Let T be the plane x-y+3z=15. Find the shortest distance d from the point p0=(-2, 4, 4) to T, and the point Q in T that is closest to p0. Use the square root symbol where needed to give an exact value for you answer.
the distance is
|-2-4+3*4-15|/√(1+1+9) = 9/√11
I'll have to get out my math book to figure out Q, but you should have some examples in your text. Also google can surely provide similar solved exercises.
To find the shortest distance from a point to a plane, we can use the formula:
d = |ax0 + by0 + cz0 + d| / √(a^2 + b^2 + c^2)
where (x0, y0, z0) is the coordinates of the point, a, b, c are the coefficients of the plane equation, and d is the constant term.
In this case, the coefficients of the plane equation x - y + 3z = 15 are a = 1, b = -1, c = 3, and d = 15. The point given is p0 = (-2, 4, 4).
Plugging in the values into the formula, we get:
d = |1*(-2) + (-1)*4 + 3*4 + 15| / √(1^2 + (-1)^2 + 3^2)
Simplifying, we have:
d = |(-2) - 4 + 12 + 15| / √(1 + 1 + 9)
= |21| / √11
= 21 / √11
So, the shortest distance d from p0 to the plane T is 21 / √11.
To find the point Q in T that is closest to p0, we need to find the coordinates (x, y, z) that satisfy the equation of the plane and minimize the distance. Let Q = (x, y, z).
Substituting x = y - 3z + 15 into the plane equation, we have:
(y - 3z + 15) - y + 3z = 15
-2z = 0
z = 0
Substituting z = 0 back into the plane equation, we have:
x - y + 3(0) = 15
x - y = 15
x = y + 15
So, the coordinates of the point Q in T that is closest to p0 is Q = (y + 15, y, 0), where y can be any real number.
Note: The point Q is not unique, and there are infinitely many points in the plane T that are equidistant from p0.