right hand
line; plane
curve (a parabola); surface (a cylinder)
a hyperbolic paraboloid
a hyperboloid of two sheets
a hyperboloid of one sheet
; ; . Yes, it is a right triangle as .
, , . The triangle is isosceles.
, , . The points lie on a line.
, , . The points do not lie on a line.
Yes, as opposite sides have equal length. ; .
Center at ; radius = 3
Center at ; radius =
closer to the surface
Interior of a sphere with radius 1 centered at the origin.
Region bounded between the planes (the coordinate plane) and .
The first octant of space along with its adjacent quarter planes; all points where each of , and are positive or zero. (Analogous to the first quadrant in the plane.)
All points in space where the value is greater than 3; viewing space as often depicted in this text, this is the region “to the right” of the plane (which is parallel to the coordinate plane.)
(a)
(b)
(b)
(a)
Answers will vary.
is a point; is a vector that describes a displacement of 1 unit in the -direction and 2 units in the -direction.
A vector with magnitude 1.
Direction
Their respective unit vectors are parallel; unit vectors and are parallel if .
It stretches the vector by a factor of 2, and points it in the opposite direction.
; ; .
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; ; .
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Sketch of shifted for clarity.
, , ,
, , ,
, , ,
, , ,
When and have the same direction. (Note: parallel is not enough.)
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The magnitude of the force on each chain is lb.
The magnitude of the force on each chain is lb.
The magnitude of the force on the chain with angle is approx. lb; the magnitude of the force on the chain with angle is approx. lb.
The magnitude of the force on each chain is 50lb.
; the weight is lifted ft (about 3.5in).
; the weight is lifted ft (about 1/16th of an inch).
; the weight is lifted ft.
; the weight is lifted ft.
Scalar
The magnitude of a vectors is the square root of the dot product of a vector with itself; that is, .
By considering the sign of the dot product of the two vectors. If the dot product is positive, the angle is acute; if the dot product is negative, the angle is obtuse.
“Perpendicular” is one answer.
not defined
Answers will vary.
Answers will vary.
Answers will vary; two possible answers are and .
Answers will vary; two possible answers are and .
Answers will vary; two possible answers are and .
Answers will vary; two possible answers are and .
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1.96lb
5lb
ft-lb
ft-lb
ft-lb
ft-lb
ft-lb
vector
right hand rule
“Perpendicular” is one answer.
T
Torque
T
18
0
36
Answers will vary.
Answers will vary.
5
21
0
5
any unit vector orthogonal to works (such as ).
ft-lb
ft-lb
ft-lb
ft-lb
With and , we have
With , we have
A point on the line and the direction of the line.
parallel
parallel, skew
Answers will vary
vector:
parametric: , ,
symmetric:
vector:
parametric: , ,
symmetric:
Answers can vary: vector:
parametric: , ,
symmetric:
Answers can vary: vector:
parametric: , ,
symmetric:
Answers can vary; here the direction is given by : vector:
parametric: , ,
symmetric:
Answers can vary; here the direction is given by : vector:
parametric: , ,
symmetric: not defined, as some components of the direction are 0.
Answers can vary; here the direction is given by : vector:
parametric: , ,
symmetric:
Answers can vary; here the direction is given by : vector:
parametric: , ,
symmetric:
vector:
parametric: ,
symmetric:
vector:
parametric: ,
symmetric: not defined
parallel
intersecting;
intersecting;
same
skew
parallel
same
skew
Since both and are on the line, is parallel to . Thus , giving a distance of .
(Note: this solution is easier once one has studied Section 11.6.) Since the two lines intersect, we can state for some scalars and . (Here we abuse notation slightly and add points to vectors.) Thus . Vector is the cross product of and , hence is orthogonal to both, and hence is orthogonal to . Thus , and the distance between lines is .
The distance formula cannot be used because since and are parallel, is and we cannot divide by .
Since and are parallel, lies in the plane formed by the two lines. Thus is orthogonal to this plane, and is parallel to the plane, but still orthogonal to both and . We desire the length of the projection of onto , which is what the formula provides.
Since the lines are parallel, one can measure the distance between the lines at any location on either line (just as to find the distance between straight railroad tracks, one can use a measuring tape anywhere along the track, not just at one specific place.) Let and as given by the equations of the lines, and apply the formula for distance between a point and a line.
A point in the plane and a normal vector (i.e., a direction orthogonal to the plane).
A normal vector is orthogonal to the plane.
Answers will vary.
Answers will vary.
Answers will vary.
Answers will vary.
Standard form:
general form:
Standard form:
general form:
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Standard form:
general form:
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Standard form:
general form:
Answers may vary;
Standard form:
general form:
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Standard form:
general form:
Answers may vary;
Standard form:
general form:
Answers may vary;
Standard form:
general form:
Answers may vary;
Standard form:
general form:
Answers may vary;
Standard form:
general form:
Answers may vary;
Standard form:
general form:
Answers may vary;
Standard form:
general form:
Answers may vary;
Standard form:
general form:
Standard form:
general form:
Answers may vary:
Answers may vary:
No point of intersection; the plane and line are parallel.
The plane contains the line, so every point on the line is a “point of intersection.”
If is any point in the plane, and is also in the plane, then lies parallel to the plane and is orthogonal to , the normal vector. Thus , giving the distance as 0.
The intersecting lines define a plane with normal vector . Since points and lie in the plane, is orthogonal to , hence , giving a distance of 0. Knowing the principles of planes, especially their normal vectors, makes this simpler.
