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I would also be posting quiz 2, but sadly it was lost by an anonymous individual named SpacemanSpiff.

1: Prove that scalar multiplication of vectors is distributive over the addition of 2-D vectors.

We can start by defining two arbitrary vectors, $\vec{u}=\langle u_1, u_2 \rangle$, and $\vec{v} = \langle v_1, v_2 \rangle$. Now let $a$ be a real number.

Then we just add our arbitrary vectors together and multiply the resulting vector by $a$, and break apart the vectors into their components, then put them back together:

$a(\vec{u} + \vec{v}) = a(\langle u_1, u_2 \rangle + \langle v_1, v_2 \rangle $

$= a\langle u_1+v_1, u_2 + v_2 \rangle $

$= \langle a(u_1+v_1), a(u_2+v_2) \rangle $

$= \langle au_1 + av_1, au_2+av_2 \rangle $

$= \langle au_1,au_2 \rangle + \langle av_1,av_2 \rangle $

$= a\langle u_1, u_2 \rangle + a \langle v_1, v_2 \rangle $

$= a\vec{u} + a\vec{v}$.

And our proof is done.

2: The intersection between the $xy$ plane and the plane $x+2y -3z = 6$ is a line. Find an $xy$ equation for that line.

The key here is to see that the intersection occurs at $z=0$, we just take the equation for the plane and set $z=0$:

$x+2y = 6$. This is our $xy$ equation.

3: Find a parametrization of the line containing the point $(1,2,3)$ and perpendicular to the plane $x+2y-3z = 6$.

The coefficients on $x$, $y$, and $z$ form a vector perpendicular to the plane, as we showed in class. So we just take this vector $\langle 1, 2, -3 \rangle $ as the direction vector for our line, and then use the point we want to be on the line as our $\vec{p_0}$:

$\vec{l}(t) = \langle 1, 2, 3 \rangle + \langle 1, 2, -3 \rangle t $

4: Find all values of $t$ such that the vector $\vec{p}(t) = \langle t^2, t \rangle $ is perpendicular to the vector $\vec{q}(t) = \langle t, -1 \rangle $.

Two vectors are perpendicular if and only if their dot product is equal to 0, so we can just set the dot product of these two vectors equal to 0:

$\langle t^2, t \rangle \cdot \langle t, -1 \rangle = t^3 - t$

$ = t(t^2-1) = 0$

This gives us solutions where $t = 0$, $t= 1$ and $t=-1$. So for those values of $t$, the vectors are perpendicular.

I would also be posting quiz 2, but sadly it was lost by an anonymous individual named SpacemanSpiff.

1: Prove that scalar multiplication of vectors is distributive over the addition of 2-D vectors.

We can start by defining two arbitrary vectors, $\vec{u}=\langle u_1, u_2 \rangle$, and $\vec{v} = \langle v_1, v_2 \rangle$. Now let $a$ be a real number.

Then we just add our arbitrary vectors together and multiply the resulting vector by $a$, and break apart the vectors into their components, then put them back together:

$a(\vec{u} + \vec{v}) = a(\langle u_1, u_2 \rangle + \langle v_1, v_2 \rangle $

$= a\langle u_1+v_1, u_2 + v_2 \rangle $

$= \langle a(u_1+v_1), a(u_2+v_2) \rangle $

$= \langle au_1 + av_1, au_2+av_2 \rangle $

$= \langle au_1,au_2 \rangle + \langle av_1,av_2 \rangle $

$= a\langle u_1, u_2 \rangle + a \langle v_1, v_2 \rangle $

$= a\vec{u} + a\vec{v}$.

And our proof is done.

2: The intersection between the $xy$ plane and the plane $x+2y -3z = 6$ is a line. Find an $xy$ equation for that line.

The key here is to see that the intersection occurs at $z=0$, we just take the equation for the plane and set $z=0$:

$x+2y = 6$. This is our $xy$ equation.

3: Find a parametrization of the line containing the point $(1,2,3)$ and perpendicular to the plane $x+2y-3z = 6$.

The coefficients on $x$, $y$, and $z$ form a vector perpendicular to the plane, as we showed in class. So we just take this vector $\langle 1, 2, -3 \rangle $ as the direction vector for our line, and then use the point we want to be on the line as our $\vec{p_0}$:

$\vec{l}(t) = \langle 1, 2, 3 \rangle + \langle 1, 2, -3 \rangle t $

4: Find all values of $t$ such that the vector $\vec{p}(t) = \langle t^2, t \rangle $ is perpendicular to the vector $\vec{q}(t) = \langle t, -1 \rangle $.

Two vectors are perpendicular if and only if their dot product is equal to 0, so we can just set the dot product of these two vectors equal to 0:

$\langle t^2, t \rangle \cdot \langle t, -1 \rangle = t^3 - t$

$ = t(t^2-1) = 0$

This gives us solutions where $t = 0$, $t= 1$ and $t=-1$. So for those values of $t$, the vectors are perpendicular.

I would also be posting quiz 2, but sadly it was lost by an anonymous individual named SpacemanSpiff.

1: Prove that scalar multiplication of vectors is distributive over the addition of 2-D vectors.

We can start by defining two arbitrary vectors, $\vec{u}=\langle u_1, u_2 \rangle$, and $\vec{v} = \langle v_1, v_2 \rangle$. Now let $a$ be a real number.

Then we just add our arbitrary vectors together and multiply the resulting vector by $a$, and break apart the vectors into their components, then put them back together:

$a(\vec{u} + \vec{v}) = a(\langle u_1, u_2 \rangle + \langle v_1, v_2 \rangle $

$= a\langle u_1+v_1, u_2 + v_2 \rangle $

$= \langle a(u_1+v_1), a(u_2+v_2) \rangle $

$= \langle au_1 + av_1, au_2+av_2 \rangle $

$= \langle au_1,au_2 \rangle + \langle av_1,av_2 \rangle $

$= a\langle u_1, u_2 \rangle + a \langle v_1, v_2 \rangle $

$= a\vec{u} + a\vec{v}$.

And our proof is done.

2: The intersection between the $xy$ plane and the plane $x+2y -3z = 6$ is a line. Find an $xy$ equation for that line.

The key here is to see that the intersection occurs at $z=0$, we just take the equation for the plane and set $z=0$:

$x+2y = 6$. This is our $xy$ equation.

3: Find a parametrization of the line containing the point $(1,2,3)$ and perpendicular to the plane $x+2y-3z = 6$.

The coefficients on $x$, $y$, and $z$ form a vector perpendicular to the plane, as we showed in class. So we just take this vector $\langle 1, 2, -3 \rangle $ as the direction vector for our line, and then use the point we want to be on the line as our $\vec{p_0}$:

$\vec{l}(t) = \langle 1, 2, 3 \rangle + \langle 1, 2, -3 \rangle t $

4: Find all values of $t$ such that the vector $\vec{p}(t) = \langle t^2, t \rangle $ is perpendicular to the vector $\vec{q}(t) = \langle t, -1 \rangle $.

Two vectors are perpendicular if and only if their dot product is equal to 0, so we can just set the dot product of these two vectors equal to 0:

$\langle t^2, t \rangle \cdot \langle t, -1 \rangle = t^3 - t$

$ = t(t^2-1) = 0$

This gives us solutions where $t = 0$, $t= 1$ and $t=-1$. So for those values of $t$, the vectors are perpendicular.