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Section 12.2 : Equations of Lines

3. Is the line through the points \(\left( {2,0,9} \right)\) and \(\left( { - 4,1, - 5} \right)\) parallel, orthogonal or neither to the line given by \(\vec r\left( t \right) = \left\langle {5,1 - 9t, - 8 - 4t} \right\rangle \)?

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Start Solution

Let’s start this off simply by getting vectors parallel to each of the lines.

For the line through the points \(\left( {2,0,9} \right)\) and \(\left( { - 4,1, - 5} \right)\) we know that the vector between these two points will lie on the line and hence be parallel to the line. This vector is,

\[{\vec v_1} = \left\langle {6, - 1,14} \right\rangle \]

For the second line the coefficients of the \(t\)’s are the components of the parallel vector so this vector is,

\[{\vec v_2} = \left\langle {0, - 9, - 4} \right\rangle \] Show Step 2

Now, from the first components of these vectors it is hopefully clear that they are not scalar multiples. There is no number we can multiply to zero to get 6.

Likewise, we can only multiply 6 by zero to get 0. However, if we multiply the first vector by zero all the components would be zero and that is clearly not the case.

Therefore, they are not scalar multiples and so these two vectors are not parallel. This also in turn means that the two lines can’t possibly be parallel either (since each vector is parallel to its respective line).

Show Step 3


\[{\vec v_1}\centerdot {\vec v_2} = - 47\]

The dot product is not zero and so these vectors aren’t orthogonal. Because the two vectors are parallel to their respective lines this also means that the two lines are not orthogonal.