find the point and intersection of a line through (1,-3,1) and (3,-12,6) and the plane given by x-y+z=6. my answers were (16/5),(-18/5),(3/4) but it was wrong.
To find the point of intersection between the line and the plane, we need to follow these steps:
Step 1: Find the direction vector of the line.
To find the direction vector, we find the vector difference between the two given points.
Direction vector d = (3-1, -12+3, 6-1) = (2,-9,5)
Step 2: Find the normal vector of the plane.
The coefficients of x, y, and z in the given equation of the plane, x-y+z=6, form the normal vector n = (1,-1,1).
Step 3: Find the point of intersection.
Let's use the point-slope form of the equation of the line to get the coordinates of a point on the line:
x-1 = 2t
y+3 = -9t
z-1 = 5t
Now substitute these values into the equation of the plane, we get:
x-y+z = 6
(2t)- (y+3)+(5t) = 6
-9t+2t+5t+3 = 6
-2t = -3
t = 3/2
Therefore, the point of intersection has coordinates:
x = 2(3/2) + 1 = 4
y = -9(3/2) - 3 = -18
z = 5(3/2) + 1 = 8.5
So the point of intersection is (4,-18,8.5).
Therefore, your answer is incorrect.