1. A line has direction angles 60, 45, 60 and passes through the point (-2,1,3). Determine the symmetric equations of the line.

For this one I know that the angles given are the direction angles, but Im not sure what to do with them, any help would be much appreciated.

2. Find the equation of a plane, every point of which is equidistant from the points A(1,1,0) and B(5,3,-2).

I've thought about this one, and I just can't wrap my head around it, would I just use a variable or something to represent the equidistance? A bit of guidance would be awesome.

2. Originally Posted by sugar_babee
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2. Find the equation of a plane, every point of which is equidistant from the points A(1,1,0) and B(5,3,-2).

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Hello,

to #2:

1. The plane passes through the midpoint M of AB:
$M\left(\frac{1+5}{2} , \frac{1+3}{2} , \frac{0+(-2)}{2} \right)$ that means: M(3, 2, -1).

2. The vector $\overrightarrow{AB}$ is the normal vector of the plane:
$\overrightarrow{AB} = \vec{B} - \vec{A} = [5, 3, -2] - [1, 1, 0] = [4, 2, -2]$

3. Plug in these values into the equation of a plane in normal form:

$p: [4, 2, -2] \cdot ([i, j, k] - [3, 2, -1]) = 0$

$p: 4x + 2y - 2z - 18 = 0$

3. Originally Posted by sugar_babee
1. A line has direction angles 60, 45, 60 and passes through the point (-2,1,3). Determine the symmetric equations of the line.

For this one I know that the angles given are the direction angles, but Im not sure what to do with them, any help would be much appreciated.

...
Hello,

I'm not certain if I understood this problem correctly:

1. I assume that the angles are between the components of the direction vector of the straight line and the coordinate axes:

Then the components are:
$i = \cos(\alpha) \Longrightarrow \cos(60^\circ) = \frac{1}{2}$
$j = \cos(\beta) \Longrightarrow \cos(45^\circ) = \frac{1}{2} \cdot \sqrt{2}$
$k = \cos(\gamma) \Longrightarrow \cos(60^\circ) = \frac{1}{2}$

2. You know the coordinates of one point and the direction of the straight line, therefore you can detrmine the equation of the line:

$[i \cdot x, j \cdot y, k \cdot z] = [-2, 1, 3] + r \cdot \left[\frac{1}{2} , \frac{1}{2} \cdot \sqrt{2}, \frac{1}{2} \right], \ \ r \in \mathbb{R}$

3. I don't know what a symmetric equation could be

4. I'm sure that the writing is not the way as you've learned it. In Germany the equation of the line would be written like this:

$l: \vec{x} = \left(\begin{array}{c}-2 \\ 1\\3\end{array}\right) + r \cdot \left(\begin{array}{c}\frac{1}{2} \\ \frac{1}{2} \cdot \sqrt{2}\\ \frac{1}{2}\end{array}\right)$