# Find the intersection of the line and plane

• Dec 12th 2006, 07:40 PM
Jenny20
Find the intersection of the line and plane
question
a) Determine whether the line
x=3+8t
y=4+5t
z=-3-t
is parallel to the plane x-3y+5z =12.

Note: I know how to solve part a) - the answer is the line and plane are not parallel.

b) Find the intersection of the line and the plane.

Note: Could you please teach me how to solve part b). Thank you very much.
• Dec 12th 2006, 07:49 PM
ThePerfectHacker
Quote:

Originally Posted by Jenny20
question
a) Determine whether the line
x=3+8t
y=4+5t
z=-3-t
is parallel to the plane x-3y+5z =12.

Note: I know how to solve part a) - the answer is the line and plane are not parallel.

The aligned vector with the line is,
$\bold{v}=8\bold{i}+5\bold{j}-\bold{k}$
The normal vector with the plane is,
$\bold{u}=\bold{i}-3\bold{k}+5\bold{k}$.
A line and plane are parallel implies that the aligned vector is perpendicular (orthogonal) to the normal vector.
To check this we look at the dot product which is not zero. Thus, they are not parallel.

Quote:

b) Find the intersection of the line and the plane.
Substitute those values of x,y,z into the equation of plane.
$(3+8t)-3(4+5t)+5(-3-t)=12$
Thus,
$3+8t-12-15t-15-5t=12$
$-12t-24=12$
$-12t=36$
$t=-3$.

Thus,
$\left\{ \begin{array}{c} x=3+8t=3+8(-3)=-21 \\ y=4+5t=4+5(-3)=-11\\ z=-3-t=-3-(-3)=0 \end{array} \right\}$
• Dec 12th 2006, 08:03 PM
Jenny20
Thank you very much ThePerfectHacker!