Question

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

Answer 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)
= (2-4+2 , 6-4-2 , 12-6-6 )
= (0,0,0)

yeah!!!