Explanation Given information : The distance from point P ( x , y , z ) to the point P 0 ( − 1 , 2 , 4 ) is same as the distance from P ( x , y , z ) to the plane 2 x − 5 y + 3 z =7 . Formula/s used : The distance AB between two points A ( x 1 , y 1 , z 1 ) and B ( x 2 , y 2 , z 2 ) is given by A B = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 + ( z 2 − z 1 ) 2 . The distance from point ( α , β , γ ) to the plane A x + B y + C z = D is given by d = | A x + B y + C z - D A 2 + B 2 + C 2 | . Calculations: We find locus (equation) of the point P ( x , y , z ) under given condition. Let Q is the foot of perpendicular from point P ( x , y , z ) to the plane 2 x − 5 y + 3 z =7 and P 0 ( − 1 , 2 , 4 ) is other point in space. The condition for the locus (curve) of point P ( x , y , z ) are shown in figure here. The distance P P 0 = ( x + 1 ) 2 + ( y − 2 ) 2 + ( z − 4 ) 2 , and length of perpendicular PQ is P Q = | 2 x − 5 y + 3 z -7 2 2 + ( − 5 ) 2 + 3 2 | = | 2 x − 5 y + 3 z -7 38 |

6th Edition

ISBN: 9781465208880

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 9.6, Problem 39PS

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Chapter 9.6, Problem 38PS

Chapter 9.6, Problem 40PS

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Chapter 9 Solutions

Calculus

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Tofind: Anequation for the set of all points P ( x , y , z ) under the given restriction or the locus of a point P ( x , y , z ) under given condition.

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