HW09-solutions

# Ax a b y b c z c 0 now i j 3 1 1 0 k 0 3

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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.

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