Vector Basics and Arithmetic
If v = <1,3> and w = <2, -2>
What is v + w and v - w?
v + w = <3,1> and v - w = <-1, 5>
Dissect this line into the x, y, and z components:
r(t) = (4, -1, 9) + t<3, 0, -5>
x = 3t+4
y = -1
z = -5t+9
Carry out the dot product between
<1,4,6> and <-3, -8, 5>
-3 - 32 + 30 = -5
A plane is formed between the points (6, 0, 1), (5, -9, 2), and (2, 7, 6)
What is the normal vector of this plane?
Create any two vectors with the points given.
(6, 0, 1) - (5, -9, 2) = <1, 9, 3>
(6, 0, 1) - (2, 7, 6) = <4, -7, -5>
Take the cross product between them to get the normal vector:
n = <1, 9, 3> x <4, -7, -5> = <-24, 17, -43>
Split the vector valued function into their respective x(t), y(t), and z(t)
r(t) = <3t2-2, cos(t+1), ln(6t)>
x = 3t2-2
y = cos(t+1)
z = ln(6t)
What is a vector of length 3 in the same direction as <4, 3>?
Take the magnitude first:
L = sqrt(42+32) = 5
Then divide the vector by the magnitude:
u = <4/5, 3/5>This is the direction of the vector with length 3, so to create the vector of length 3, just multiply the unit vector by 3.
v = 3u = <12/5, 9/5> <====== Answer
At what point does this line intersect the xy-plane?
r(t) = (3,4,5) + t<9, -1, -4>
We are in the xy-plane, so z = 0.
Therefore: z = -4t+5 = 0 ==> t = 5/4
The line intersects the xy-plane at point
(57/4, 11/4, 0)
Carry out the cross product between
<1, 5, 8>
<6, -2, 0>
<16, 48, -32>
Or if you want to scale it down (its still orthogonal to the other two vectors) you can write <1, 3, -2> as well.
What is the relationship between these two planes?
3x - 6y + 4z = 7
4x + 4y + 3z = 4
Orthogonal
The dot products of their normal vectors are 0
3(4) -6(4) + 4(3) = 0
What is the PROJECTION of the vector valued function:
r(t) = <2t3, 8/t, e2t>
on the xz-plane?
Bonus question, does this VVF intersect that plane?
xz-plane? That means y = 0
Therefore:
r(t) = <2t3, 0, e2t>
And no, it does not intersect the xz-plane.
The vector made of points (1,3,4) and (3, -5, 6) and the vector made of points (6, 2, 9) and point P are parallel.
What is point P?
v = (3, -5, 6) - (1, 3, 4) = <2, -8, 2>
Therefore:
w = <2, -8, 2> = (p1, p2, p3) - (6, 2, 9)
p1 - 6 = 2 ==> p1 = 8
p2 - 2 = -8 ==> p2 = -6
p3 - 9 = 2 ==> p3 = 11
Therefore, P = (8, -6, 11)
Determine whether the lines
(0,1,1) + t<1,1,2>
and
(2,0,3) + s<1,4,4>
intersect. If so, what point?
Dissect the lines first
x = t = s+2
y = t+1 = 4s
z = 2t+1 = 4s+3
Then choose a variable to replace. I chose t because t = s+2
And put this replacement in the other two equations:
(s+1) + 1 = 4s ==> s = 2/3
2(s+1) +1 = 4s+3 ==> s = 0
Since these s values don't match, the lines don't intersect.
The dot product between two unit vectors is 1/2.
What is the angle between them?
Remember the equation
v * w = ||v|| ||w|| cos(theta)
(Assume * means dot product)
Since we are dealing with two unit vectors, the magnitude of both of them is 1.
So the equation looks like:
1/2 = cos(theta)
Therefore, theta = pi/3
The following plane and line intersect at one point.
9x - 7z = 16
r(t) = (4, 0, 2) + t(1, 2, 3)
Find that point
First dissect the line:
x = t+4
y = 2t
z = 3t+2
There is no y in the plane equation, so you only have to plug in x and z.
9(t+4) - 7(3t+2) = 16
Therefore: t = 1/2
The point of intersection is: (9/2, 1, 7/2)
What is the parameterization of a circle with radius 2 centered at (2, 1, 5)
and it's parallel to the xy-plane?
If this circle is parallel to the xy-plane, then that means that the z-value is constant.
To parameterize a circle, you need to use POLAR CONVERSIONS.
But since this circle is centered at a point other than the origin:
x = 2cos(theta) + 2
y = 2sin(theta) + 1
Therefore: r(t) = <2cos(theta) + 2, 2 sin(theta) + 1, 5>
Write me a unit vector e that makes an angle of 4pi/7 with the positive x-axis?
Remember the signs: magnitude, angle, and positive x-axis.
What does that tell you? Polar conversion!
x = rcos(theta) and y = rsin(theta)
Your r is 1 because you're dealing with a unit vector, and your theta is 4pi/7.
Therefore: e = <cos(4pi/7), sin(4pi/7)>
There are two lines in 3-space:
(1,2,4) + t<2,1,-1>
(6,3,-1) + s<-5, 2, c>
Find the value of c that makes these lines intersect.
Dissect the lines
x = 2t+1 = -5s+6
y = t+1 = 2s+3 ==> t = 2s+2
z = -t+4 = cs - 1
Choose a variable to replace (I chose t) in the other equation.
