Find the vector equation of the line that passes through the point (2,-1,7) and is parallel to the line of intersection of the planes x + 2y - 3z = -6 and 3x - y + 2z = 4
We need the direction of their line of intersection....
double the 2nd plane equation, then add that to the 1st
x+2y-3z = -6
6x - 2y + 4z = 8
7x + z = 2
z = 2-7x
let x=1
then z = -5
back in the 1st
1 + 2y + 15 = -6
y = -11 -----------> point (1,-11,-5)
let x = 0
then z = 2
back in 1st
0+2y -6 = -6
y = 0 ------------> point (0,0,2)
direction vector of line of intersection of the planes = (1, -11, -7)
so one such vector equation is
r = (2,-1,7) + t(1,-11,-7)
(the question should have said, "Find a vector equation...."
since the answer I obtained is not unique)
We need the direction of their line of intersection? That's all the question mentioned... so is that still the correct answer? :/
To find the vector equation of the line that is parallel to the line of intersection of the given planes, we need to find the direction vector of the intersecting line first.
Let's find the direction vector of the intersecting line by taking the cross product of the normal vectors of the two planes. The normal vectors are the coefficients of x, y, and z in the equations of the planes.
For the first plane: x + 2y - 3z = -6, the normal vector is <1, 2, -3>.
For the second plane: 3x - y + 2z = 4, the normal vector is <3, -1, 2>.
Taking the cross product of these two normal vectors, we get:
<1, 2, -3> × <3, -1, 2> = (2 * 2 - (-1) * (-3))i + ((-3) * 3 - 2 * 2)j + (1 * (-1) - 2 * 3)k
= 4i + (-5)j - 7k
So, the direction vector of the line of intersection is <4, -5, -7>.
Now that we have the direction vector, we can write the vector equation of the line. Let's call (x₀, y₀, z₀) the coordinates of the given point (2, -1, 7).
The vector equation of a line passing through a point and parallel to a given vector is given by:
r = r₀ + tv
where r is the position vector of any point on the line, r₀ = <x₀, y₀, z₀> is the position vector of the given point, t is a parameter, and v = <a, b, c> is the direction vector.
In this case, the position vector r₀ = <2, -1, 7> and the direction vector v = <4, -5, -7>. Therefore, the vector equation of the line is:
r = <2, -1, 7> + t<4, -5, -7>
Simplifying this equation, we get:
r = <2 + 4t, -1 - 5t, 7 - 7t>
So, the vector equation of the line passing through the point (2, -1, 7) and parallel to the line of intersection of the given planes is r = <2 + 4t, -1 - 5t, 7 - 7t>.