Expert: Scott A Wilson - 6/21/2009

Question
So out of 5 assignments these were the only ones i got stuck on. Can you show me the steps??

1. Find the length of v.
Initial point of v:(0,-1,0)
Terminal point of v;(1,2,-2)

2. Find the angle theta between the two vectors
u=10i+40j
v=-3j+8k

3. Find u x v and show that it is orthogonal to both u and v.  u=6k  v=-i+3j+k

4. Find the area of the parallelogram that has vectors as adjacent sides. u=i+3j+k   v=i+k

Answer
1. Find the length of v given that the initial point of
v is (0,-1,0) and the final point of v is (1,2,-2)

Let the start of the v vector be (x1,y1,z1) and
the end of the vector v be (x2,y2,z2).

The length in three dimensional rectangular coordinates is
√((x2-x1)² + (y2-y1)² + (z2-z1)²).

The values are (1-0)², (2+1)², and (-2-0)²,
which are, respectively, 1, 9, and 4.

That gives us the length is √(1+9+4) = √14



2. Find the angle theta between the two vectors u=10i+40j and
v=-3j+8k.  This means that u is (10, 40, 0) and v is (0, -3, 8).

Call the angle between the vectors Θ.

The formula is then cosΘ = u·v / ||u|| x ||v||
where u·v = ux*vx + uy*vy + uz*vz.

It can be seen that u·v = 10*0 + 40(-3) + 0*8 = -120.

Given v = (vx, vy, vz), it is known that ||v|| = √(vx² + vy² + vz²).

That means ||u|| = √(100 + 1600 + 0) = √1700 = 10√17 and
||v|| = √(0 + (-3)² + 8²) = √(9+64) = √73.

Putting them all into the formula gives
cosΘ = -120 / ((10√17)(√73)) = -12/√1241.

Now the denominator should be a number without a square root in it, so what we have is cosΘ = -12√1241 / 1241.

So that answer would be Θ = arccos(-12√1241 / 1241).

Note that the angle between two vectors is always between 0° and 180°.  To take the arccos of a negative number puts the angle between 90° and 180°.  The best way to take the arccos(-x) is to compute 180° - arccos(x).



3. Find u x v and show that it is orthogonal to both u and v
where u=6k, v=-i+3j+k

Let u = (ux, uy, uz) and v = (vx, vy, vz).  

I can now say that u x v = (x, y, z) where
x = uy*vz – uz*vy, y = uz*vx – ux*vz, and z = ux*vy – uy*vx.

It was given that u = (0, 0, 6) and v = (-1, 3, 1).

That makes the cross product equal to
(0*1 – 6*3, 6(-1)-0*1, 0*3 - –1*0) = (-18, -6, 0) = b.

The dot product of this vector with u and v should be 0 to be orhtogonal to each of them.

It can be seen that u·b = 0(-18) + 0(-6) + 6(0) = 0 and
v·b = -18(-1) – 6(3) + 1*0 = 18 – 18 = 0.



4. Find the area of the parallelogram that has vectors as adjacent sides. u=i+3j+k and  v=i+k.

Given the sides have lengths x and y, with x being that base and y the upward angled line, it is known that the height of a parallelogram h = y sinΘ where Θ is the angle between the x and y.

Now how do we find Θ?

It is know that cosΘ = u·v/(||u|| ||v||).

Using this, it is known that ||u|| = √(1² + 3² + 1²) = √11 and
||v|| = √(1² + 1² + 0) = √2.

It can also be seen that u·v = 1*1 + 3*0 + 1*1 = 2.

This tells us that cosΘ = 2/(√11√2).
It can be seen that 2/√2 = √2, so we have cosΘ = √2/√11.

Since sin²Θ + cos²Θ = 1, sin²Θ + 2/11 = 1,
so sin²Θ = 1 – 2/11 = 9/11.
This means that sinΘ = √9/√11 = 3/√11.

||u|| = √11 and ||y|| = √2,
so the answer is ||u|| ||v|| sinΘ = (√11)(√2)(3√11).

Since there are 2 √11’s, combine them and multiply by 3,
giving an answer of 33√2.