Let A = (0,0,0), B = (9,8,12), and C = (6,2,3). Find coordinates for the point on line AB that is closest to C.
GOOD GRIEF, Just realized I did this same question a week ago.
At least I ended up with the same answer.
http://www.jiskha.com/display.cgi?id=1484278204
Let the perpendicular distance from C to AB meet AB at D
vector AB = [9,8,12]
equation of plane through C and D having AB as a normal:
9x + 8y + 12z = c
but (6,2,3) lies on it , so the plane is
9(6)+8(2)+12(3) = c = 106
plane equation: 9x+8y+12z = 106
equation of line AB:
x = 0+9t
y= 0+8t
z= 0+12t
where does AB cut the plane?
9(9t) + 8(8t) + 12(12t) = 106
289t = 106
t = 106/289
so point D is (954/289 , 848/289 , 1272/289)
check my arithmetic
To find the point on line AB that is closest to C, we need to find the projection of vector AC onto the line AB.
Let's start by finding the direction vector of line AB.
Direction vector of line AB = B - A = (9, 8, 12) - (0, 0, 0) = (9, 8, 12).
Next, let's find the vector from point A to C.
Vector AC = C - A = (6, 2, 3) - (0, 0, 0) = (6, 2, 3).
The projection of vector AC onto line AB can be found using the formula:
projection of AC onto AB = (AC dot AB) / (AB dot AB) * AB.
Now, let's plug in the values to find the projection:
AC dot AB = (6, 2, 3) dot (9, 8, 12) = 6*9 + 2*8 + 3*12 = 54 + 16 + 36 = 106.
AB dot AB = (9, 8, 12) dot (9, 8, 12) = 9*9 + 8*8 + 12*12 = 81 + 64 + 144 = 289.
projection of AC onto AB = (AC dot AB) / (AB dot AB) * AB
= 106 / 289 * (9, 8, 12)
= (0.367, 0.323, 0.488) * (9, 8, 12)
= (3.303, 2.584, 5.853).
Finally, we add the projection vector to point A to find the coordinates of the point on line AB that is closest to C:
Coordinates of point on line AB closest to C = A + (3.303, 2.584, 5.853)
= (0, 0, 0) + (3.303, 2.584, 5.853)
= (3.303, 2.584, 5.853).
Therefore, the coordinates of the point on line AB that is closest to C are approximately (3.303, 2.584, 5.853).
To find the coordinates for the point on line AB that is closest to point C, we need to follow these steps:
Step 1: Calculate the vector AB
To find the vector AB, we subtract the coordinates of point A from the coordinates of point B.
AB = B - A = (9-0, 8-0, 12-0) = (9, 8, 12)
Step 2: Calculate the vector AC
To find the vector AC, we subtract the coordinates of point A from the coordinates of point C.
AC = C - A = (6-0, 2-0, 3-0) = (6, 2, 3)
Step 3: Calculate the projection of vector AC onto vector AB
We can find the projection of vector AC onto vector AB by using the dot product formula and dividing it by the magnitude of vector AB squared.
Projection of AC onto AB = (AC · AB) / ||AB||^2
where · represents the dot product and ||AB|| represents the magnitude of vector AB.
Step 4: Calculate the closest point coordinates
The coordinates for the point on line AB that is closest to point C can be found by adding the projection (from Step 3) of vector AC onto vector AB to the coordinates of point A.
Closest Point on AB = A + (Projection of AC onto AB) * AB
Let's calculate it:
AB = (9, 8, 12)
AC = (6, 2, 3)
|AB|^2 = 9^2 + 8^2 + 12^2 = 81 + 64 + 144 = 289
(AC · AB) = 6*9 + 2*8 + 3*12 = 54 + 16 + 36 = 106
Projection of AC onto AB = (AC · AB) / ||AB||^2 = 106 / 289 ≈ 0.367
Closest point on AB = A + (Projection of AC onto AB) * AB
= (0, 0, 0) + 0.367 * (9, 8, 12)
= (0, 0, 0) + (3.303, 2.936, 4.404)
= (3.303, 2.936, 4.404)
Therefore, the coordinates for the point on line AB that is closest to point C are approximately (3.303, 2.936, 4.404).