Thread: Tough Vector Question with plane + lines

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2. Originally Posted by noland0
A line (L1) is parallel to a Plane (P)

L1: (4,-8,7) + t(2,3,2)
P: x -2y + 2z = 7

(a) Find the distance between the line and the plane

(b) find the the cartesian equation(s) for the straight line which is the reflection of L1 in the plane (P)
to a): The plane is given in normal form with $\displaystyle \vec n = (1, -2, 2)$. Re-write the equation of the plane into Hesse normal form by dividing by the value of the normal vector:

$\displaystyle |\vec n| = \sqrt{1^2+2^2+2^2} = 3$. Thus

$\displaystyle P_H:\frac13 x- \frac23 y + \frac23 z - \frac73 = 0$

Now choose a point of L1 and calculate the distance of this point to the plane. If t = 0 then Q(4, -8, 7) is a point of the line. Plug in the coordinates of Q into the equation of $\displaystyle P_H$ to calculate the distance QP:

$\displaystyle d(Q,P) = \frac13 \cdot 4- \frac23 \cdot (-8) + \frac23 \cdot 7 - \frac73 = 3$

The positive sign indicates that the point Q and the origin are on different sides of the plane P.

3. Originally Posted by noland0
A line (L1) is parallel to a Plane (P)

L1: (4,-8,7) + t(2,3,2)
P: x -2y + 2z = 7

(a) Find the distance between the line and the plane

(b) find the the cartesian equation(s) for the straight line which is the reflection of L1 in the plane (P)
to b) If I understand your question correctly then you are looking for a line which is parallel to L1, is placed on the other side of P as L1 and has the same distance to P as L1. If so:

A line through A(4, -8, 7) and perpendicular to P cuts the plane in F(1, -2, 1). F is the mid point of the segment AA'. Use the midpoint formula toget the coordinates of A'(-2, 4, -5).

Thus the line L1' has the equation:

$\displaystyle \vec r = (-2,4,-5)+s\cdot (2,3,2)~\implies~\begin{array}{lr}x=&-2+2s\\y=&4+3s\\z=&-5+2s\end{array}$

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5. $\displaystyle$
$\displaystyle \vec r = (-2,4,-5)+s\cdot (2,3,2)~\implies~\begin{array}{lr}x=&-2+2s\\y=&4+3s\\z=&-5+2s\end{array}$
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