Calculus 3 help? and please tell me how you know?

Question

1)Describe in words the region of R^3 represented by the equation or inequality:
x=10
x>3
x^2+z^2 (less than or equal to) 9

2)Find a unit vector that has the same direction as the given vector
8i - j +4k

3)A woman walks due west on the deck of a ship at 3 mph. the ship is moving north at 22mph. find the speed and direction of the woman relative to the surface of the water.

4) Find a vector orthogonal to the plane through the points P, Q, & R, and find the area of the triangle PQR

P(1,0,0), Q(0,2,0), R( 0,0,3)

Answer 1

At first glance, I thought you were refering to spherical coords. Now, I realize that by R^3, you mean the standard 3 dimensional system.

So, if you fix x at 10, you get a plane that is perpendicular to the x axis and passes through the point (10,0,0).

If you restrict x>3, you get you get a half-space, like a half-plane or a ray, only in 3-space. It can be thought of as all points to the right of x=3.

x^2 + z^2 <= 9 is a solid cylinder centered on the y-axis.

2) To get the unit vector, divide the given vector by sqrt(i^2+j^2+k^2) or sqrt(81). This would be 8/sqrt(81) * i - 1/sqrt(81) * j + 4/sqrt(81) * k

3) Just draw the vectors with a vector of length 3 pointing west(left) and another from the head of this one of length 22 pointing north(up). The length is easy as it is sqrt(3^2 + 22^2) =sqrt(493). You get the direction using trig. The reference angle can be found from InvTan(22/3) But, this angle isn't the real direction. What ever number you get from InvTan(22/3), that is the angle above due west that your heading will be.

4) To get the orthoganal vector, take the cross product of any 2 vectors forming the triangle. This will always give you a vector that is perp to the plane containing the other 2. The side from P to Q would be the same as Q-P or (-1, 2, 0). The side from P to R would be the same as R-P or (-1, 0, 3). Now, take the cross product of these two to get the vector perp to the triangle = (6, 3, 2)
And, the answer to your last question is: A = magnitude of [1/2(PxQ + QxR + RxP)] = mag of 1/2[(0,0,2) + (6,0,0) + (0, -3, 0)] = mag(6,-3,2)/2 = 1/2 sqrt(36+9+4) = 7/2.

Answer 2

1 R^3 and y-axis are not mentioned. Strange. You have a circle on xy plane touching a line (x>3 and x=10).
2. If V=8i - j +4k then unit vector =V/|V|
3. You mean find the velocity. Speed=|velocity| ( speed is magnitude of the velocity).
You have V=3i + 22j

4. Generate two vector using P(1,0,0), Q(0,2,0), R( 0,0,3) say A and B. Find their cross or vector product N=AxB. Guess what N is the vector perpendicular to the plane!

Area of a triangle
Compute an edge (say in xy-plane) then compute a perpendicular line from the z-axis onto this edge. Area= .5 (edge)( perpendicular line from the z-axis).
Hint use dot product AdotB=|A||B|Cos(angle between them) so A dot B= 0 since cos(90)=0

Answer 3

Only part 1 is unanswered and the question is not clear. If all three conditions are to be satisfied simultaneously, there is no such region. x^2 + z^2 <=9 implies x <= 3.

Separately, x = 10 is a plane parallel to the y-z plane passing through (10,0,0)

x> 3 is the set of all points in R^3 strictly to the left of the plane x = 3.


x^2+z^2 < = 9

is a solid cylinder of radius 3, infinite in both directions, parallel to the y-axis and whose axis is the y-axis.

Answer 4

1.) not sure

2.) sqrt( 8^2 + (-1)^2 + 4^2) = sqrt(81) = 9
and the unit vector is 8/9i- 1/9j+4/9k

3.) her speed is sqrt(3^2 + 22^2) whichc is approx 22.204 and diraction is -3/22.204 i + 22/22.204 j

4.) the ortho vector is 1 i + 2 j + 3 k
not sure about the area if I get it i'll up date this

Answer 5

There's a fine line between asking to learn and asking others to do your homework for you.

A big, fat, fine line.

Answer 6

no way!