Hi, I'm having trouble answering this question.
The lines l1 and l2 are defined as
l1: (x-1)/3 = (y-5)/2 = (z-12)/-2
l2: (x-1)/8 = (y-5)/11 = (z-12)/6
The plane j contains both l1 and l2. Find the Cartesian equation of j.
Take the cross product and the normal to both lines is <2,-2,1>
The plane is then just
2(x-1)-2(y-5)+1(z-12) = 0
2x-2y+z = 4
To find the Cartesian equation of plane j, we need to find two direction vectors for the plane. These direction vectors can be found by taking the cross product of the direction vectors of lines l1 and l2.
The direction vector of line l1 is given by the coefficients of x, y, and z in the equation. So, the direction vector of l1 is (3, 2, -2).
Similarly, the direction vector of line l2 is (8, 11, 6).
Now, we can find the cross product of these two direction vectors:
v = (3, 2, -2) × (8, 11, 6)
To compute the cross product, we can use the formula:
v = (a2 * b3 - a3 * b2, a3 * b1 - a1 * b3, a1 * b2 - a2 * b1)
Substituting the values, we get:
v = (2 * 6 - (-2) * 11, (-2) * 8 - 3 * 6, 3 * 11 - 2 * 8)
= (12 + 22, -16 - 18, 33 - 16)
= (34, -34, 17)
So, the vector v is (34, -34, 17).
The equation of plane j can be written as:
ax + by + cz = d
To find the values of a, b, c, and d, we can substitute the coordinates of a point on line l1 into the equation.
Let's use the point (1, 5, 12) from line l1:
1a + 5b + 12c = d
We can now find the value of d by substituting the point (1, 5, 12) and the direction vector (34, -34, 17) into the equation:
1 * 34 + 5 * (-34) + 12 * 17 = d
34 - 170 + 204 = d
68 = d
So, the Cartesian equation of plane j is:
34x - 34y + 17z = 68