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Find the terminal point Q of a nonzero vector u = −−−→ P Q u=PQ→ with initial point P(−1, 3, −5) and such that a. u has the same direction as v = (6, 7, −3). b. u is oppositely directed to v = (6, 7, −3).

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okpalawalter8

Answer:

a

the terminal point Q of a nonzero vector u is

             [tex]Q = (5,4,-8)[/tex]

b

  Hence the terminal point Q of a nonzero vector u is

              [tex]Q = (-7,-10,-2)[/tex]    

Explanation:

From the question we are given that

           The initial point is  [tex]P(-1,3,-5)[/tex]

So the vector u can be expressed like this

        [tex]u =(x+1,y-3,z+5)[/tex]

From the question we are told that u has the same direction as v = (6,7,-3)

         Hence

                       [tex](x+1,y-3,z+5) = (6,7,-3)[/tex]

  => [tex]x+1 =6[/tex]

                 [tex]x=5[/tex]

=>   [tex]y-3 = 7[/tex]

             [tex]y =4[/tex]

=>  [tex]z+5 = -3[/tex]

              [tex]z =-8[/tex]

Hence the terminal point Q of a nonzero vector u is

             [tex]Q = (5,4,-8)[/tex]

 In the case that u is moving in the opposite direction of [tex]v(6,7,-3)[/tex]

         This means that

                                [tex](x+1,y-3,z+5) = (-6,-7,3)[/tex]

       => [tex]x+1 =-6[/tex]

                 [tex]x = -7[/tex]

    =>   [tex]y-3 = -7[/tex]

                   [tex]y =-10[/tex]

  =>   [tex]z+5 = 3[/tex]

               [tex]z =-2[/tex]

      Hence the terminal point Q of a nonzero vector u is

              [tex]Q = (-7,-10,-2)[/tex]    

                   

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