[SOLVED] triangle problem using vectors

[Dx]

Hi!
question: Two sides of a triangle are formed by vectors i - 4j-k and -2i - j+k. The area is ?

A=1/2bh so I know to multiply the two vectors as such 1/2(vector1 x vector2) But what do I substitute foe i, j and k? Its not given in the problem.
so far...
1/2(-2i^2-ij+ik+8ji+4j^2-4jk+2ki+kj-k^2)
so what do I substitute foe i, j and k?
Dx [;)]

[StephenPrivitera]

You shouldn't concern yourself with i's and j's for an area problem. A= (1/2)(|v_1|)(|v_2|)sinO. So find the magnitudes of the vectors and find the angle between them.

[HallsofIvy]

How in the world could you be doing a problem like this if you don't know what i,j, i are?

It's not a matter of "what to substitute for i, j, k". They are not numbers. i is the unit vector in the x direction, j is the unit vector in the y direction, k is the unit vector in the z direction.

Also, you do not multiply vectors the way you seem to be trying.

Here, "u x v" is the cross product. It can be defined as "the vector whose length is |u||v|sin(theta) (where theta is the angle between the two vectors) and whose direction is perpendicular to both u and v in the "right hand rule" sense.

It can also be calculated as a determinant:

| i j k|
| 1 -4 -1|
|-2 -1 1|
which equals i((-4)(1)-(-1)(-1))- j(1(1)-(-1)(-2)+ k(1(-1)-(-2)(-4))
or -5i+ j- 9 k. It's length is [sqrt](25+ 1+ 91)= [sqrt](117)
Half of that is the area of the triangle.