I am really lost and confused!

Consider the lines r = (1,-1,1) + t(3,2,1) and r = (-2,-3,0) + u(1,2,3).
a) Find their point of intersection.
b) Find a vector equation for the line perpendicular to both of the given lines that passes through their point of intersection.

For a I got the equations
3t-1u+3 = 0
2t-2u+2 = 0
t-3u+1 = 0

Then I put 2 and 3 together and got u = -4.

In the back of the book the answers says (-2, -3, 0). So am I already wrong?

2) Show that the lines r = (4,7,-1) + t(4,8,-4) and r = (1,5,4) + u(-1,2,3) intersect at right angles and find the point of intersection.

Can someone please walk me through these?

When I solved equations 2 and 3 I got u=0 and t=-1, giving me r = (-2,-3,0) in both r = ... equations.

check your arithmetic

for the second, if they are perpendicular, then the dot product of their directions vectors should be zero

(4,8,-4)∙(-1,2,3) = -4 + 16 - 12 = 0
so they are perpendicular.
For the intersection point do exactly the same thing you did for the first one without making arithmetic errors.

BTW, your equations for the first one were correct.

Sure, I'm here to help! Let's start with the first problem:

1) To find the point of intersection between the lines, you need to solve the system of equations formed by equating the two given equations. You correctly obtained the following equations:

3t - u + 3 = 0
2t - 2u + 2 = 0
t - 3u + 1 = 0

Now, let's solve this system of equations. We can begin by solving the first two equations simultaneously:

3t - u + 3 = 0 (equation 1)
2t - 2u + 2 = 0 (equation 2)

To eliminate the variable 'u,' we can multiply equation 1 by 2 and equation 2 by -1, then add the resulting equations:

6t - 2u + 6 = 0 (equation 1 multiplied by 2)
-2t + 2u - 2 = 0 (equation 2 multiplied by -1)

Adding these equations eliminates the 'u' term:

4t + 4 = 0

Now, solve for 't':

4t = -4
t = -1

Next, substitute 't' back into one of the original equations (you can choose any one) to solve for 'u':

t - 3u + 1 = 0 (equation 3)

(-1) - 3u + 1 = 0

-3u = 0
u = 0

Now that you have the values of 't' and 'u,' substitute them back into one of the original equations to find the corresponding point of intersection. Let's use the second given equation:

r = (-2, -3, 0) + u(1, 2, 3)

Substituting u = 0:
r = (-2, -3, 0)

So, the point of intersection between the lines is (-2, -3, 0).

Regarding your second question:

2) To show that the two lines intersect at right angles, you need to demonstrate that the direction vectors of the lines are perpendicular to each other. If two vectors are perpendicular, their dot product is zero.

The direction vector of the first line is (4, 8, -4), and the direction vector of the second line is (-1, 2, 3). Let's take the dot product of these two vectors:

(4, 8, -4) · (-1, 2, 3) = 4(-1) + 8(2) + (-4)(3) = -4 + 16 - 12 = 0

Since the dot product is zero, the lines are perpendicular to each other.

Now, to find the point of intersection, you can follow the same steps as in the first problem. Solve the system of equations formed by equating the two given equations (r1 and r2) and find the values of 't' and 'u'. Once you have the values, substitute them back into one of the original equations to find the point of intersection.

I hope this explanation helps you understand the process of solving these problems. If you have any more questions, feel free to ask!