Question

*calculus 3* FOR u^ = i^ - 2k^ & v^ = 2i^ - j^ - k^, compute II 2( u^ X v^) - 3v^ II.

ALSO, find the intersection of the plan x+y+z-6=0 and the line containing the points (1,0,1) and (3,-1,1).

NOTE: the symbol ^ is supposed to represent the vector hat(or line), I just couldn't find any other symbol to represent it haha
and the X symbol between u^ and v^ is supposed to represent cross product multiplication.

Answer 1

Oh btw I've already computed the magnitude of 2( u^ X v^) - 3v^, it's particularly the second part that is troubling me.

Answer 2

Well, I didn't study this formally, so I guess I may not be using formal terminology and many things can be simplified.
If you don't mind I prefer using \(\hat{\mathbf{x}}, \hat{\mathbf{y}}, \hat{\mathbf{z}}\) here for the axes unit vectors.

Ok, so first let's find the vector between those points
$$
\begin{array}{rrr}
A = (1,&0,&1) \\
B = (3,&-1,&1)
\end{array} \\
\vec{ \mathbf{v} } = B - A = (3-1)\hat{ \mathbf{x} } + (-1 - 0) \hat{\mathbf{y}} + (1-1) \hat{\mathbf{z}} = 2\hat{\mathbf{x}} - \hat{\mathbf{y}}$$Now let's normalize the vector into a unit vector:
$$
||\vec{\mathbf{v}}|| = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}\\
\hat{\mathbf{v}} = \frac{\vec{\mathbf{v}}}{ ||\vec{\mathbf{v}}|| } = \frac{2\hat{\mathbf{x}} - \hat{\mathbf{y}}}{\sqrt{5}} = \frac{2}{\sqrt{5}} \hat{ \mathbf{x} } - \frac{1}{\sqrt{5}} \hat{\mathbf{y}}
$$We can now take one of the points and use the unit vector \( \hat{ \mathbf{v}} \) to figure out how many 'steps' in that direction will hit the plane.
Notice that the vector's direction can be in the opposite direction and in that case we'll just get a negative number of 'steps'.
So we say that \(t\) steps along the unit vector correspond to:
$$
t \cdot \hat{ \mathbf{v} } = \frac{2t}{\sqrt{5}} \hat{ \mathbf{x} } - \frac{t}{\sqrt{5}} \hat{ \mathbf{y}}
$$So \(t\) steps in the vector's direction correspond to \(\frac{2t}{\sqrt{5}}\) steps in the x direction and \(\frac{-t}{\sqrt{5}}\) steps in the y direction.
If we step \(t\) steps in the direction of our vector from point \(A\), we get to the point \(P\) such that:
$$
A = (1,0,1) \\
P = (1 + \frac{2t}{\sqrt{5}}, - \frac{t}{\sqrt{5}}, 1)
$$The coordinates of every point on the plane will cause the plane equation to be true. We want to find the \(t\) for which the point \(P\) will be on the plane, so we solve the plane equation for \(P\):
$$
x+y+z - 6 = 0 \\
\left( 1 + \frac{2t}{\sqrt{5}} \right) + \left(- \frac{t}{\sqrt{5}} \right) + (1) - 6 = 0 \\
\frac{2t - t}{\sqrt{5}} = 4 \\
t = 4\sqrt{5} \\
P = (1 + \frac{2 (4\sqrt{5})}{\sqrt{5}}, - \frac{(4\sqrt{5})}{\sqrt{5}}, 1) = (1 + 8, - 4, 1) = (9,-4,1)
$$Notice we'll get the exact same result using point \(B\) instead.
Hope it helps \( \Large ☺\)

Answer 3

Don't worry about the terminology I totally understood what you were saying! Thanks, that was a very good run down of how to solve the problem :))
@pitamar do you think you could also double check to see if my magnitude (or norm) for the first part is right? I got II 2(u^ X v^) - 3v^ II = 74^1/2

Answer 4

Glad I could help
Sure lemme try

Answer 5

Hmm.. I'm getting something very different.. let me check again

Answer 6

Let me ask you something. you wrote in the question \( \hat{ \mathbf{u}} \) and \( \hat{ \mathbf{v}} \). but they are defined that way. So I guess you don't mean their normalized form? because that's what I know the hat is used for:
http://en.wikipedia.org/wiki/Unit_vector

Answer 7

What i'm asking is basically, is it:
$$ || 2( \vec{ \mathbf{u} } \times \vec{ \mathbf{v} }) - 3\vec{ \mathbf{v} } ||
$$Or
$$ || 2( \hat{ \mathbf{u} } \times \hat{ \mathbf{v} }) - 3\hat{ \mathbf{v} } ||$$

Answer 8

Yes they aren't normalized, it should be the first one

Answer 9

well in that case I get \(\sqrt{110}\), it is not what you got, right?

Answer 10

No, do you mind showing me how you got that?

Answer 11

I got radical 74

Answer 12

Well, at first I tried through wolframalpha.. but man that site can be so annoying.. it knows anything but english.
So I quit and did it by hand, then I used Sage (know that program? it's like opensource version of matlab) and got the same result.
So here is what I did:

Answer 13

omg, my cat..

