To determine Find the distance between the given line to the plane. Explanation Given information: The line equation is, x = 2 + t , y = 1 + t , z = − 1 2 − 1 2 t The plane equation is, x + 2 y + 6 z = 10 Formula used: The formula to calculate the parametric equations for a line through P 0 ( x 0 , y 0 , z 0 ) parallel to nonzero vector v = v 1 i + v 2 j + v 3 k is, x = x 0 + t v 1 (1) y = y 0 + t v 2 (2) z = z 0 + t v 3 (3) The vector form for the two points P 1 ( x 1 , y 1 , z 1 ) and P 2 ( x 2 , y 2 , z 2 ) is calculated as follows, P 1 P 2 → = ( x 2 − x 1 ) i + ( y 2 − y 1 ) j + ( z 2 − z 1 ) k (4) The formula to calculate the distance from a point S to a plane through a point P with a normal n is, d = | P S ⇀ ⋅ n | n | | (5) Calculation: Compare the given line equation with the equations (1), (2) and (3) to find v 1 , v 2 , v 3 of non-zero vector. v 1 = 1 , v 2 = 1 , v 3 = − 1 2 Therefore, non-zero vector v = v 1 i + v 2 j + v 3 k becomes v = 1 i + 1 j − 1 2 k . Generally, the vector n = A i + B j + C k is normal to the plane A x + B y + C z = D . Therefore, the non-zero vector of the given plane equation x + 2 y + 6 z = 10 is expressed as, n = 1 i + 2 j + 6 k Given line equation is parallel to the plane since, v ⋅ n = ( 1 i + 1 j − 1 2 k ) ⋅ ( 1 i + 2 j + 6 k ) = ( 1 + 2 − 3 ) = 0 Assume the value of t to obtain the point S ( x , y , z ) that lies on the line equation

University Calculus: Early Transcendentals (4th Edition)

4th Edition

ISBN: 9780134995540

Author: Joel R. Hass, Christopher E. Heil, Przemyslaw Bogacki, Maurice D. Weir, George B. Thomas Jr.

Publisher: PEARSON

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Chapter 11.5, Problem 46E

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Find the distance between the given line to the plane.

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Chapter 11.5, Problem 45E

Chapter 11.5, Problem 47E

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