If a plane passes through two points:
$A=(1,0,-2)$ and $B=(3,-1,1)$
and is parallel to the line:
$r=(3,-1,5)^T + a(0,2,-1)^T$,
then what is the equation of the plane?
I have obtained that the plane must be of the form:
$p=(1,0,-2)^T +\delta_1((1,0,-2)^T-(3,-1,1)^T)+\delta_2((3,-1,1)^T-(x,y,z)^T)$
but I can't determine what direction vectors of the plane are parallel to the direction vector of the line?
can this be solved using the dot product?
Best Answer
Step 1: The line’s direction vector d = (0,2,−1). Since the plane is parallel to the line, we can get a vector on the plane u = d = (0,2,−1).
Step 2: Vector AB is on the plane, let v = AB = A – B = (1, 0, −2) - (3, −1, 1) = (-2, 1, -3). We can verify that u and v are nonparallel vectors.
Step 3: Because the question does not specify which form of equation we need to find, the easiest one in this case is Vector Form (from this, we can easily get Parametric Form):
Method 1 – Vector Form: x = p + su + tv → x = (1, 0, −2) + s (0,2,−1) + t (-2, 1, -3) OR x = (3, −1, 1) + s (0,2,−1) + t (-2, 1, -3)
Method 2 – Parametric Form: can easily get from Method 1.
Method 3 – Normal Form: the cross product of u and v will be plane’s normal vector: n = u x v = (0,2,−1) x (-2, 1, -3) = (-5, 2, 4). So, the normal form equation is n · x = n · p → (-5, 2, 4) · x = (-5, 2, 4) · (1, 0, −2) = -13 OR (-5, 2, 4) · x = (-5, 2, 4) · (3, −1, 1) = -13