Final answer:
To find the point P on the y-axis that is on both planes, set x = 0 and z = 0 for each plane and solve for y. The unique point P is (0, 1, 0). To find a unit vector u parallel to both planes, find the normal vectors of the planes and normalize one of them. The unit vector u with a positive first coordinate is (2/sqrt(30), 1/sqrt(30), 5/sqrt(30)). The vector equation for the line of intersection of the two planes is r(t) = (0, 1, 0) + t(5, -10, -2).
Step-by-step explanation:
To find the unique point P on the y-axis that is on both planes, we need to find the values of x, y, and z that satisfy the equations of both planes.
For the first plane, 2x + y + 5z = 1, if we set x = 0 and z = 0, we can solve for y:
0 + y + 0 = 1
y = 1
Therefore, the point P on the y-axis that is on the first plane is (0, 1, 0).
For the second plane, 2x + 5z = 0, since there is no y term, any point on the y-axis will satisfy this equation. So, we can choose any value for y and set x = 0 and z = 0. Let's choose y = 1:
2(0) + 5(0) = 0
Therefore, the point P on the y-axis that is on the second plane is (0, 1, 0).
Since both points have the same x and z coordinates, the unique point P on the y-axis that is on both planes is (0, 1, 0).
To find a unit vector u with a positive first coordinate that is parallel to both planes, we can find the normal vectors of the planes and then normalize one of them.
For the first plane, the normal vector is (2, 1, 5). For the second plane, the normal vector is (2, 0, 5). Both normal vectors are parallel to the planes.
Let's normalize the first vector:
||u|| = sqrt(2^2 + 1^2 + 5^2) = sqrt(30)
So, the unit vector u with a positive first coordinate that is parallel to both planes is (2/sqrt(30), 1/sqrt(30), 5/sqrt(30)).
To find a vector equation for the line of intersection of the two planes, we can take the cross product of the normal vectors of the planes to get a vector that is perpendicular to both planes.
Let's calculate the cross product:
(2, 1, 5) x (2, 0, 5) = (5, -10, -2)
Now, we can use one of the points on the line of intersection, which is (0, 1, 0), and the direction vector (5, -10, -2) to form the vector equation:
r(t) = (0, 1, 0) + t(5, -10, -2)