A few questions - Planes

Way Below Xero · **Joined:** Mon, 26 May 2003 01:25:40 UTC **Posts:** 32

Just a few questions. Muchly thanx for the help in advance.

Show that the lines x-1=(y-2)/2=z+3 and x+1=y-3=2z+5 are coplanar and find the equation of the plane which contains them.

Also...

Q is any point in the plane Ax+By+Cz = D
d is the distance from P(x1, y1, z1) to the given plane.

Explain why d = (l QP * n l)/l n l

(where * means dot product and l are modulus signs, oh and n is the normal vector)

Thanx guys

andyistic · **Posted:** Sun, 1 Jun 2003 06:12:23 UTC

Way Below Xero wrote:

Show that the lines x-1=(y-2)/2=z+3 and x+1=y-3=2z+5 are coplanar and find the equation of the plane which contains them.

Line A: $x-1=\frac{y-2}{2}=z+3$
Line B: x+1=y-3=2z+5

Line A:

Line B:

In both line A and B:
y=2z+8

This is the plane which contains both lines.

Way Below Xero · **Joined:** Mon, 26 May 2003 01:25:40 UTC **Posts:** 32

Thanx andy

Could someone have a look at the second bit as well???

(The proof)

Soroban · **Posted:** Mon, 2 Jun 2003 02:46:43 UTC

Hello, Way Below Zero!

This is a "projection" problem - which has a standard proof.
I'll modify it for this particular problem.

Draw a horizontal line (the plane).
Draw a vector upward from the line (the normal n) through point P.
Let the foot of the perpendicular be R. Then |PR| = d.
On the horizontal line select point Q, and draw QP.

Let angle QPR = $\theta$ .

In the right triangle: $d = |QP|\cos\theta$

The angle between QP and n is given by: $cos\theta = \frac{|QP\cdot{n}|}{|QP||n|}$

So, we have: $d = |QP|\frac{|QP\cdot{n}|}{|QP||n|}= \frac{|QP\cdot{n}|}{|n|}$

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A few questions - Planes

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