# Thread: Finding equation of plane?

1. ## Finding equation of plane?

Hi

How would i do the following question:

Find the equation of the plane containing the point A(1,3,2) and the line which has Cartesian equation:

$\frac{x+1}{7} = \frac{y-2}{4} = \frac{z}{-1}$

P.S

2. Find two points on the line. Take the vectors from those two points to A, and take the cross product. That'll be your normal vector. Use point A, or one of the other points you found as a point in the plane.

3. ok this is what i have done:

took the cross product of:

i+3j+2k
-i+2j

and

i+3j+2k
7i+4j-k

4. Hmm. I don't think the point 7i+4j-k is on the line. I'd agree that -i+2j+0k is on the line.

5. so i take the cross product of:

i+3j+2k
-i+2j

then i sub the value from point A

6. Originally Posted by Paymemoney
so i take the cross product of:

i+3j+2k
-i+2j

then i sub the value from point A
This really makes no sense. Substitute the "value form point A" into what? And what are you using that cross product for?

Ackbeet suggested initially that you "Find two points on the line". Have you done that? What two points did you get?

7. that's my problem what do mean by finding two points on the line? How can you find two lines when you only have one equation to solve from.

8. wait, do use the equation similar to n=AB * AC

9. You do have two equations:

$\dfrac{x+1}{7} = \dfrac{y-2}{4}$ and

$\dfrac{x+1}{7}= \dfrac{z}{-1}.$

To find two different points, simply pick an x-coordinate, and then solve for y and z.

I have not seen two different points on the line, only one: (-1,2,0). You need another point on the line, because you have to take the cross product of two linearly independent vectors in the plane. (Incidentally, it would probably impress your teacher if you showed that point A was not on the line.)

10. Originally Posted by Paymemoney
Find the equation of the plane containing the point A(1,3,2) and the line which has Cartesian equation:
$\frac{x+1}{7} = \frac{y-2}{4} = \frac{z}{-1}$
In this sort of problem it is always to look the parametric form:
$\left\{ \begin{gathered}
x = - 1 + 7t \hfill \\
y = 2 + 4t \hfill \\
z = 0 - t \hfill \\
\end{gathered} \right.$

Now pick any two values for $t$ to get two points.

11. Originally Posted by Paymemoney
that's my problem what do mean by finding two points on the line? How can you find two lines when you only have one equation to solve from.
You were given the line as
$\frac{x+1}{7} = \frac{y-2}{4} = \frac{z}{-1}$
That's NOT one equation, it is two : $\frac{x+1}{7} = \frac{y-2}{4}$ and $\frac{y-2}{4} = \frac{z}{-1}$
(strictly speaking, there is a third equaton, $\frac{x+1}{7} = \frac{z}{-1}$ but that's obviously not independent of the first two.)

Take x= 6 (so that $\frac{x+1}{7}= \frac{7}{7}= 1$ is easy) and then you have $\frac{y- 2}{4}= 1$ so that y= 6 and $\frac{z}{-1}= 1$ so that z= -1. One point on the line is (6, 6, -1).
Now take x= -8 so that $\frac{x+ 1}{7}= \frac{-7}{7}= -1$. Then $\frac{y- 2}{4}= -1$ so that y= -2 and $\frac{z}{-1}= -1$ so that z= 1. Another point on the line is (-8, -2, 1).

Can you find the equation of the plane that contains the points (6, 6, -1), (-8, -2, 1), and (1, 3, 2)?

12. Originally Posted by Plato
In this sort of problem it is always to look the parametric form:
$\left\{ \begin{gathered}
x = - 1 + 7t \hfill \\
y = 2 + 4t \hfill \\
z = 0 - t \hfill \\
\end{gathered} \right.$

Now pick any two values for $t$ to get two points.
Do you see how Plato got those three parametric equations? Set t equal to each of the three fractions in the original form and solve for x, y, and z in terms of t. Also I disagree that it is easier to do that. See my solution for x, y, z above.

13. Originally Posted by HallsofIvy
You were given the line as
$\frac{x+1}{7} = \frac{y-2}{4} = \frac{z}{-1}$
That's NOT one equation, it is two : $\frac{x+1}{7} = \frac{y-2}{4}$ and $\frac{y-2}{4} = \frac{z}{-1}$
(strictly speaking, there is a third equaton, $\frac{x+1}{7} = \frac{z}{-1}$ but that's obviously not independent of the first two.)

Take x= 6 (so that $\frac{x+1}{7}= \frac{7}{7}= 1$ is easy) and then you have $\frac{y- 2}{4}= 1$ so that y= 6 and $\frac{z}{-1}= 1$ so that z= -1. One point on the line is (6, 6, -1).
Now take x= -8 so that $\frac{x+ 1}{7}= \frac{-7}{7}= -1$. Then $\frac{y- 2}{4}= -1$ so that y= -2 and $\frac{z}{-1}= -1$ so that z= 1. Another point on the line is (-8, -2, 1).
so did you just randomly choose any coordinate for x, y and z?

Can you find the equation of the plane that contains the points (6, 6, -1), (-8, -2, 1), and (1, 3, 2)?
yes i can find it now

14. Originally Posted by HallsofIvy
Do you see how Plato got those three parametric equations? Set t equal to each of the three fractions in the original form and solve for x, y, and z in terms of t. Also I disagree that it is easier to do that. See my solution for x, y, z above.
yes i understand this part.

15. so i could make it t=1 and t=2 to give me two points

$(6,6,-1)$, $(13,10,-2)$

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