Posted by **Jenny** on Monday, April 9, 2012 at 2:46am.

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

- Calculus -
**Reiny**, Monday, April 9, 2012 at 8:58am
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!!!

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