# Thread: planes, intersections and angles

1. ## planes, intersections and angles

hey... i have worked out the cross product vector and i get (-1, 3, -3)

and the vector equation of the line that i got is (x,y,z)= (1,2,-1) + t(-1,3,-3)
which i think is correct....

am i correct??? and if so or not..how do i go on about working out question c and d??

thanks

2. Originally Posted by b00yeah05

hey... i have worked out the cross product vector and i get (-1, 3, -3)

and the vector equation of the line that i got is (x,y,z)= (1,2,-1) + t(-1,3,-3)
which i think is correct....

am i correct??? and if so or not..how do i go on about working out question c and d??

thanks
Hi,

1. I've got as the result of the cross product: [-1, -3, -7]

2. Therefore the equation of the common line would be: [x,y,z] = [1,2,-1] + t[-1, -3, -7]

3. The angle between 2 planes is equal to the angle between the normal vectors of the planes. Use the the definition of the dot product to calculate the angle.
If the angle between the planes is called $\alpha$ then I've got:

$\cos(\alpha) = \frac{[1,2,-1] \cdot [2,-3,1]}{\sqrt{6} \cdot \sqrt{14}}~\implies~ \alpha \approx 123.06^\circ$

3. hey..i made a mistake on the cross product so i got -3 but i fixed it and now have -7 also...but does ur working answer the question to part c??? as it says use this information and answer from part b to find vector equation to find line of intersection of plane 1 and plane 2??? why have you only used one plane??? in this case plane 2??

4. Originally Posted by b00yeah05
hey..i made a mistake on the cross product so i got -3 but i fixed it and now have -7 also...but does ur working answer the question to part c??? as it says use this information and answer from part b to find vector equation to find line of intersection of plane 1 and plane 2??? why have you only used one plane??? in this case plane 2??
1. The cross product of the 2 normal vector will yield the direction vector of the line of intersection. That means we know the direction of the line.

2. If the point $x_0$ lies on both planes it must lie on the line of intersection. With one point and the direction vector the line is determined completely.

3.
why have you only used one plane??? in this case plane 2??
I don't understand what you mean here.

5. yeah..sorry..i was confused a bit myself after i read that, but never the less..i understand what you did and i tried doing it again and i ended up with teh same answer..so thanks