Exam 11: Vectors; Lines, Planes, and Quadric Surfaces in Space

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The symmetric equations of the line through (3,2,1) in the direction of 3ij+k3 \mathbf { i } - \mathbf { j } + \mathbf { k } is

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In the three-dimensional Euclidean space, the graph of x2+y216z2=1x ^ { 2 } + \frac { y ^ { 2 } } { 16 } - z ^ { 2 } = - 1 is

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In the three-dimensional Euclidean space, the graph of z=x24+y29z = \frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 9 } is

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The center and radius of the sphere x2+y2+z24x=0x ^ { 2 } + y ^ { 2 } + z ^ { 2 } - 4 x = 0 are

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The angle, in degrees and correct to two decimal places, between v=2i3j+k\mathbf { v } = 2 \mathbf { i } - 3 \mathbf { j } + \mathbf { k } and w=ij+k\mathbf { w } = \mathbf { i } - \mathbf { j } + \mathbf { k } is

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The equation of the plane through (2,-1,3) parallel to the xy-plane is

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The symmetric equations of the line through (1,2,-1) perpendicular to each of the vectors 2i+4j2k2 \mathbf { i } + 4 \mathbf { j } - 2 \mathbf { k } and 3i2j+k- 3 \mathbf { i } - 2 \mathbf { j } + \mathbf { k } is

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The parametric equations of the line of intersection of the planes x2y+4z=2x - 2 y + 4 z = 2 and x+y2z=5x + y - 2 z = 5 are

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The point of intersection of the line x=2t+1,y=5t+1,z=3tx = 2 t + 1 , y = 5 t + 1 , z = 3 t and the plane x+y+z=2x + y + z = 2 is

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The distance between P1=(0,4,1)P _ { 1 } = ( 0,4,1 ) and P2=(1,8,2)P _ { 2 } = ( 1,8,2 ) is

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In the three-dimensional Euclidean space, the graph of 4x2+y2=164 x ^ { 2 } + y ^ { 2 } = 16 is

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The value of a that makes v=i+jk\mathbf { v } = \mathbf { i } + \mathbf { j } - \mathbf { k } and w=5i+aj+k\mathbf { w } = 5 \mathbf { i } + a \mathbf { j } + \mathbf { k } orthogonal is

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In the three-dimensional Euclidean space, the graph of z=x216y29z = \frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 9 } is

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Let , and u=1,1,6\mathbf { u } = \langle 1,1 , - 6 \rangle Then v=3,0,8\mathbf { v } = \langle 3,0,8 \rangle is

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The equation of the plane through (2,3,-1) normal to the line x12=y13=z+12\frac { x - 1 } { 2 } = \frac { y - 1 } { 3 } = \frac { z + 1 } { - 2 } is

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The equation of the plane through (-3,1,4) parallel to the yz-plane is

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Let , and u=1,1,6,\langle \mathbf { u } = 1,1 , - 6 , \rangle Then v=3,0,8\mathbf { v } = \langle 3,0,8 \rangle is

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The distance between P1=(5,6,1)P _ { 1 } = ( 5,6 , - 1 ) and P2=(3,8,2)P _ { 2 } = ( 3,8,2 ) is

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Let P1=(3,2,5)P _ { 1 } = ( 3 , - 2,5 ) and P2=(10,4,3)P _ { 2 } = ( 10,4,3 ) If v is P1P2\overrightarrow { P _ { 1 } P _ { 2 } } in standard position, then v is

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The distance from the point (2,3,4)( 2 , - 3,4 ) to the plane x+2y+2z=13x + 2 y + 2 z = 13 is

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