Find the vector equation for a line that passes through $P$ and intersects $L$












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I just need to know if my answer is right. If it isn't please tell me what the answer is and what I did wrong.



Question:




Let $L$ be the line with parametric equations
begin{align*}
x & = 5−t
\ y &= −10+2t
\ z & = −4−2t
end{align*}
Find the vector equation for a line that passes through the point $P=(3, −8, −5)$ and intersects L at a point that is distance 5 from the point $Q=(5, −10, −4)$. Note that there are two possible correct answers. Use the square root symbol '$sqrt{}$' where needed to give an exact value for your answer.




⟨x,y,z⟩=(?,?,?)+t⟨?,?,?⟩



My solution:
begin{align*}
a & = 5 - t
\ b & = -10 + 2t
\ c & = -4 -2t
\ (a-5)^2 + (b + 10)^2 + (c + 4)^2 & = 25
\ t^2 + 4t^2 + 4t^2 &= 25
\ t &= pm 5/3
\ a & = 5 pm 5/3
\ b & = -10 pm 10/3
\ c & = -4 pm 10/3
end{align*}



Answer: $langle x,y,zrangle = (3,-8,-5) + tlangle 2 pm 5/3, 2 pm 10/3, 1 pm 10/3rangle$






linear-algebra vector-spaces vectors linear-transformations vector-analysis







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    0












    $begingroup$


    I just need to know if my answer is right. If it isn't please tell me what the answer is and what I did wrong.



    Question:




    Let $L$ be the line with parametric equations
    begin{align*}
    x & = 5−t
    \ y &= −10+2t
    \ z & = −4−2t
    end{align*}
    Find the vector equation for a line that passes through the point $P=(3, −8, −5)$ and intersects L at a point that is distance 5 from the point $Q=(5, −10, −4)$. Note that there are two possible correct answers. Use the square root symbol '$sqrt{}$' where needed to give an exact value for your answer.




    ⟨x,y,z⟩=(?,?,?)+t⟨?,?,?⟩



    My solution:
    begin{align*}
    a & = 5 - t
    \ b & = -10 + 2t
    \ c & = -4 -2t
    \ (a-5)^2 + (b + 10)^2 + (c + 4)^2 & = 25
    \ t^2 + 4t^2 + 4t^2 &= 25
    \ t &= pm 5/3
    \ a & = 5 pm 5/3
    \ b & = -10 pm 10/3
    \ c & = -4 pm 10/3
    end{align*}



    Answer: $langle x,y,zrangle = (3,-8,-5) + tlangle 2 pm 5/3, 2 pm 10/3, 1 pm 10/3rangle$






    linear-algebra vector-spaces vectors linear-transformations vector-analysis







    share|cite|improve this question











    $endgroup$















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      0





      $begingroup$


      I just need to know if my answer is right. If it isn't please tell me what the answer is and what I did wrong.



      Question:




      Let $L$ be the line with parametric equations
      begin{align*}
      x & = 5−t
      \ y &= −10+2t
      \ z & = −4−2t
      end{align*}
      Find the vector equation for a line that passes through the point $P=(3, −8, −5)$ and intersects L at a point that is distance 5 from the point $Q=(5, −10, −4)$. Note that there are two possible correct answers. Use the square root symbol '$sqrt{}$' where needed to give an exact value for your answer.




      ⟨x,y,z⟩=(?,?,?)+t⟨?,?,?⟩



      My solution:
      begin{align*}
      a & = 5 - t
      \ b & = -10 + 2t
      \ c & = -4 -2t
      \ (a-5)^2 + (b + 10)^2 + (c + 4)^2 & = 25
      \ t^2 + 4t^2 + 4t^2 &= 25
      \ t &= pm 5/3
      \ a & = 5 pm 5/3
      \ b & = -10 pm 10/3
      \ c & = -4 pm 10/3
      end{align*}



      Answer: $langle x,y,zrangle = (3,-8,-5) + tlangle 2 pm 5/3, 2 pm 10/3, 1 pm 10/3rangle$






      linear-algebra vector-spaces vectors linear-transformations vector-analysis







      share|cite|improve this question











      $endgroup$




      I just need to know if my answer is right. If it isn't please tell me what the answer is and what I did wrong.



      Question:




      Let $L$ be the line with parametric equations
      begin{align*}
      x & = 5−t
      \ y &= −10+2t
      \ z & = −4−2t
      end{align*}
      Find the vector equation for a line that passes through the point $P=(3, −8, −5)$ and intersects L at a point that is distance 5 from the point $Q=(5, −10, −4)$. Note that there are two possible correct answers. Use the square root symbol '$sqrt{}$' where needed to give an exact value for your answer.




      ⟨x,y,z⟩=(?,?,?)+t⟨?,?,?⟩



      My solution:
      begin{align*}
      a & = 5 - t
      \ b & = -10 + 2t
      \ c & = -4 -2t
      \ (a-5)^2 + (b + 10)^2 + (c + 4)^2 & = 25
      \ t^2 + 4t^2 + 4t^2 &= 25
      \ t &= pm 5/3
      \ a & = 5 pm 5/3
      \ b & = -10 pm 10/3
      \ c & = -4 pm 10/3
      end{align*}



      Answer: $langle x,y,zrangle = (3,-8,-5) + tlangle 2 pm 5/3, 2 pm 10/3, 1 pm 10/3rangle$






      linear-algebra vector-spaces vectors linear-transformations vector-analysis




      linear-algebra vector-spaces vectors linear-transformations vector-analysis






      share|cite|improve this question















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      edited Jan 16 at 21:37







      DreamVision2017

















      asked Jan 16 at 20:30









      DreamVision2017DreamVision2017

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          Your solution is correct. However, you may add a bit of explanation, e.g. "Let $(a,b,c)$ be the intersection point on $L$ with distance $5$ to $Q$. Then one has..."






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            Your solution is correct. However, you may add a bit of explanation, e.g. "Let $(a,b,c)$ be the intersection point on $L$ with distance $5$ to $Q$. Then one has..."






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              Your solution is correct. However, you may add a bit of explanation, e.g. "Let $(a,b,c)$ be the intersection point on $L$ with distance $5$ to $Q$. Then one has..."






              share|cite|improve this answer









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                Your solution is correct. However, you may add a bit of explanation, e.g. "Let $(a,b,c)$ be the intersection point on $L$ with distance $5$ to $Q$. Then one has..."






                share|cite|improve this answer









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                Your solution is correct. However, you may add a bit of explanation, e.g. "Let $(a,b,c)$ be the intersection point on $L$ with distance $5$ to $Q$. Then one has..."







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                answered Jan 22 at 12:55









                bruderjakob17bruderjakob17

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