Write a vector equation and scalar equation of the plane containing lines
[x,y,z]= [-2,3,12]+s[-2,1,5]
[x,y,z]= [1,-4,4]+t[-6,3,15]
Did you notice that our two lines have the same direction?
(the second vector is a multiple of the first)
So we need a second direction vector, we can obtain that by using the two given points (-2,3,12) and (1,-4,4)
that direction vector is [3,-7,-8]
so a vector equation is
[x,y,z] = (-2,3,12) + s[-2,1,5] + t[3,-7,-8]
for the scalar equation we will take the cross-product of our two direction vectors to get a normal, which is
[27,-1,11]
http://www.analyzemath.com/vector_calculators/vector_cross_product.html
so our equation is
27x - y + 11z = c
plug in (1,-4,4) to get c
27 + 4 +44 = c = 75
So the scalar equation is
27x – y + 11z = 75
To find the vector equation and scalar equation of the plane containing the given lines, we first need to find the vector that is perpendicular to both lines. This can be done by finding the cross product of the direction vectors of the lines.
The direction vector of the first line is [-2, 1, 5], and the direction vector of the second line is [-6, 3, 15]. To find a vector that is perpendicular to both, we take the cross product:
n = [-2, 1, 5] x [-6, 3, 15]
n = (1*-15 - 5*3)i - (-2*15 - 5*-6)j + (-2*3 - 1*-6)k
n = -15i - 18j + 0k
n = -15i - 18j
The vector n is the normal vector to the plane. Now, we can find a point on the plane by taking any point that lies on one of the lines. Let's choose the point [-2, 3, 12] from the first line.
Now we can write the vector equation of the plane:
[r - [-2, 3, 12]] * [-15, -18] = 0
Simplifying this equation, we have:
[-15(r1 + 2) - 18(r2 - 3)] = 0
This is the vector equation of the plane containing the given lines.
To find the scalar equation of the plane, we can rewrite the vector equation by expanding the scalar product:
-15r1 - 30 - 18r2 + 54 = 0
Combining like terms, we have:
-15r1 - 18r2 + 24 = 0
This is the scalar equation of the plane containing the given lines.