2(2s+2) + 1 = -5s + 6 ==> s = 1/9
Now replace this s value in the last equation to get c.
-(2s+2) + 4 = cs-1
-2(1/9) + 2 = (1/9)c - 1
Therefore: c = 25
The cross product of vectors v and w is <1, 2, 7>.
Compute this:
(2v - w) x (v + w)
*x* means cross product.
We use the distributive property of cross products for this problem. I suggest distributing the first set of parenthesis.
==> 2v x (v + w) - w x (v + w)
Now we use the distributive property again!
==> 2v x v + 2v x w - w x v - w x w
Since the cross product of a vector by itself is 0, we can remove any terms where a vector is in a cross product with itself
We are left with:
==> 2v x w - w x v
Another cross product property is that the cross product is not commutative. So v x w = -w x v
Therefore:
==> 2v x w + v x w = 3v x w = 3(v x w)
= <3, 6, 21>
The following line and point reside on a plane:
r(t) = (2, 0, -4) + t<-1, -1, 3>
(0, -9, 2)
Write the equation of that plane.
You need two vectors to get the normal vector of a plane. You already have one, which is <-1, -1, 3>
You can use the starting point of the line and the other point to get the second vector.
(2, 0, -4) - (0, -9, 2) = <2, 9, -6>
Now to get the normal vector, take the cross product.
n = <-1, -1, 3> x <2, 9, -6> = <-21, 0, -7>
If you want to scale it down, you can also do <3, 0, 1>
Now, create the plane equation:
3(x-x0) + 1(z - z0) = 0
You can plug in any point you want, I plugged in (2, 0, -4) and got:
3x + z = 2
What is the VVF of the curve traced at the intersection between the plane y = 1/2 and the sphere
x2 + y2 + z2 = 1
Plug y = 1/2 into the sphere equation.
x2 + 1/4 + z2 = 1
Therefore: x2 + z2 = 3/4
Now, solve for one variable with respect to another.
Therefore: r(t) = <t, 1/2, +/- sqrt(3/4 - t2)>
Or if you want polar conversions:
r(t) = <sqrt(3)/2 cos(theta), 1/2, sqrt(3)/2 sin(theta)>
Give me a vector with magnitude 3 forming an angle of pi/3 from the negative y-axis.
This angle is pi/3 radians from the negative y-axis. Therefore, you are in the fourth quadrant.
If you remember the phrase: All Silver Tea Cups (1- All, 2- Sine, 3-Tangent, 4-Cosine), then you would know that in the fourth quadrant, only cosines are positive while the rest are negative.
Therefore: v = <-3cos(pi/3), 3sin(pi/3)>
The line r(t) = (3,1,-4) + t<-2,-2,3> intersects the sphere
(x-1)2 + (y+3)2 + z2 = 8
at two points. Determine those two points.
Dissect the line:
x = -2t+3
y = -2t+1
z = 3t-4
Plug them into the sphere equation:
(-2t+3-1)2 + (-2t+1 +3)2 + (3t-4)2 = 8
(2-2t)2 + (4-2t)2 + (3t-4)2 = 8
You are going to have to expand the terms out, but your final quadratic should be:
17t2 - 48t + 36 = 8
==> 17t2 - 48t + 28 = 0
Therefore, t = 2 and 14/17.
Now we plug in. The points are:
(-1, -3, 2) and (23/17, -11/17, 26/17)
Let's see if you remember this:
Three vertices of a parallelopiped is made of the following points:
(1, -2, 5), (4, 0, -2), (8, -1, -1), and (-7, 0, 0)
Find the volume of the parallelopiped.
I chose (-7, 0, 0) so:
u = (-7, 0, 0) - (1, -2, 5) = <-8, 2, -5>
v = (-7, 0, 0) - (4, 0, -2) = <-11, 0, 2>
w = (-7, 0, 0) - (8, -1, -1) = <-15, 1, 1>
The volume of a parallelopiped is calcuated by a triple scalar product, so u * (v x w)
First take the cross product of v and w, which is:
<-3, -19, -11>
Then take the dot product with u
-8(-3) + 2(-19) - 5(-11) = 41
The volume of the parallelopiped is 41.
What line does the plane
2x - 7y + z = 2
make with the yz-plane?
The yz-plane means that x = 0. So the equation of the yz-plane is x = 0.
If x = 0, its normal vector should be n = <1, 0, 0>
Therefore, to get the direction vector of the line of intersection, perform the cross product between the normal vectors of both planes (order doesn't matter).
d = <2, -7, 1> x <1, 0, 0> = <0, 1, 7>
Now to get the point.......
As long as your starting point agrees with the equation z = 2+7y, your point is valid.
What is the VVF of the curve traced by the intersection of these two surfaces?
z = x2 - y2
z = x2 + xy - 1
Since there are two different functions equal to z, set those two function equal to each other:
x2 - y2 = x2 + xy - 1
==> 0 = y2 + xy - 1
Solve for one variable with respect to the other
1 = y(y + x)
1/y = y + x
x = y + 1/y
Plug back into z:
z = y2 + 2 + 1/y2 - y2 ==> z = 2 + 1/y2
Now, replace all the y's with t's.
Therefore: r(t) = <t + 1/t, t, 2 + 1/t2>