The distance between two points $P (x_1,y_1,z_1)$ and $Q (x_2,y_2,z_2)$ may be written as $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$. Recognizing that this distance is also the length of the difference vector $\vec{PQ}$, write an expression for the distance between $P$ and $Q$ that does not express the components of the difference vector. How would you modify your expression if $P$ and $Q$ were each points in 6-dimensional space?
Let $\vec u = (-2,0,0)$, $\vec v=(3,4,0)$, and $\vec w=(2,4,-1)$.
- Compute $\text{proj}_{\vec u}\vec v$. Draw $\vec u$, $\vec v$, and $\text{proj}_{\vec v}\vec u$ in the $xy$ plane.
- Compute $\text{proj}_{\vec v}\vec u$. Draw $\vec u$, $\vec v$, and $\text{proj}_{\vec u}\vec v$ in the $xy$ plane.
- Compute the work done by $\vec w$ to move an object through the displacement $\vec v$.
- Compute the cross product $\vec v\times \vec w$.
- Compute the cross product $\vec w\times \vec v$.
- Compute the area of the parallelogram whose edges are parallel to $\vec v$ and $\vec w$.
- Place each of the vectors in 3D with the base at the origin, and then find the the area of the triangle whose vertices are the heads of $\vec u$, $\vec v$, and $\vec w$.
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