Let T be the plane −x+2y = 4. Find the shortest distance d from the point P0=(2, 2, 1) 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 your answer.
I know I have posted this many times but I don't know what to do.. please help
google is your friend. Try searching for
distance from point to plane
for some helpful discussions. For example,
mathinsight.org/distance_point_plane
it's just an extension of the usual formula for the distance from a point to a line.
Is this the final answer
d=sqrt5
Q:
x=-30/7
y=-18/7
z=-8/7
No worries, I'm here to help! Finding the shortest distance from a point to a plane involves a few steps. Let's break it down:
Step 1: Convert the equation of the plane to vector form:
- Take two points on the plane (x1, y1, z1) and (x2, y2, z2).
- Subtract one point from the other to get a vector that lies on the plane.
- The equation of the plane can be written as (x, y, z) = (x1, y1, z1) + s * (x2 - x1, y2 - y1, z2 - z1) + t * (a, b, c), where s and t are real numbers, and (a, b, c) is a vector normal to the plane.
Step 2: Find a point Q on the plane that is closest to P0:
- Substitute the coordinates of P0 into the equation of the plane.
- Solve for s and t to find the coordinates of Q.
Step 3: Calculate the shortest distance d from P0 to the plane using the distance formula:
- The distance formula between two points in space is given by:
d = √((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2)
Now, let's apply these steps to the given problem:
Step 1: Convert the equation of the plane to vector form:
- In this case, the equation of the plane is -x + 2y = 4.
- Let's choose two points on the plane, (0, 2, 2) and (2, 3, 2), to get a vector on the plane.
- Subtracting these points, we obtain a vector on the plane: (-2, 1, 0).
So, the equation of the plane in vector form is (x, y, z) = (0, 2, 2) + s * (-2, 1, 0) + t * (a, b, c), where s and t are real numbers, and (a, b, c) is a vector normal to the plane.
Step 2: Find a point Q on the plane that is closest to P0:
- Substituting the coordinates of P0, we have:
2 = 2 + s * (-2) + t * a (for the x-coordinate)
2 = 2 + s * 1 + t * b (for the y-coordinate)
1 = 2 + s * 0 + t * c (for the z coordinate)
Solving these equations, we get s = -1 and t = 0. Therefore, the coordinates of Q are (1, 1, 2).
Step 3: Calculate the shortest distance d from P0 to the plane:
- Using the distance formula, we have:
d = √((2 - 1)^2 + (2 - 1)^2 + (1 - 2)^2)
d = √(1 + 1 + 1)
d = √3
So, the shortest distance from P0=(2, 2, 1) to the plane −x+2y=4 is √3, and the point Q in the plane closest to P0 is (1, 1, 2).