Question

Find the distance the point P(3, 1, 7), is to the plane through the three points

Q(5, 0, 3), R(0, -1, 7), and S(7, 5, 0).

Answer 1

To find the distance from a point P to a plane defined by three points Q, R, and S, we can use the following formula:

distance = |(A * x + B * y + C * z + D) / √(A^2 + B^2 + C^2)|

where (A, B, C) are the coefficients of the normal vector of the plane, (x, y, z) are the coordinates of the point P, and D is the constant in the equation of the plane.

To find the coefficients of the normal vector, we can use the cross product of two vectors lying in the plane, say, QR and RS.

Let's first find the vectors QR and RS:

QR = (5 - 0, 0 - (-1), 3 - 7) = (5, 1, -4)

RS = (7 - 0, 5 - (-1), 0 - 7) = (7, 6, -7)

Next, we find the cross product of QR and RS:

cross product of QR and RS = (B, C, A) = (7, -30, 15)

So, the normal vector of the plane is (15, -30, 7).

Now we substitute the values for the coefficients and the coordinates of the point P into the formula:

distance = |(15 * 3 + (-30) * 1 + 7 * 7 + D) / √(15^2 + (-30)^2 + 7^2)|

To find the value of D, we can use any of the three points on the plane and substitute their coordinates into the equation of the plane in the form Ax + By + Cz + D = 0.

Let's use the point Q:

15 * 5 + (-30) * 0 + 7 * 3 + D = 0

Expanding and solving for D, we get:

75 + 21 + D = 0

D = -96

Finally, substituting the values back into the formula for distance, we get:

distance = |(15 * 3 + (-30) * 1 + 7 * 7 + -96) / √(15^2 + (-30)^2 + 7^2)|

distance = |(-24 + -30 + 49 - 96) / √(15^2 + (-30)^2 + 7^2)|

distance = |(-101) / √(15^2 + (-30)^2 + 7^2)|

distance = |(-101) / √(15^2 + (-30)^2 + 7^2)|

distance = |(-101) / √(225 + 900 + 49)|

distance = |(-101) / √1274|

distance = |(-101) / 35.6|

distance = 2.85

So, the distance between the point P and the plane is approximately 2.85.