Consequently the line intersects the xy plane at q1 1

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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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