, 4, -2) and b = (5, -3, 3). Solution Since |a| = V42 + 42 + (-2)² = and |b| = V52 + (-3)² + 32 = and since a ·b = (4)(( + (4)(-3) + (-2)( O) we have from this corollary a ·b cos(0) |a||b| So the angle between a and b is as follows. (Round your final answer to two decimal places.) 1 e = COS |rad
Transcribed Image Text:Example Video Example Find the angle between the vectors a = (4, 4, -2) and b = (5, -3, 3). Solution Since |a| = V42 + 42 + (-2)² = and |b| = V52 + (-3)² + 32 = and since a ·b = (4)(( + (4)(-3) + (-2)( O) we have from this corollary a ·b cos(0) |a||b| So the angle between a and b is as follows. (Round your final answer to two decimal places.) 1 e = COS |rad
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