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Unformatted text preview: ving direction vector v is
given parametrically by
r(t) = a + tv , a= a, b, c . r(t) = a + tv ,
Now −
−
→
PQ = a = a, b, c . 1, 4, 3 is a direction vector for the given line, so
a = 4, 1, 2 , v= 1, 4, 3 . Thus But for ℓ,
a = 1, 4, 4 , v = 2, 3, 4 . Thus r(t) = 4 + t, 1 + 4t, 2 + 3t . Consequently,
r(t) = 1 + 2t, 4 + 3t, 4 + 4t , so z = 0 when t = −1. Consequently, the line
ℓ intersects the xy plane at
Q(−1, 1, 0) . x = 4 + t, y = 1 + 4t, z = 2 + 3t
are parametric equations for the line.
009 10.0 points geyer (dag2798) – HW09 – rusin – (55735)
Find an equation for the plane passing
through the points
Q(2, −1, 1) , 5 Consequently, the plane R(−1, −2, 1) , 3x − 9y + z − 16 = 0 S (3, −1, −2) .
passes through Q, R and S .
1. 3x + y + 9z + 16 = 0
keywords: plane, cross product, plane determined by three points, dot product 2. 3x − 9y + z + 16 = 0
3. 3x + y + 9z − 16 = 0 010
4. 3x − y + 9z + 16 = 0 Find an equation for the plane passing
through the origin and parallel to the plane 5. 3x − 9y + z − 16 = 0 correct 3x + 2y − 5z = 3 . 6. 3x + 9y − z − 16 = 0
Explanation:
Since the points Q, R, and S lie in the
plane, the displacement vectors
−
−
→
QR = −3, −1, 0 ,
−
→
QS = 10.0 points lie in the plane. Thus the cross product
j 0 0 −3 4. 2x − 5y − 2z + 3 = 0 k −1 1 2. 2x + 5y +...
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 Spring '07
 Sadler
 Determinant, Multivariable Calculus

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