Explanation Given: x + 2 2 = y − 1 1 = z 4 A has coordinates ( 1 , − 1 , 2 ) Length ofline segment B C ¯ =5 Calculation: Consider a triangle ABC whose vertices B and C lie along the line, x + 2 2 = y − 1 1 = z 4 ...... ( 1 ) The co-ordinates of point A are ( 1 , − 1 , 2 ) and the length of line-segment BC is 5 units. Use the formula for the distance from a point to a line, “The distance from the point P to the line L is given by, d = ‖ v → × Q P → ‖ ‖ v → ‖ ...... ( 2 ) Where v → vector is parallel to L and Q is any point on L ." Compare equation (1) with symmetric form of a line x − x 0 A = y − y 0 B = z − z 0 C , a point on the given line is Q ( − 2 , 1 , 0 ) and a vector parallel to the given line is v → = 2 i ^ + j ^ + 4 k ^ . The vector Q A → will be, Q A → = O A → − O Q → = ( i ^ − j ^ + 2 k ^ ) − ( − 2 i ^ + 1 j ^ + 0 k ^ ) = ( 1 + 2 ) i ^ + ( − 1 − 1 ) j ^ + ( 2 − 0 ) k ^ = 3 i ^ − 2 j ^ + 2 k ^ Compute the cross product of vectors v → = 2 i ^ + j ^ + 4 k ^ and Q A → = 3 i ^ − 2 j ^ + 2 k �

7th Edition

ISBN: 9781524916817

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

Publisher: Kendall Hunt Publishing

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Coordinate Geometry

Coordinate geometry, also known as analytic geometry or Cartesian geometry in classical mathematics, is a type of geometry that is studied using a coordinate system. Coordinate geometry is a branch of mathematics that describes the position of points on …

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Chapter 9, Problem 52SP

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To find: the area of the triangle

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Chapter 9, Problem 51SP

Chapter 9, Problem 53SP

Chapter 9 Solutions

Calculus

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the area of the triangle

Solution Summary: The author explains how to find the required area of the triangle by comparing the equation with symmetric form of a line.

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Chapter 9 Solutions