Let T be the plane −y+2z = −8. Find the shortest distance d from the point P0=(−5, 4, 2) 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.
d = 0
Q = (0, 0, 0)
Please help!!
To find the shortest distance, d, from the point P0 to the plane T, we need to find the perpendicular distance between the point and the plane.
1. Start by converting the equation of the plane T into point-normal form. Point-normal form is given by:
A(x - x1) + B(y - y1) + C(z - z1) = 0,
where (x1, y1, z1) is a point on the plane and (A, B, C) is the normal vector to the plane.
In this case, the plane T can be rewritten as:
0x + (-1)(y - 0) + 2(z - 0) = -8,
which simplifies to:
-y + 2z + 8 = 0.
So the normal vector to the plane T is (0, -1, 2), and a point on the plane can be taken as (0, 0, 0).
2. Next, find the vector that connects the given point P0 to the point on the plane T. This can be done by subtracting the coordinates of P0 from the coordinates of any point on the plane T. In this case, we can subtract (0, 0, 0) from (−5, 4, 2) to get the vector:
P0 - P = (-5 - 0, 4 - 0, 2 - 0) = (-5, 4, 2).
3. Now, we need to find the projection of the vector P0 - P onto the normal vector of the plane T. This is obtained by taking the dot product of the two vectors and dividing by the magnitude of the normal vector. Let's call the projection vector as proj.
proj = ((-5, 4, 2) · (0, -1, 2)) / ||(0, -1, 2)||,
where · represents the dot product and || || represents the magnitude.
The dot product is given by:
(-5, 4, 2) · (0, -1, 2) = -5(0) + 4(-1) + 2(2) = -4.
The magnitude of the normal vector is given by:
||(0, -1, 2)|| = √(0^2 + (-1)^2 + 2^2) = √(1 + 4) = √5.
Therefore,
proj = -4 / √5.
4. The shortest distance, d, from the point P0 to the plane T is the length of the projection vector, which can be obtained by taking the absolute value of the projection:
d = |proj| = |-4 / √5| = 4 / √5.
5. To find the point Q in the plane T that is closest to P0, we need to find the coordinates of Q using the projection vector.
Q = P0 - proj * (0, -1, 2),
where * represents scalar multiplication.
Q = (-5, 4, 2) - (4 / √5)(0, -1, 2) = (-5, 4, 2) - (0, -4 / √5, 8 / √5).
Simplifying,
Q = (-5, 4 + 4 / √5, 2 + 8 / √5) = (-5, (4√5 + 4) / √5, (2√5 + 8) / √5).
Therefore, the shortest distance d from the point P0=(-5, 4, 2) to the plane T is 4 / √5, and the point Q in the plane T that is closest to P0 is Q = (-5, (4√5 + 4) / √5, (2√5 + 8) / √5).