find a vector equation of the line, which passes through the point (1,3,11) and is perpendicular to the yz-plane
if it's perpendicular to the y-z plane, then it has only an x component. So, it would be
r = (1+t)i + 3i + 11i
where t can be any real number.
make that
r = (1+t)i + 3j + 11k
To find the vector equation of the line passing through the point (1, 3, 11) and perpendicular to the yz-plane, we need to find a vector that is parallel to the line.
Since the line is perpendicular to the yz-plane, it means it is parallel to the x-axis. The x-axis can be represented as the vector <1, 0, 0>.
Now, we can use the point (1, 3, 11) and the parallel vector <1, 0, 0> to write the vector equation of the line.
The vector equation of a line passing through a point (a, b, c) with a parallel vector <p, q, r> is given by:
r = (a, b, c) + t(p, q, r)
In this case, a = 1, b = 3, c = 11 and the parallel vector <p, q, r> is <1, 0, 0>.
Therefore, the vector equation of the line passing through (1, 3, 11) and perpendicular to the yz-plane is:
r = (1, 3, 11) + t(1, 0, 0)