find the distance from the point Q to the line "l"
Q=(0,1,0)
l with the equation [x,y,z] = [1,1,1,] + t[-2,0,3]
To find the distance from a point to a line in 3D space, you can use the formula:
distance = |(Q - P) · n| / |n|
Where Q is the point, P is a point on the line, n is the normal vector of the line, and | | indicates the magnitude or length of a vector.
In this case, Q = (0, 1, 0) and the line l can be parametrically defined as P = (1, 1, 1) + t(-2, 0, 3). We need to find a point P on the line and the normal vector n.
To find a point P on the line, we can substitute t = 0 into the parametric equation:
P = (1, 1, 1) + 0(-2, 0, 3) = (1, 1, 1).
The vector PQ = Q - P is then (0, 1, 0) - (1, 1, 1) = (-1, 0, -1).
To find the normal vector n, we take the direction vector (-2, 0, 3) of the line and find a vector perpendicular to it by taking the cross product with another vector. For simplicity, we can choose the cross product with the vector (1, 0, 0):
n = (-2, 0, 3) × (1, 0, 0)
= (0, -3, 0) + (0, 0, 2)
= (0, -3, 2).
Now we can calculate the distance:
distance = |(-1, 0, -1) · (0, -3, 2)| / |(0, -3, 2)|
= |(-1)(0) + (0)(-3) + (-1)(2)| / √(0^2 + (-3)^2 + 2^2)
= |-2| / √(0 + 9 + 4)
= 2 / √13.
Therefore, the distance from the point Q(0, 1, 0) to the line l is 2 / √13.