Final answer:
The unique point on the y-axis on both planes is P (0, 1/5, 0). The unit vector u parallel to both planes with a positive first coordinate is (25, -5, -5) normalized. The vector equation for the line of intersection is r(t) = (0, 1/5, 0) + t(25, -5, -5)/√(675).
Step-by-step explanation:
To find the unique point P on the y-axis which is on both planes given by 1x+5y+5z=1 and 1x+5z=0, we need to set the x and z coordinates to 0 since the y-axis is defined by x=0 and z=0. Hence, the point P has the coordinates (0, y, 0). Substituting x=0 and z=0 into the first equation gives us 5y=1, which means y=1/5. Therefore, point P is (0, 1/5, 0).
To find the unit vector u with a positive first coordinate that is parallel to both planes, we need to find the cross product of the normals of the two planes. The normal to the first plane is (1, 5, 5) and to the second plane is (1, 0, 5). The cross product of these two vectors is u = (25, -5, -5). The magnitude of this vector is √(25² + (-5)² + (-5)²) = √(675), and the unit vector u is (25, -5, -5) / √(675). To ensure a positive first coordinate, we simply normalize this vector without changing the sign of the first coordinate.
Using parts (A) and (B), we can find a vector equation for the line of intersection of the two planes. A vector equation for a line can be written as r(t) = p + tu, where p is a point on the line and u is the direction vector of the line. The point P (0, 1/5, 0) and the unit vector u will be used to construct this equation, resulting in r(t) = (0, 1/5, 0) + t(25, -5, -5)/√(675).