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 $\displaystyle \alpha$ then I've got:
$\displaystyle \cos(\alpha) = \frac{[1,2,-1] \cdot [2,-3,1]}{\sqrt{6} \cdot \sqrt{14}}~\implies~ \alpha \approx 123.06^\circ$
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 $\displaystyle x_0$ 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.I don't understand what you mean here.why have you only used one plane??? in this case plane 2??