gerardoblk7653 gerardoblk7653
  • 19-08-2019
  • Mathematics
contestada

Find a vector that has the opposite direction of u = (1,-2,4), but which has length square root of 3.

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LammettHash
LammettHash LammettHash
  • 19-08-2019

Normalize [tex]\vec u[/tex] by dividing [tex]\vec u[/tex] by its magnitude:

[tex]\dfrac{\vec u}{\|\vec u\|}=\dfrac{(1,-2,4)}{\sqrt{1^2+(-2)^2+4^2}}=\dfrac{(1,-2,4)}{\sqrt{21}}[/tex]

Multiply by -1 to reverse its direction, and again by [tex]\sqrt3[/tex] to ensure this new vector has the magnitude we want. Notice that [tex]\sqrt{\dfrac3{21}}=\dfrac1{\sqrt7}[/tex].

[tex]-\sqrt3\dfrac{\vec u}{\|\vec u\|}=\dfrac{(-1,2,-4)}{\sqrt7}[/tex]

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