Anyone can help me with Vector Eqn? thanks?

Question

1. Find the cartesian equation of the line of intersection of the two planes 2x - 3y - z = 1 and 1 and 3x + 4y + 2z = 3


2. The four points A,B,C and D have position vectors ( 4 2 -1) (1 2 1 ) (-3 0 3) and (5 -4 1) respectively.

The perpendicular from D to the containing A,B and C meets the plane at E. Find

(a) the scalar product vector equation of the plane containing A,B and C.

(b) the vector equation of the straight line through D and E.
(c) the position vector of the point E.

3. show that the line with vector equation r = (6 -5 1) + (1 -2 3) is perpendicular to the plane with vector equation r. (1 -2 3) = -9. Find the position vector of the point of intersection of the line and plane and the distance from the point with position vector (1 1 -11) to this point of intersection.

My ans for this is 7units can anyone help me with working thanks.

Answer 1

2x-3y-z = 1
3x + 4y + 2z = 3
4x - 6y - 2z = 2.........(* 1st eq by 2)
add 2 and 3 to eliminate z
7x - 2y = 5...........hence x = (5+2y)/7
z = 2x - 3y - 1 = (3-17y)/7
hence the cartesian equation is
(7x-5)/2 = y = (7z-3)/-17
to be continued 2 and 3
I am going out for a couple of hours, so I will work on it when I get back, unless someone beats me to it.

2) BA = (3, 0, -2)...............CA = (7, 2, -4)
the normal for the plane is BA * CA = -2i + j - 3k
equation of plane is r. (-2, 1, -3) = -3
(b) since the line DE is perpendicular to the plane, the normal of the plane is a direction vector for the line and the equation is
r = (5, -4, 1) + t ( -2, 1, 3)
(c) plug the line into plane to find point of intersection
-2( 5-2t) + (-4 + t) - 3(1 + 3t) = -3
-10 + 4t - 4 + t - 3 - 9t = -3
t = -7/2
E( 12, -7.5, -19/2)
3) The direction vector of the line is (1, -2, 3) which is the same as the normal of the plane, hence they are perpendicular.
(6 + t, -5 -2t, 1 + 3t) . (1, -2, -3) = -9
solve for t , you get 11/2
plug t into equation of line to find co-ordinates of point of intersection (11.5, -16, 17.5)
Use the distance formula to find distance between points
sqrt (10.5^2 + 17^2 + 28.52)
I do not get your answer so check my calculations, but the methodology is correct.