3d vectors. Does a point lie on the line?

AI Thread Summary
To determine if the point (16, -3, -2) lies on the line defined by the vector equation r = 2i + 3j - 5k + t(4i - 2j + k), the point can be expressed as 16i - 3j - 2k. The line can be represented in parametric form as x = 2 + 4t, y = 3 - 2t, and z = -5 + t. By substituting values for t, one can check if the coordinates match those of the point in question. The method involves solving the equations for t and checking for a consistent solution across all three dimensions. This approach will confirm whether the point lies on the line or not.
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Homework Statement



L1 has the vector equation r=2i+3j-5k + t(4i-2j+k)
Does the point (16,-3,-2) lie on the line?

Homework Equations



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The Attempt at a Solution



I have a simple(i think) vector question but i just don't know the method to work it out.


I have searched on the internet and through my books and i can't find a similar problem anywhere. Any help would be much appreciated.

Thanks
 
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The point (16, -3, -2) can also be written as 16i - 3j - 2k. Does that help? Try writing the line in the similar form r = Ai + Bj + Ck where A, B, and C may be functions of t.
 
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Conversely, the vector equation for the line r=2i+3j-5k + t(4i-2j+k) is the same as the parametric equation x= 2+ 4t, y= 3- 2t, z= -5+ k.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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