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Let the direction vector of the line be u = (p, q, r). To find u and a second point on the line, compare the x - X1 equation of the line 2 y - 4 = z with the standard form of the equation y - Y1 z - z1 -1 p r (X1, Y1, Z4) ,4, 0 2 -1, 1) u = Step 2 Let v be the vector from the point (0, 4, 0) to the point (12, -10, 6), then v can be given in the compon form as (8,0, – 16) × v = (12, – 14, 6) Step 3 You have found u = (2, -1, 1) and v = (12, -14, 6). Find the normal vector for the planen = u x v. i 1 2 n = u x v = -1 12 -14 V -14 = 8i - 16 16 k Step 4 The plane containing the point (x2, Y2, Z2) = (12, -10, 6) and having the normal vector (8, 0, –16) = (a, b, c) can be represented in the standard form as, n = 8i – 16k = a(x — х2) + b(y — У2) + c(z — z2) 3 о. = Substitute in the equation of the plane and simplify. 8(х — ) + (y + 10) – 16(z – 6) = 0

Let the direction vector of the line be u = (p, q, r). To find u and a second point on the line, compare the x - X1 equation of the line 2 y - 4 = z with the standard form of the equation y - Y1 z - z1 -1 p r (X1, Y1, Z4) ,4, 0 2 -1, 1) u = Step 2 Let v be the vector from the point (0, 4, 0) to the point (12, -10, 6), then v can be given in the compon form as (8,0, – 16) × v = (12, – 14, 6) Step 3 You have found u = (2, -1, 1) and v = (12, -14, 6). Find the normal vector for the planen = u x v. i 1 2 n = u x v = -1 12 -14 V -14 = 8i - 16 16 k Step 4 The plane containing the point (x2, Y2, Z2) = (12, -10, 6) and having the normal vector (8, 0, –16) = (a, b, c) can be represented in the standard form as, n = 8i – 16k = a(x — х2) + b(y — У2) + c(z — z2) 3 о. = Substitute in the equation of the plane and simplify. 8(х — ) + (y + 10) – 16(z – 6) = 0

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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(p, q, r). To find u and a second point on the line, compare the x - X1 Let the direction vector of the line be u = equation of the line 2 У — 4 = z with the standard form of the equation y - Y1 z - Z1 -1 r (X1, Y1, Z1) = (0 4, 0 (2 -1, 1) u = Step 2 Let v be the vector from the point (0, 4, 0) to the point (12, -10, 6), then v can be given in the component form as (8,0, – 16) × | (12, – 14, 6) V = Step 3 You have found u = (2, –1, 1) and v = (12, –14, 6). Find the normal vector for the plane n = u x v. j k 1 v -1 1 n = u x v = 12 |-14 -14 6 = 8i 16 16 k Step 4 The plane containing the point (x2, Y2, z2) = (12, -10, 6) and having the normal vector n = 8i – 16k = (8, 0, –16) = (a, b, c) can be represented in the standard form as, a(x — х2) + b(y - У2) + с(z — z2) %3D 0. Substitute in the equation of the plane and simplify. 8(х — ) + (y + 10) – 16(z – 6) = 0
Transcribed Image Text:(p, q, r). To find u and a second point on the line, compare the x - X1 Let the direction vector of the line be u = equation of the line 2 У — 4 = z with the standard form of the equation y - Y1 z - Z1 -1 r (X1, Y1, Z1) = (0 4, 0 (2 -1, 1) u = Step 2 Let v be the vector from the point (0, 4, 0) to the point (12, -10, 6), then v can be given in the component form as (8,0, – 16) × | (12, – 14, 6) V = Step 3 You have found u = (2, –1, 1) and v = (12, –14, 6). Find the normal vector for the plane n = u x v. j k 1 v -1 1 n = u x v = 12 |-14 -14 6 = 8i 16 16 k Step 4 The plane containing the point (x2, Y2, z2) = (12, -10, 6) and having the normal vector n = 8i – 16k = (8, 0, –16) = (a, b, c) can be represented in the standard form as, a(x — х2) + b(y - У2) + с(z — z2) %3D 0. Substitute in the equation of the plane and simplify. 8(х — ) + (y + 10) – 16(z – 6) = 0
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