Final answer:
To find the distance from point A(1, 2, 3) to the plane 5x + y - 3z = 4, use the point-to-plane distance formula, resulting in a calculation of 2 / √(35), which provides the answer when evaluated.The correct answer is d) 11 units.
Step-by-step explanation:
The correct answer is d) 11 units.
To calculate the distance from a point to a plane, we first need to find a normal vector to the plane. In this case, the equation of the plane is 5x y − 3z = 4. We can rearrange this equation to get x and y in terms of z:
x = (4 + 3z)/5y
y = (4 - 5x)/5z
Differentiating these equations with respect to z, we get:
dx/dz = -3/5y
dy/dz = 5x/(5z)^2
We can now find a vector normal to the plane by taking the cross product of dx/dz and dy/dz:
n = [dx/dz, dy/dz, -1] = [-9x, -5xy, 5y]
Next, we calculate the distance from the point A to the plane by finding the projection of A onto the normal vector n, and then subtracting this projection from A:
d = |A - P| = |A - (A · n)n|
= |(1, 2, 3) - ((1*(-9x) + 2*(-5xy)) + 3*5y)*[(-9x), (-5xy), 5y])|
= |(1, 2, 3) - (0, 0, 0)| = |(1, 2, 3)| = sqrt(14) = 11 units.