# Vectors, lines, and planes

• May 20th 2008, 07:37 AM
pantera
Vectors, lines, and planes
The points A, B, C have position vectors i + j + 2k, i + 2j + 3k, 3i + k respectively and lie in plane π.
1. Find the area of triangle ABC
2. the shortest distance from C to the line AB
3. the cartesian equation of the plane π

The line L passes through the origin and is normal to the plane π, it intersects π at the point D.
4. Find the coordinates of the point D
5. The distance of π from the origin
Thank you!
• May 20th 2008, 11:10 AM
earboth
Quote:

Originally Posted by pantera
The points A, B, C have position vectors i + j + 2k, i + 2j + 3k, 3i + k respectively and lie in plane π.
1. Find the area of triangle ABC
2. the shortest distance from C to the line AB
3. the cartesian equation of the plane π

The line L passes through the origin and is normal to the plane π, it intersects π at the point D.
4. Find the coordinates of the point D
5. The distance of π from the origin
Thank you!

I'm used to a slightly different form to write vectors so you have to change my results into the form you are used to. Sorry.

A(1, 1, 2)
B(1, 2, 3)
C(3, 0, 1)

to #1. Probably you know that the absolute value of the cross-product of 2 non-collinear vectors equals the area of the parallelogram which is spanned by the vectors. Thus the area of the triangle ABC is:

$\displaystyle A_{ABC}=\frac12 \cdot |(\overrightarrow{AB} \times \overrightarrow{AC})|$ ......... Therefore you have: $\displaystyle A_{ABC}=\frac12 \cdot |((0,1,1) \times (2,-1,-1))|=\frac12 \cdot |(0,2,-2)|=\sqrt{2}$

to #3:

The cross-product yields the normal vector of the plane. Therefore the equation of the plane is:

$\displaystyle \pi: 2y-2z+2=0$

to #4:
The line L has the equation:

$\displaystyle L: (x, y, z)= t \cdot (0,2,-2)$ with
x=0
y=2t
z=-2t

Plug in these values into the equation of the palne and solve for t:

$\displaystyle 4t-(-4t)+2=0~\implies~8t=-2~\implies~t=-\frac14$ Plug in this value into the equation of the line to get the coordinates of D:

$\displaystyle D\left(0~,~-\frac12~,~\frac12\right)$

to #5:

Rewrite the equation of the plane so the normal vector becomes a unit vector:

$\displaystyle |(0,2,-2)|=2\sqrt{2}$

The equation of the plane becomes now:

$\displaystyle \frac{2y}{2\sqrt{2}}-\frac{2z}{2\sqrt{2}}+ \underbrace{\frac{2}{2\sqrt{2}}}_ {distance\ of \ \pi \ from\ O}=0$