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Anonymous, I approached this problem exactly as you described. I also found the point of intersection to be (1,2,3).

I then rewrote the linear equations as follows:$$\vec{p}(t) = \langle3,1,4\rangle + \langle2,-1,1\rangle(t)$$$$\vec{q}(t) = \langle-2,0,2\rangle + \langle3,2,1\rangle(t)$$

From here I took the cross products of the directional vectors since I know they are both on the plane in order to get the normal vector. $$\langle2,-1,1\rangle \times \langle3,2,1\rangle = \langle-3,1,7\rangle$$

plugging what I have into the equation for a plane I get $$\langle-3,1,7\rangle\cdot\langle x-1, y-2, z-3\rangle = 0$$ $$-3(x-1) + (y-2) + 7(z-3) = 0$$ Or $$-3x + y + 7z = 20$$

As usual, I would love comments, critiques, etc. Thanks!

Anonymous, I approached this problem exactly as you described. I also found the point of intersection to be (1,2,3).

I then rewrote the linear equations as follows:$$\vec{p}(t) = \langle3,1,4\rangle + \langle2,-1,1\rangle(t)$$$$\vec{q}(t) = \langle-2,0,2\rangle + \langle3,2,1\rangle(t)$$

From here I took the cross products product of the directional vectors since I know they are both on the plane in order to get the normal vector. $$\langle2,-1,1\rangle \times \langle3,2,1\rangle = \langle-3,1,7\rangle$$

plugging what I have into the equation for a plane I get $$\langle-3,1,7\rangle\cdot\langle x-1, y-2, z-3\rangle = 0$$ $$-3(x-1) + (y-2) + 7(z-3) = 0$$ Or $$-3x + y + 7z = 20$$

As usual, I would love comments, critiques, etc. Thanks!

Anonymous, I approached this problem exactly as you described. I also found the point of intersection to be (1,2,3).

I then rewrote the linear equations as follows:$$\vec{p}(t) = \langle3,1,4\rangle + \langle2,-1,1\rangle(t)$$$$\vec{q}(t) = \langle-2,0,2\rangle + \langle3,2,1\rangle(t)$$

From here I took the cross product of the directional vectors since I know they are both on the plane in order to get the normal vector. $$\langle2,-1,1\rangle \times \langle3,2,1\rangle = \langle-3,1,7\rangle$$

plugging what I have into the equation for a plane I get $$\langle-3,1,7\rangle\cdot\langle x-1, y-2, z-3\rangle = 0$$ $$-3(x-1) + (y-2) + 7(z-3) = 0$$ Or $$-3x + y + 7z = 20$$

As usual, I would love comments, critiques, etc. Thanks!

Anonymous, I approached this problem exactly as you described. I also found the point of intersection to be (1,2,3).

I then rewrote the linear equations as follows:$$\vec{p}(t) = \langle3,1,4\rangle + \langle2,-1,1\rangle(t)$$$$\vec{q}(t) = \langle-2,0,2\rangle + \langle3,2,1\rangle(t)$$

From here I took the cross product of the directional vectors since I know they are both on the plane in order to get the normal vector. $$\langle2,-1,1\rangle \times \langle3,2,1\rangle = \langle-3,1,7\rangle$$

plugging what I have into the equation for a plane I get $$\langle-3,1,7\rangle\cdot\langle x-1, y-2, z-3\rangle = 0$$ $$-3(x-1) + (y-2) + 7(z-3) = 0$$ Or $$-3x + y + 7z = 20$$

As usual, I would love comments, critiques, etc. Thanks!

ADDED for Tiffany:

The first thing to do to find point of intersection is to change the parameter of one equation, lets say change $\vec{q}(t)$ to $\vec{q}(s)$, making it $$\vec{q}(s) = \langle-2+3s, 2s, 2+s\rangle$$

Now you set the components of $\vec{p}$ and $\vec{q}$ equal to each other.

3 $$3 + 2t = -2 + 3s

1 3s$$

$$1 - t = 2s

4 2s$$

$$4 + t = 2 + ss$$

If you solve for one variable then back substitute, you should end up with t = -1 and s = 1. Using these values to plug back into the vector equations gives you $$\vec{p}(-1) = \langle1,2,3\rangle$$$$\vec{q}(1) = \langle1,2,3\rangle$$

Thus the intersection...Does this sound right?