1. ## Normal To Plane

The plane P1 passes through the P, with position vector i + 2jk, and is perpendicular to the line L with equation

r = 3i – 2k +l(-i + 2j + 3k)

They got the normal to the plane to be : – i + 5j + 3k

How did they get that? Thanks in advance.

(Sorry for the strange font, I can't be bothered to edit the font.)

2. Originally Posted by Air
The plane P1 passes through the P, with position vector i + 2jk, and is perpendicular to the line L with equation

r = 3i – 2k +l(-i + 2j + 3k)

They got the normal to the plane to be : – i + 5j + 3k

How did they get that? Thanks in advance.

(Sorry for the strange font, I can't be bothered to edit the font.)
Looks like they made a typo ....

Since the line is perpendicular to the plane and a vector in the direction of the line is -i + 2j + 3k, a normal to the plane will be -i + 2j + 3k, NOT – i + 5j + 3k ......

3. Originally Posted by mr fantastic
Looks like they made a typo ....

Since the line is perpendicular to the plane and a vector in the direction of the line is -i + 2j + 3k, a normal to the plane will be -i + 2j + 3k, NOT – i + 5j + 3k ......
So a line in the form:

$\displaystyle r = a + \lambda b$ where 'a' is position vector and 'b' is direction vector. If this is perpendicular to a plane, does that always mean that the 'n' for the plane is 'b'.

So r.n=a.n < Equation of plane. n is normal which the direction vector of a perpendicular line.

Is that correct?

4. Originally Posted by Air
So a line in the form:

$\displaystyle r = a + \lambda b$ where 'a' is position vector and 'b' is direction vector. If this is perpendicular to a plane, does that always mean that the 'n' for the plane is 'b'.

So r.n=a.n < Equation of plane. n is normal which the direction vector of a perpendicular line.

Is that correct?

EDIT: It can't be a typo as they wanted us to show that the cartesian equation of the plane is x-5y-3z=6 which wouldn't work if we used -i+2j+3k.
May as well post the entire question, I agree with Mr F. Looks like they typoed in the question.

5. Originally Posted by Air
So a line in the form:

$\displaystyle r = a + \lambda b$ where 'a' is position vector and 'b' is direction vector. If this is perpendicular to a plane, does that always mean that the 'n' for the plane is 'b'. Mr F says: Yes.

[snip]

EDIT: It can't be a typo as they wanted us to show that the cartesian equation of the plane is x-5y-3z=6 which wouldn't work if we used -i+2j+3k.
They are wrong. I'd say the typo is in the given equation of the line .... it should be $\displaystyle r = 3i - 2k + \alpha (-i + 5j + 3k)$ ......

6. Another Question:
The Points A, B and C lie on the plane P and, relative to a fixed origin O, they have position vectors
a = 3i - j + 4k, b = -i + 2j, c = 5i - 3j + 7k

Find an equation of Pin the form r.n = p.

^ I worked out direction vector of AB and AC but how do I work out n?

7. Originally Posted by Air
Another Question:
The Points A, B and C lie on the plane P and, relative to a fixed origin O, they have position vectors
a = 3i - j + 4k, b = -i + 2j, c = 5i - 3j + 7k

Find an equation of Pin the form r.n = p.

^ I worked out direction vector of AB and AC but how do I work out n?
$\displaystyle \overrightarrow{AC} \times \overrightarrow{AB}$

Bobak

8. Originally Posted by Air
Another Question:
The Points A, B and C lie on the plane P and, relative to a fixed origin O, they have position vectors

a = 3i - j + 4k, b = -i + 2j, c = 5i - 3j + 7k

Find an equation of Pin the form r.n = p.

^ I worked out direction vector of AB and AC but how do I work out n?
n will be the cross (vector) product of AB and AC:

n = AB x AC.