# Vector equation of line with angle to a plane

• Jun 28th 2010, 01:02 PM
nvwxgn
Vector equation of line with angle to a plane
Problem is: Find the vector equation of the line r that contains the point P (1, 1, 1), is parallel to the plane x + 2y - z = 0 and presents a pi/3 radians angle with the plane x - y + 2z - 1 = 0

I let the direction vector of r be (a, b, c) and consider that (a, b, c).(1, 2, -1) = 0, since it's parallel to the first plane. Also, sqrt(3)/2 = (a, b, c).(1, -1, 2) / |a, b, c| times sqrt(6). But then, I'm stuck at a three-variable system with two equations and also a modulus.

How am I supposed to do that? Thanks in advance.
• Jun 28th 2010, 03:01 PM
gmatt
Consider the information that you are given carefully, I will help.

The line $\displaystyle r$ is parallel to the plane $\displaystyle P_1$ defined by $\displaystyle \vec{x}\cdot (1,2,-1) = 0.$ This means that it lies in $\displaystyle P_1$ and as you have correctly deduced, if we let the direction vector of $\displaystyle r$ be $\displaystyle \vec{a}=(a,b,c)$ then we require that $\displaystyle \vec{a} \cdot (1,2,-1) = 0$. Let $\displaystyle P_2$ be the plane defined by $\displaystyle \vec{x} \cdot(1,-1,2) = 1$.

Now consider the intersection of $\displaystyle P_1$ and $\displaystyle P_2$, $\displaystyle P_1 \cap P_2.$ Clearly $\displaystyle P_1 \cap P_2$ is a line with direction vector $\displaystyle \vec{b}=(1,2,-1) \times (1,-1,2)$, and furthermore $\displaystyle P_1 \cap P_2$ is parallel with both $\displaystyle P_1$ and $\displaystyle P_2$. For $\displaystyle r$ to have $\displaystyle \pi / 3$ angle with $\displaystyle P_2$ we must have that $\displaystyle r$ and $\displaystyle P_1 \cap P_2$ have angle $\displaystyle \pi / 3$ since $\displaystyle r$ is parallel to $\displaystyle P_1$, and $\displaystyle P_1 \cap P_2$ is parallel with both $\displaystyle P_1$ and $\displaystyle P_2$.

Thus, we require that $\displaystyle \frac{\vec{a} \cdot \vec{b}}{\parallel \vec{a} \parallel \parallel \vec{b} \parallel} = \cos ( \pi / 3 ).$

You have two conditions so far:

1) $\displaystyle \vec{a} \cdot (1,2,-1) = 0$
2) $\displaystyle \frac{\vec{a} \cdot \vec{b}}{\parallel \vec{a} \parallel \parallel \vec{b} \parallel} = \cos ( \pi / 3 )$.

The final (third) condition on $\displaystyle \vec{a}$ is rather artificial. Lets put the condition on $\displaystyle \vec{a}$ so that we find a unit vector, that is the 3 conditions are then

1) $\displaystyle \vec{a} \cdot (1,2,-1) = 0$
2) $\displaystyle \frac{\vec{a} \cdot \vec{b}}{ \parallel \vec{b} \parallel} = \cos ( \pi / 3 )$.
3) $\displaystyle \left\parallel \vec{a} \right\parallel = 1.$

I think that what I wrote above is correct; however, I am not sure why you are given the point through which the line $\displaystyle r$ must pass through, I believe its a red herring.