1. ## system of equalities

I need help with these questions.....they're from grade 12 geometry and discrete math.

1. Analyse the normal vectors of the following planes and describe the situation of each system. Include a geometric interpretation of each solution:

7x-5y+4z-9=0

-5x+4y+z-3=0

x+2y-z+1=0

2. Analyse the normal vectors of the following planes and describe the situation of each system. Include a geometric interpretation of each solution:

7x-5y+4z-9=0

-x+3y+5z-2=0

5x+y+14z-20=0

3. Determine the parametric equations for a line that is perpendicular to the
yz-plane and passes through X(-2,4,3).

4. If a line and its normal vector with tail at (0,0) intersect at N(-2,5), determine the equation of the line.

5. A line has a scalar equation 5x+2y-10=0. Express the equation of the line in symmetric form.

6. Explain how knowing the normal vectors of two non-parallel planes can be
used to determine the scalar equation of another plane that is perpendicular to the two given planes.

7. Determine whether the two lines intersect and, if so, determine the
point(s) of intersection.

2. Originally Posted by Raiden_11
...

3. Determine the parametric equations for a line that is perpendicular to the
yz-plane and passes through X(-2,4,3).

...
Hello,

the normal vector of the yz-plane is $\overrightarrow{n_{yz}} = \left( \begin{array}{c}1 \\ 0 \\ 0 \end{array} \right)$

Therefore the (vector) equation of the line is:

$L:\vec{x} = \left( \begin{array}{c}-2 \\ 4 \\ 3 \end{array} \right) + k \cdot \left( \begin{array}{c}1 \\ 0 \\ 0 \end{array} \right)$

which will give the system of equations:
$x = -2 + k, ~ y = 4,~ z = 3$

3. Use Pythagoras and you can see what the hypoteneuses of the respectives triangles will be. There is a pattern.

$\sqrt{1^{2}+1^{2}}=\sqrt{2}$

$\sqrt{1^{2}+(\sqrt{2})^{2}}=\sqrt{3}$

$\sqrt{1^{2}+(\sqrt{3})^{2}}=\sqrt{4}$

.
.
.
$\sqrt{1^{2}+(\sqrt{n})^{2}}=\sqrt{n+1}$

See the pattern?.

4. Originally Posted by Raiden_11
...

4. If a line and its normal vector with tail at (0,0) intersect at N(-2,5), determine the equation of the line.

...
Hello,

Let L be the name of the line. Then:

1. $N(-2, 5) \in L$

2. The normal vector of the line is: $\vec{n} = \left( \begin{array}{c}-2 \\ 5 \end{array} \right)$

If $\vec{x}$ is the position vector of every point of the line and $\vec{a}$ is the position vector to a fixed point A on the line then

$L: \vec{n}(\vec{x} - \vec{a}) = 0$

Use this equation:

$L: \left( \begin{array}{c}-2 \\ 5 \end{array} \right) \left(\vec{x} - \left( \begin{array}{c}-2 \\ 5 \end{array} \right) \right) = 0$

Expand and you'll get: -2x + 5y -29 = 0 <-- equation of the line

5. Originally Posted by Raiden_11
...

6. Explain how knowing the normal vectors of two non-parallel planes can be
used to determine the scalar equation of another plane that is perpendicular to the two given planes.

...
Hello,

if $\overrightarrow{n_1}$ and $\overrightarrow{n_2}$ are the normal vectors of the 2 given planes then

$\overrightarrow{n_3} = \overrightarrow{n_1} \times \overrightarrow{n_2}$

6. Originally Posted by Raiden_11
...

5. A line has a scalar equation 5x+2y-10=0. Express the equation of the line in symmetric form.

...
Hello,

I don't know a "symmetric form" of an equation. If you are looking for an equation of the line using vectors you calculate first the coordinates of 2 points which represent the position vectors of thesepoints:

$5x + 2y - 10 =0 \Longleftrightarrow y = -\frac{5}{2} x + 5$

A(2, 0)
B(0, 5)

The direction vector of the line is: $\vec{u} = \left( \begin{array}{c} 2 \\ 0\end{array} \right) - \left( \begin{array}{c} 0 \\ 5\end{array} \right) = \left( \begin{array}{c} 2 \\ -5\end{array} \right)$

Now you have the fixed point A and the direction of the line. Therefore the equation is:

$L: \vec{x} = \left( \begin{array}{c} 2 \\ 0\end{array} \right) + r \cdot \left( \begin{array}{c} 2 \\ -5\end{array} \right)$

7. ## Can you help me with the others?

What about number 1 and 2