Determine the vector equation of the line l in R^3 through the points P

and Q with position vectors p = (2, −1, 6) and q = (1, 0, 11).
Determine the equation of the plane perpendicular to the line PQ which
passes through the origin.

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To determine the vector equation of the line through points P and Q in R^3, we need to find a direction vector and a point on the line.

1. Direction vector:
To find the direction vector of the line, we subtract the position vectors of the two points. Let's call this direction vector d.
d = q - p = (1, 0, 11) - (2, -1, 6) = (-1, 1, 5)

2. Point on the line:
We can choose either point P or Q as a point on the line. Let's choose point P. So, the position vector of the point on the line, r, is the same as the position vector of point P: r = (2, -1, 6)

Therefore, the vector equation of the line L through points P and Q is:
r = p + t * d
where r is the position vector of any point on the line, t is a parameter that can take any real value, p is the position vector of point P, and d is the direction vector we found earlier.

So, the vector equation of the line L is:
r = (2, -1, 6) + t * (-1, 1, 5)

Next, to determine the equation of the plane perpendicular to line PQ and passes through the origin, we need to find a normal vector to the plane.

The normal vector will be the same as the direction vector of the line PQ because any vector perpendicular to the line will also be normal to the plane. So, the normal vector is:
n = (-1, 1, 5)

The equation of a plane passing through the origin with a normal vector n is given by:
n · r = 0
where n is the normal vector, r is the position vector of any point on the plane, and · denotes the dot product.

Substituting the values, we get:
(-1, 1, 5) · r = 0

So, the equation of the plane perpendicular to the line PQ and passing through the origin is:
-x + y + 5z = 0.