HW09-solutions

# HW09-solutions - moore(jwm2685 – HW09 – gilbert...

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Unformatted text preview: moore (jwm2685) – HW09 – gilbert – (55485) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Which of the following statements are true for all lines and planes in 3-space? I. two lines parallel to a third line are par- allel , II. two planes parallel to a third plane are parallel , III. two lines perpendicular to a plane are parallel . 1. I and II only 2. II only 3. all of them correct 4. III only 5. none of them 6. I and III only 7. II and III only 8. I only Explanation: I. TRUE: each of the two lines has a direc- tion vector parallel to the direction vector of the third line, so must be scalar multiples of each other. II. TRUE: each of the two planes has a normal vector parallel to the normal vector of the third plane, and so are parallel, hence the planes are parallel. III. TRUE: the two lines will have direction vectors parallel to the normal vector of the plane, and so be parallel, hence the two lines are parallel. 002 10.0 points Which of the following surfaces is the graph of 6 x + 4 y + 3 z = 12 in the first octant? 1. x y z 2. x y z correct 3. x y z moore (jwm2685) – HW09 – gilbert – (55485) 2 4. x y z 5. x y z 6. x y z Explanation: Since the equation is linear, it’s graph will be a plane. To determine which plane, we have only to compute the intercepts of 6 x + 4 y + 3 z = 12 . Now the x-intercept occurs at y = z = 0, i.e. at x = 2; similarly, the y-intercept is at y = 3, while the z-intercept is at z = 4. By inspection, therefore, the graph is x y z 003 10.0 points Find parametric equations for the line pass- ing through the point P (2 , − 3 , 4) and parallel to the vector ( 4 , 1 , − 4 ) . 1. x = 4 + 2 t, y = 1 − 3 t, z = − 4 + 4 t 2. x = 2 + 4 t, y = − 3 + t, z = 4 − 4 t correct 3. x = 4 − 2 t, y = − 1 + 3 t, z = − 4 + 4 t 4. x = 4 + 2 t, y = 1 + 3 t, z = 4 − 4 t 5. x = 2 − 4 t, y = 3 − t, z = 4 − 4 t 6. x = − 2 + 4 t, y = 3 + t, z = − 4 − 4 t Explanation: A line passing through a point P ( a, b, c ) and having direction vector v is given para- metrically by r ( t ) = a + t v , a = ( a, b, c ) . Now for the given line, a = ( 2 , − 3 , 4 ) , v = ( 4 , 1 , − 4 ) . Thus r ( t ) = ( 2 + 4 t, − 3 + t, 4 − 4 t ) . Consequently, x = 2 + 4 t, y = − 3 + t, z = 4 − 4 t are parametric equations for the line. keywords: line, parametric equations, direc- tion vector, point on line 004 10.0 points moore (jwm2685) – HW09 – gilbert – (55485) 3 A line ℓ passes through the point P (3 , 2 , 1) and is perpendicular to the plane x + 3 y + z = 6 . At what point Q does ℓ intersect the xy- plane?...
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HW09-solutions - moore(jwm2685 – HW09 – gilbert...

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