webwork questions

Can you help with this problem: Find the angle of intersection of the plane -1 x+4y+2z = -4 with the plane 4x-1y+5z=-3
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The angle of intersection of two planes is the angle between the two normal vectors, in this case <-1,4,2> and <4,-1,5>.  We have Cos(θ)=(<-1,4,2>.<4,-1,5>)/(|<-1,4,2>|*|<4,-1,5>|)
=(-4 - 4 + 10)/(√(1+16+4)√(16+1+25))=2/(√21√42)=2/(21√2)=√2/21, so θ~86.1 degrees
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Hello, so for question #8 I continue to get -11/2 for the second y value, I
have no idea why it is wrong, could you explain

Last answer:
        AnSwEr0001: 0
        AnSwEr0002: 1/2
        AnSwEr0003: 11/2
        AnSwEr0004: -4

Find the vector equation for the line of intersection of the planes 5x−4y−2z=1 and 5x + 5 z = 0 

r=<______, ______, 0>+t <-20, ______, ______> 

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Well....it's hard to tell, but it looks like you are making arithmetic errors....
First of all, we note that two planes with normal vectors that are not parallel will intersect in a line.  Here the normal vectors are <5, -4, -2> and <5, 0, 5> which are clearly not on the same line (otherwise one would have to be a constant times the other).  Our next step is to find a point
 <______, ______, 0> 
that satisfies both equations: since z=0  we have 0 = 5x + 5z = 5x +0 => x=0 too, so the first vector must look like  <0, ______, 0>. In order to make the first equation true,  now the equation 1 = 5*0-4y-2*0 = -4y must be true, so y= -1/4, and the first vector must be <0, -1/4, 0>.  To get the second vector,  we must find a vector perpendicular to both normal vectors.  Such a vector is parallel to, i.e. a scalar times, the cross product <5, -4, -2>x<5, 0, 5> = <-20, -35, 20>.  Since this vector already has the form <-20, ______, ______>, the scalar must be 1 and we're done.