Consider the planes 3x+5y+2z=1 and 3x+2z=0,(a)find the unique point P on the y-axis which is on both planes.(b)find a unit vector v with positive first coordinate that is a parallel to both planes.(c)use parts(a) and (b) to find a vector form(with parameter t)for the line of intersectionof the two planes.

(a) Find the unique point P on the y-axis which is on both planes.

On the y-axis, x = 0 and z = 0. Substituting these values into the equations of the planes gives:

For the first plane, 3x + 5y + 2z = 1 becomes 5y = 1, so y = 1/5. For the second plane, 3x + 2z = 0 is already satisfied because x = 0 and z = 0.

So, the unique point P on the y-axis which is on both planes is P(0, 1/5, 0).

(b) Find a unit vector v with positive first coordinate that is a parallel to both planes.

The direction of the line of intersection of the planes is given by the cross product of the normal vectors of the planes. The normal vector of the first plane is (3, 5, 2) and of the second plane is (3, 0, 2).

The cross product of these vectors is (52 - 20, 23 - 23, 30 - 53) = (10, 0, -15).

To make this a unit vector, we divide by its magnitude, which is sqrt(10^2 + 0^2 + (-15)^2) = sqrt(325).

So, the unit vector v with positive first coordinate that is parallel to both planes is (10/sqrt(325), 0, -15/sqrt(325)).

(c) Use parts (a) and (b) to find a vector form (with parameter t) for the line of intersection of the two planes.

The line of intersection of the two planes passes through the point P and has direction vector v. So, its vector form is given by r = P + tv.

Substituting the values we found, we get r = (0, 1/5, 0) + t(10/sqrt(325), 0, -15/sqrt(325)).

This simplifies to r = (10t/sqrt(325), 1/5, -15t/sqrt(325)).

So, the vector form (with parameter t) for the line of intersection of the two planes is r = (10t/sqrt(325), 1/5, -15t/sqrt(325)).