In my video (link to be updated) I have worked out examples of how to use projection vectors to find the foot of perpendicular (for both lines and planes). The benefits of this method

1. it is faster,
2. it uses a similar concept for both lines and planes with slight modifications.

The downsides, however, is that it involves a rather unwieldy projection vector formula $(\mathbf{a}\cdot\mathbf{\hat{b}}) \mathbf{\hat{b}}$ and that minor details make a huge difference to whether we get the correct answer (e.g. where are our vectors pointing, $\overrightarrow{AB}$ or $\overrightarrow{BA}$? Is our final projection vector $\overrightarrow{AF}$ or $\overrightarrow{FB}$?).
In this post I will offer an alternative method to find foot of perpendicular using earlier concepts that some students find easier to understand.

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Finding foot of perpendicular to a line

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In the video (link to be updated), we went through how to find the line of intersection between two planes using a GC. In this post, we will discuss an alternative method to accomplish this without the use of technology. Just like in our video, we will use the following example:

Find the equation of the line of intersection between $$ p_1: \mathrm{r} \cdot \begin{pmatrix} 1 \\ 0 \\ 5 \end{pmatrix} = 7, \quad p_2: \mathrm{r} \cdot \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} = -3.$$

The two important things we need to find an equation for a line are (1) a point on the line and (2) the direction vector parallel to the line. We will thus tackle the two separately:

Finding the direction vector

The key observation here is that the direction vector of the line of intersection is perpendicular to the normal vectors of both planes (try to visualize it!). Thus, to find the direction vector of the line of intersection, we can use the cross product. Hence

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