find parametric equations of the line of intersection of planes x-2y 7z=9 and 3x 5y-z=5 B)If u and v are perpendicular vectors such as lul=6 and lvl=4, find the projection of the 2u -3v on u 2v. (the letters have little right arrows on each of them.) Write clearly and BIG Solution P1= x-2y+7z=9 P2= 3x+5y-z= 5 The cross product of the normal vectors of the two planes will be the directional vector \'a\', of the line of intersection. normal vector of P1(1i-2j+7k) normal vector of P2=(3i+5j-k) vector \'a\'= -33i+22j+11k Now find a point on the line. It will be a point in both planes. Let z = 0 and solve for x and y. from P1 x-2y=9 x=9+2y from P2 3x+5y=5 substituting value of x in above equation 3(9+2y)+5y=5 11y=5-27 11y=-22 y= -2 x=9+2(-2) x=5 so point =p=(5,-2,0) equation of line L= p+at L=(5i-2i)+t( -33i+22j+11k) B) IuI=6 and IvI=4 vector a= 2u-3v vector b= u+2v projection on b is = a.b/I bI =(2u-3v).(u+2v)I u 2 +(2v) 2 I =(2u 2 -6v 2 )/36+64 =(2x36-6x16)/10 =I72-96I/10 =24/10 projection magnitude =2.4 .

1. find parametric equations of the line of intersection of planes
x-2y 7z=9
and
3x 5y-z=5
B)If u and v are perpendicular vectors such as lul=6 and lvl=4,
find the projection of the 2u -3v on u 2v.
(the letters have little right arrows on each of them.)
Write clearly and BIG
Solution
P1= x-2y+7z=9
P2= 3x+5y-z= 5
The cross product of the normal vectors of the two planes will be the directional vector 'a', of
the line of intersection.
normal vector of P1(1i-2j+7k)
normal vector of P2=(3i+5j-k)
vector 'a'= -33i+22j+11k
Now find a point on the line. It will be a point in both planes. Let z = 0 and solve for x and y.
from P1
x-2y=9
x=9+2y
from P2
3x+5y=5
substituting value of x in above equation
3(9+2y)+5y=5
11y=5-27
11y=-22
y= -2
x=9+2(-2)
x=5
so point =p=(5,-2,0)

2. equation of line L= p+at
L=(5i-2i)+t( -33i+22j+11k)
B)
IuI=6 and IvI=4
vector a= 2u-3v
vector b= u+2v
projection on b is = a.b/I bI
=(2u-3v).(u+2v)I u 2
+(2v) 2
I
=(2u 2
-6v 2
)/36+64
=(2x36-6x16)/10
=I72-96I/10
=24/10
projection magnitude =2.4