Thread: Vectors!

HI I want to check if my answers are right here, I can't do the last part so I believe my answers in the previous parts must be wrong..

Points A,B,C,D have the following coordinates
A1,3,1) B1,2,4) C2,3,6) D5,-2,1)

ai.) Evaluate the vector product, giving your answers in terms of the unit vectors i,j,k.

I found AB and AC and got -5i, 3j and 1k from doing the cross product.

aii.) Find the area of the triangle ABC.

I took the magnitude of the cross product result in ai.) and halved it, getting (1/2)sqrt(35)units^2

b.) The plane containing the points A,B, C is denoted by pi and the line passing through D perpendicular to pi is denoted by L. The point of intersection of L and pi is denoted by P.

bi.) FInd the cartesian equation of pi.

AB is the direction vector of pi, I thought. And then I'll need a point so
I got x=1, y=3-t, z=1+3t

bii.) FInd the cartesian equation of L.

The result of the cross product in ai.) is the direction vector of L, I thought. Dotting it by the direction vector of pi, I will get a zero.
So my answer was
x=5-5t, y=-2+3t, z=1+t

c.) Find the coordinates of P.

This is where I know my previous answers are wrong. I though to equate all the x, y and z values of L and pi. I get different values for t each time...

Please help me out!

Thank you!

Did you get a satisfactory answer to your question? I can help you on this - but I'm not going to if you already have a solution. The post has been here a while and I just joined the site.

Originally Posted by Tutu

HI I want to check if my answers are right here, I can't do the last part so I believe my answers in the previous parts must be wrong..

Points A,B,C,D have the following coordinates
A1,3,1) B1,2,4) C2,3,6) D5,-2,1)

ai.) Evaluate the vector product, giving your answers in terms of the unit vectors i,j,k.

I found AB and AC and got -5i, 3j and 1k from doing the cross product.

aii.) Find the area of the triangle ABC.

I took the magnitude of the cross product result in ai.) and halved it, getting (1/2)sqrt(35)units^2

Yes, this is correct.

b.) The plane containing the points A,B, C is denoted by pi and the line passing through D perpendicular to pi is denoted by L. The point of intersection of L and pi is denoted by P.

bi.) FInd the cartesian equation of pi.

AB is the direction vector of pi, I thought. And then I'll need a point so
I got x=1, y=3-t, z=1+3t

I don't undertand what you are saying here. You were asked for an equation of the plane, pi. This is the parametric equations for a line, not a plane.

bii.) FInd the cartesian equation of L.

The result of the cross product in ai.) is the direction vector of L, I thought. Dotting it by the direction vector of pi, I will get a zero.
So my answer was
x=5-5t, y=-2+3t, z=1+t

Yes, this is correct.

[qote]c.) Find the coordinates of P.

This is where I know my previous answers are wrong. I though to equate all the x, y and z values of L and pi. I get different values for t each time...

Please help me out!

Thank you![/QUOTE]
The equation of a plane, with normal vector <A, B, C> and containing point [tex](x_0, y_0, z_0)[tex] is .
You can use that to answer (bi) and then replace x, y, and z in the equation of the plane with the x, y, and z from (bii) to get a single equation to solve for t.