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Engineering Calculus Notes 73

# Engineering Calculus Notes 73 - Q P Q x,y,z −→ N P P...

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1.5. PLANES 61 For example the plane given by the equation 6 x + 9 y 3 z = 12 has normal vector −→ N = 6 −→ ı + 9 −→ 3 −→ k = 3 2 (4 −→ ı 6 −→ + 2 −→ k ) so multiplying the equation by 2 / 3, we get an equivalent equation 4 x 6 y + 2 z = 8 which shows that this plane is parallel to (and does not intersect) the plane specified earlier by 4 x 6 y + 2 z = 8 (since 8 negationslash = 8). We can also use vector ideas to calculate the distance from a point Q ( x,y,z ) to the plane P given by an equation Ax + By + Cz = D. proj −→ N −−→ P 0 Q dist( Q, P ) Q ( x,y,z ) −→ N P P 0 ( x 0 ,y
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Unformatted text preview: Q, P ) Q ( x,y,z ) −→ N P P ( x ,y ,z ) Figure 1.32: dist( Q, P ) I± P ( x ,y ,z ) is any point on P (see Figure 1.32 ) then the (perpendicular) distance ±rom Q to P is the (length o± the) projection o± −−→ P Q = △ x −→ ı + △ y −→ + △ z −→ k in the direction o± −→ N = A −→ ı + B −→ + C −→ k dist( Q, P ) = v v v proj −→ N −−→ P Q v v v...
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