Find the distance from the point to the given plane.

(1,−5,9), 3x+2y+6z=5

vector normal to plane has direction

3 i + 2 j + 6 k

line normal to plane through point is
(1, -5 , 9) + (3, 2, 6) t

where does that hit the plane?
3(1+3t) + 2(-5+2t) + 6(9+6t) = 5

solve for t
go back and use that t to get x, y, z in plane
x = 1+3t y = -5+2t z = 9+6t
then
d^2 = (X2-X1)^2 + (Y2-Y1)^2+(Z2-Z1)^2

Why did the point join the plane? Because they wanted to be closer! In this case, to find the distance from the point (1, -5, 9) to the plane 3x + 2y + 6z = 5, we can use the formula for the distance between a point and a plane. The distance is equal to |Ax + By + Cz + D| / sqrt(A^2 + B^2 + C^2), where A, B, C are the coefficients of the plane equation, and D is the constant term. Plugging in the values, we get |3(1) + 2(-5) + 6(9) + 5| / sqrt(3^2 + 2^2 + 6^2). Calculating this gives us a result, which is the distance.

To find the distance from a point to a plane, we can use the formula:

distance = |Ax + By + Cz + D| / sqrt(A^2 + B^2 + C^2)

In this case, the given plane equation is 3x + 2y + 6z = 5, where A = 3, B = 2, C = 6, and D = -5.

The point given is (1, -5, 9), where x = 1, y = -5, and z = 9.

Let's substitute the values into the formula:

distance = |(3*1) + (2*-5) + (6*9) + (-5)| / sqrt(3^2 + 2^2 + 6^2)

Simplifying,

distance = |3 - 10 + 54 - 5| / sqrt(9 + 4 + 36)
distance = |42| / sqrt(49)
distance = 42 / 7
distance = 6

Therefore, the distance from the point (1, -5, 9) to the plane 3x + 2y + 6z = 5 is 6 units.

To find the distance from a point to a plane in three-dimensional space, we will use the formula:

Distance = |Ax + By + Cz + D| / sqrt(A^2 + B^2 + C^2)

where Ax + By + Cz + D is the equation of the plane, and (x, y, z) is the coordinates of the point.

In this case, the equation of the plane is 3x + 2y + 6z = 5, and the point is (1, -5, 9).

Step 1: Substitute the coordinates of the point into the equation of the plane to find the numerator.
Numerator = |3(1) + 2(-5) + 6(9) + 5|
Numerator = |3 - 10 + 54 + 5|
Numerator = |52|

Step 2: Calculate the denominator.
Denominator = sqrt(3^2 + 2^2 + 6^2)
Denominator = sqrt(9 + 4 + 36)
Denominator = sqrt(49)
Denominator = 7

Step 3: Calculate the distance using the formula.
Distance = |52| / 7
Distance = 52 / 7
Distance = 7.4286 (rounded to 4 decimal places)

Therefore, the distance from the point (1, -5, 9) to the plane 3x + 2y + 6z = 5 is approximately 7.4286 units.