Calculus
posted by Jenny on .
Which value of k will make the plane r=(k,3k,12) +s(2,2,3)+t(1,1,3), s,teR passing through the origin?
a) 2
b) 1
c)0
d) 1

to pass through the origin, r = (0,0,0)
so
k + 2s  t = 0
3k + 2s + t = 0
12 + 3s + 3t = 0 > 4 + s + t = 0
add 1st and 2nd > 4k + 4s = 0 or k = s
add 1st and 3rd > k+4 + 3s = 0
sub in k=s
s + 4 + 3s = 0
s =  2
then k = 2
check:
if k= 2 and s =  2, then in 3rd
4  2 + t = 0
t = 2
r = (2, 6, 12) 2(2,2,3) 2(1,1,3)
= (24+2 , 642 , 1266 )
= (0,0,0)
yeah!!!