# Which value of k will make the plane r=(k,3k,12) + s(2,2,3) + t(-1,1,3), pass through the origin?

The answer is supposedly 2. How do you get this?

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• KFr
Lv 4
9 years ago

You have a linear combination of 3 vectors, which make up your plane. In order to the plane to pass through the origin, it must exist a linear combination of your vectors that gives the zero vector (0,0,0).

You see that the third coordinate of the first vector is 12. In order to have zero as final result, you see that it works if you put s=-2 and t=-2, straight away, so that 12-6-6=0. The third coordinate is already zero.

Doing the same with the other coordinates, you get

3k-4-2=0 and thus k=2

and

k-4+2=0 and thus k=2.

If you prefer, more rigorously, you can create a system of equations with 3 variables k, s and t and the 3 equations that result from putting the result of the sum equal to zero for each coordinate.

There is a solution to this particular case and this solution is 2.

• Anonymous
9 years ago

doh doh doh doh doh