Question

There are two vectors, u and v, which I will write below:

Answer 1

u = (p^2, -2, 6).
v = (2, p - 1, 2p + 1)

Answer 2

Given that p is a constant, what are its values when u is perpendicular to v. Now I know the scalar product of u and v must equal 0, and then I must complete the quadratic equation which I will have of p to get the answers. But what I don't know is the first steps I must take to work out the scalar product.

Answer 3

stack them, multiply down the columns, and add the resulting row

Answer 4

(a,b,c)
(p,q,r)
---------
ap+bq+cr

if that is equal to 0, they are perps

Answer 5

and its the inner product, not the scalar product

Answer 6

unless im nor recalling if they are synonomous names :)

Answer 7

u = (p^2, -2 , 6 )
v = ( 2 , p - 1, 2p + 1)
-----------------------
2p^2 -2p+2 +12p+6 = 0

Answer 8

Of course yes, thanks, I got it. Its working now. Then p^2 + 5p + 4 = 0, therefore p = -1 or -4. Right, thanks.

Answer 9

2p^2 -2p+2 +12p+6 = 0

p^2 -p+1 +6p+3 = 0

p^2 +5p +4 = 0

4 and 1 look appropriate, so -4 and -1 are fine