lncognlto
  • lncognlto
There are two vectors, u and v, which I will write below:
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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lncognlto
  • lncognlto
u = (p^2, -2, 6). v = (2, p - 1, 2p + 1)
lncognlto
  • lncognlto
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.
amistre64
  • amistre64
stack them, multiply down the columns, and add the resulting row

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amistre64
  • amistre64
(a,b,c) (p,q,r) --------- ap+bq+cr if that is equal to 0, they are perps
amistre64
  • amistre64
and its the inner product, not the scalar product
amistre64
  • amistre64
unless im nor recalling if they are synonomous names :)
amistre64
  • amistre64
u = (p^2, -2 , 6 ) v = ( 2 , p - 1, 2p + 1) ----------------------- 2p^2 -2p+2 +12p+6 = 0
lncognlto
  • lncognlto
Of course yes, thanks, I got it. Its working now. Then p^2 + 5p + 4 = 0, therefore p = -1 or -4. Right, thanks.
amistre64
  • amistre64
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

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