Answer 14

First we have
$$
\vec{ \mathbf{u}} = \hat{ \mathbf{i} } -2\hat{ \mathbf{k}} \\
\vec{ \mathbf{v}} = 2\hat{ \mathbf{i}} - \hat{ \mathbf{j}} - \hat{ \mathbf{k}}$$
Right? I didn't make a mistake so far

Answer 15

yup that's correct

Answer 16

Ok, sorry about that.. it can be really hard when my cats walks on my keyboard..
Anyway, let's first calculate the cross product:
First we have
$$
\vec{ \mathbf{u}} \times \vec{ \mathbf{v}} =
\left| \begin{matrix}
u_2 & u_3 \\
v_2 & v_3
\end{matrix}\right| \hat{ \mathbf{i} }
-
\left| \begin{matrix}
u_1 & u_3 \\
v_1 & v_3
\end{matrix}\right| \hat{ \mathbf{j} }
+
\left| \begin{matrix}
u_1 & u_2 \\
v_1 & v_2
\end{matrix}\right| \hat{ \mathbf{k} }
$$Right?

Answer 17

In our case it is (I write them as rows instead of columns if you don't mind):
$$
\vec{ \mathbf{u}} = \begin{bmatrix} 1 & 0 & -2\end{bmatrix}\\
\vec{ \mathbf{v}} = \begin{bmatrix} 2 & -1 & -1\end{bmatrix}\\
\vec{ \mathbf{u}} \times \vec{ \mathbf{v}} =
\left| \begin{matrix}
0 & -2 \\
-1 & -1
\end{matrix}\right| \hat{ \mathbf{i} }
-
\left| \begin{matrix}
1 & -2 \\
2 & -1
\end{matrix}\right| \hat{ \mathbf{j} }
+
\left| \begin{matrix}
1 & 0 \\
2 & 1
\end{matrix}\right| \hat{ \mathbf{k} }
$$

Answer 18

Ah yeah, i see my mistake - i accidentally multiplied 0 and -2 haha
it should be 0 multiplied by -1 and -1 multiplied by 2

Answer 19

oh but there is an issue with the matrix after the k vector, it should be 1,0,2, and -1

Answer 20

I wrote above \(v_2 = 1\) instead of -1 accidently for \(\hat{ \mathbf{k}}\)
$$
\begin{align*}
\vec{ \mathbf{u}} \times \vec{ \mathbf{v}} = &(0 \cdot (-1) - (-2) \cdot (-1)) \hat{\mathbf{i}} \\
-&(1 \cdot (-1) - (-2) \cdot 2)\hat{\mathbf{j}} \\
+&(1 \cdot (-1) - 0 \cdot 2)\hat{\mathbf{k}} \\
\end{align*}
$$
$$
\begin{align*}
\vec{ \mathbf{u}} \times \vec{ \mathbf{v}} = &-2 \hat{\mathbf{i}} \\
&-3\hat{\mathbf{j}} \\
&-\hat{\mathbf{k}} \\
\end{align*}\\
2 (\vec{ \mathbf{u}} \times \vec{\mathbf{v}}) = -4\hat{\mathbf{i}} - 6 \hat{\mathbf{j}} -2\hat{\mathbf{k}} $$$$
\begin{align*}
\vec{\mathbf{t}} = 2 (\vec{ \mathbf{u}} \times \vec{\mathbf{v}}) -3\vec{\mathbf{v}} =& -4\hat{\mathbf{i}} - 6 \hat{\mathbf{j}} -2\hat{\mathbf{k}}\\
&- 3(2\hat{\mathbf{i}} - \hat{\mathbf{j}} -\hat{\mathbf{k}}) \\
=& -4\hat{\mathbf{i}} - 6 \hat{\mathbf{j}} -2\hat{\mathbf{k}} \\
&-6\hat{\mathbf{i}} +3 \hat{\mathbf{j}} +3\hat{\mathbf{k}} \\
=& -10\hat{\mathbf{i}} - 3 \hat{\mathbf{j}} + 1\hat{\mathbf{k}}
\end{align*}\\
|| \vec{\mathbf{t}}|| = \sqrt{(-10)^2 + (-3)^2 + 1^2} = \sqrt{100 + 9 + 1} = \sqrt{110}
$$

Answer 21

Corrected:
$$
\vec{ \mathbf{u}} = \begin{bmatrix} 1 & 0 & -2\end{bmatrix}\\
\vec{ \mathbf{v}} = \begin{bmatrix} 2 & -1 & -1\end{bmatrix}\\
\vec{ \mathbf{u}} \times \vec{ \mathbf{v}} =
\left| \begin{matrix}
0 & -2 \\
-1 & -1
\end{matrix}\right| \hat{ \mathbf{i} }
-
\left| \begin{matrix}
1 & -2 \\
2 & -1
\end{matrix}\right| \hat{ \mathbf{j} }
+
\left| \begin{matrix}
1 & 0 \\
2 & \color{red}{-1}
\end{matrix}\right| \hat{ \mathbf{k} }
$$

Answer 22

I got the same results with Sage, so I'm pretty sure I didn't do any mistake.
I really like those environments, they make it all significantly less painful...

Answer 23

At least not computational mistake

Answer 24

Yup, that's definitely it! Thanks!

Answer 25

You're welcome =)