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Unformatted text preview: 2z = 0
3. 3x + 2y − 5z = 0 correct 1, 0, −3 , i
−
−
→−
→
n = QR × QS = −3 1. 5x − 3y + 2z = 0 is normal to the plane.
On the other hand, if P (x, y, z ) is an arbitrary point on the plane, then the displacement vector
−
−
→
v = P Q = x − 2, y + 1, z − 1
lies in the plane, so 5. 3x − 2y + 5z + 3 = 0
6. 5x + 3y − 2z + 3 = 0
Explanation:
The scalar equation for the plane through
P (a, b, c) with normal vector
n = Ai + B j + C k
is v · n = 0. A(x − a) + B (y − b) + C (z − c) = 0 . Now
i j −3 −1
1 0 k
0 = 3, −9, 1 . −3 But P (a, b, c) = (0, 0, 0) if the plane passes
through the origin, while
n = 3i + 2j − 5k But then
v · n = 3(x − 2) − 9(y + 1) + (z − 1)
= 3x − 9y + z − 16 = 0 . if the plane is parallel to
3x + 2y − 5z = 3 geyer (dag2798) – HW09 – rusin – (55735)
since parallel planes have the same normal
vector.
Consequently, the plane has equation 10.0 points For which of the following quadratic relations is its graph a hyperbolic paraboloid? 3x + 2y − 5z = 0 .
011 012 6 1. x2 + y 2 − z 2 = 1 10.0 points Express in vector form the line passing
through the point P (4, −2, 4) and perpendicular to the plane 2. z 2 = x2 + y 2 2x + y − 2z = 6 . 4. z = x2 + y 2
5. z = y 2 − x2 correct 1. r(t) = 2 − 4t, −1 + 2t, −2 + 4t 6. z 2 − x2 − y 2 = 1 2. r(t) = 2 + 4t, 1 − 2t, −2 + 4t
3. r(t) = −4...
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This note was uploaded on 05/12/2013 for the course M 408d taught by Professor Sadler during the Spring '07 term at University of Texas at Austin.
 Spring '07
 Sadler
 Determinant, Multivariable Calculus

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