Cross Products
What is the cross product of vectors
u = <5, 2, 3>
v = <1, 2, 8>
u x v = <10, -37, 8>
u = <1, 2, 3>
v = <0, 9, 3>
What is the projection of the vector u onto vector v?
Projvu = 27/90 * <0, 9, 3> = 3/10 * <0, 9, 3>
= <0, 27/10, 9/10>
What is the vector parameterization of
r(t) = <1, 0, 2> + t<3, 9, 11>x = 3t + 1
y = 9t
z = 11t + 2
What vector is orthogonal to every single vector on a plane?
The plane's normal vector.
The triple scalar product gives you the volume of what?
A paralellopiped
What is the value of
u * (u x v)?
* indicates dot product
0.
The cross product will give a vector orthogonal to u and v, so if you dot u with it, you will get 0.
When I write:
Projvu
What vector is doing the projecting?
u
Find a vector parameterization that passes through P(3,2,3) and is parallel to the line
r(t) = <1,1,-1> + t<2,3,-3>
Same vector, different starting point, so:
rn(t) = <3,2,3> + t<2,3,-3>
x = 2t+3
y = 3t+2
z = -3t + 3
What is the normal vector to the plane:
3(x-2) + 5(y-7) + z-3 = 0
n = <3,5,1>
The Flash and Superman try to race to two different coffee shops in Central City. The Flash runs in the direction of f = <3, 5, 6> and Superman runs in the direction of s = <5, 4, 0>.
After Superman reaches his coffee shop, how far would Superman now have to run if his new path would create a right angle with the path The Flash took?
*Hint: use projections!*
spar =35/70 * <7, 2, 3> = <7/2, 1, 3/2>
s - spar = <5, 4, 0> - <7/2, 1, 3/2>
= <3/2, 3, -3/2> = snew
||snew|| = Square root of 54/4
Which one of these is a legal statement?
a) v x w = w x v
b) (u * v) x w
c) v x v = 0
d) None of these are legal
d) is right.
a) the cross product isn't commutative
b) you can't cross a scalar with a vector
c) A vector crossed with itself gives you 0 VECTOR not just 0. There is a difference
If the angle between two vectors is obtuse, does the projection from one vector onto the other exist? If so, how?
It does exist, the projection vector just goes in the opposite direction of the vector it is being projected onto.
Find a set of scalar parametric equations that pass through the point (1,2,-3) and is perpendicular to the xz plane.
nxz = <0,1,0>
x = 1
y = t+2
z = -3
A plane contains points (8, 1, 2), (2, 5, 1), and (4, 0, 0).
Find the normal vector to the plane.
v = (8,1,2) - (2,5,1) = <6, -4, 1>
w = (2,5,1) - (4,0,0) = <-2,5, 1>
n = v x w = <-9, -14, 38>
Give me an example of the symmetric form of a line that is perpendicular to the plane:
3x + 4y - 5z = 10
Your answer must have the components of the normal vector <3,4,-5> in the denominator of all three terms.
For example:
(x-2)/3 = (y-4)/4 = z/-5
Assuming that:
u x v = <1,1,0> u x w = <0,3,1> v x w = <2,-1,1>
Find:
w x (u + v)
w x (u + v) = w x u + w x v = -(u x w) - (v x w)
= <0,-3, -1> + <-2, 1, -1> = <-2, -2, -2>
Give me the components of a vector that has a 60 degree angle with the positive x-axis and has magnitude 5.
*This should be a 2-D vector*
v = <5cos(60*), 5sin(60*)>
* means degrees here
Check if these two lines intersect:
r(s) = <3,-7,5> + s<1,-2,4>
s(t) = <-7,-8,4> + t<3,1,-1>
If they do, give the coordinate point they intersect at.
t = 3
s = -1
So they intersect at (2,-5,1)
2x - 3y + z = 5
-x + 4y - 2z = -7
What is the direction vector of the line that forms when these two planes intersect?
Shortcut is to do the cross product between the two normal vectors.
d = n1 x n2 = <2, -3, 1> x <-1, 4, -2>
= <2, 3, 5>
If I used the right hand rule for determining the directions of my axes, if my z-axis was facing towards me and the yz plane was facing north, what direction is the x-axis in?
North
Thumb towards you is z-axis, and since yz plane is facing north, y-axis is facing west, therefore x axis is facing north.
But also good to note that the positive yz plane faces the same direction the x axis faces.
Using the same cross products as 400:
u x v = <1,1,0> u x w = <0,3,1> v x w = <2,-1,1>
Find:
(u - 2v) x (u + 2v)
<4,4,0>
(u - 2v) x (u + 2v) = (u - 2v) x u + (u - 2v) x 2v
= u x u - 2v x u + u x 2v + 2v x 2v
= 0 - 4v x u= 4(u x v) = <4,4,0>
When Leonard found out about Sheldon Cooper's wall safe, Sheldon yanked Leonard's arm in the direction of
<3, 8, 2>
with a force of 60N. What are the components of the force vector?
*both vectors are going in the same direction*
Answer: 60*<3,8,2> / sqrt(77)
Find unit vector of the direction vector
||<3,8,2>|| = sqrt(77)
u = <3,8,2> / sqrt(77)F
Funit = u
F = Funit*||F|| = 60*<3,8,2> / sqrt(77)
r(t) = <2,7,6> + t<1,2,3>
r(u) = <5,2,3> + u<6,0,-2>
Why?
The dot product of their direction vectors equals 0
dt * du = (1*6) +(2*0) +(3*-2) = 6 + 0 - 6 = 0
The plane x - y + 5x = 10 and the line r(t) = <1,0,1> + t<-2,1,1> intersects at what point?
(-3,2,3)
Parameterize r(t) into
x = -2t+1, y = t, and z = t + 1
Plug into the plane equation and solve for t:
(-2t + 1) - (t) + 5(t+1) = 10
-2t + 1 - t + 5t + 5 = 10
2t + 6 = 10
2t = 4
t = 2
Then plug t back into r(t) and then solve for x,y, and z:
x = -2(2) + 1 = -3
y = 2
z = 2+1 = 3
Therefore, they intersect at (-3,2,3)
What is the angle between these two planes?
2x - y + 3z = 7
x + 2y - z = 4
This one is fun >:)
Answer: 71 degrees or 1.24 rad
The angle between two planes is the same as the angle between their normal vectors.
You can try to use dot product method or cross product method. Dot product is easier :)
<2,-1,3> * <1,2,-1> = 2 - 2 - 3 = -3
Remember the formula:
v * w = ||v|| || w|| cos(theta)Time to compute the magnitudes of both normal vectors:
||v|| = sqrt(4 + 1 + 9) = sqrt(14)
||w|| = sqrt(1 + 4 + 1) = sqrt(6)
Therefore:
cos(theta) = 3/[sqrt(14)*sqrt(6)]
You will get 71 degrees or 1.24 rad as your answer.