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I'm trying to find the angle between the vectors

\begin{align} r_{1} &: 2j -4k + \lambda (i+k) \\ r_{2} &: 7i + k + \mu (2i-j+k) \end{align}

The equation for the angle between them is $$\cos(\theta)= \frac{r_{1} \cdot r_{2}}{\lvert r_{1} \rvert \lvert r_{2} \rvert }$$

What throws me off is that the vectors have both a position vector and a direction vector, can I use either?

Edit: fixed huge mistake with the D.V. of the 2nd line

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    $\begingroup$ Presumably these are not vectors, but lines, parametrized by $\lambda$ and $\mu$ respectively. To find the angle between the lines, use the direction vector. $\endgroup$ – Wouter Oct 22 '17 at 12:14
  • $\begingroup$ Are the two lines supposed to intersect? Based on the latest versions of the formulas, they do not. $\endgroup$ – David K Oct 22 '17 at 12:30
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at first i would write both vectors in the form $r_1$:$$\lambda\vec{i}+2\vec{j}+\vec{k}(\lambda-4)$$ $r_2$:$$\vec{i}(7+\mu)+\vec{k}(1+\mu)$$ if These vectors reprent line then we gat $$\cos(\theta)=\frac{2+1}{\sqrt{2}\sqrt{6}}$$ and we get $$\theta=30^{\circ}$$

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  • $\begingroup$ Plug these into the angle formula and you get an expression with $\lambda$ and $\mu,$ that is, the "angle" becomes a function of the two parameters. I don't think this is what is wanted. $\endgroup$ – David K Oct 22 '17 at 12:27
  • $\begingroup$ ok but so is the question $\endgroup$ – Dr. Sonnhard Graubner Oct 22 '17 at 12:29
  • $\begingroup$ ok it can be that these vectors are lines $\endgroup$ – Dr. Sonnhard Graubner Oct 22 '17 at 12:30
  • $\begingroup$ The final answer looks correct, the question is how we get there. $\endgroup$ – David K Oct 22 '17 at 12:52
  • $\begingroup$ i have used the formula $$\cos(\theta)=\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}$$ $\endgroup$ – Dr. Sonnhard Graubner Oct 22 '17 at 12:59
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I do not view either $r_1$ or $r_2$ as a "vector." Instead, each of them is a line in a three-dimensional space. In order to describe each line, we have arbitrarily chosen one point on each line (identified by the position vector of that point) and parameterized the rest of the line by a vector parallel to the line.

The choice of point for the position vector really is arbitrary: we could just as well write $$ r_1{:\ } i + 2j - 3k + \xi(i+k) $$ and it would describe exactly the same line as in the question.

We can also substitute a scalar multiple of the direction vector: $$ r_1{:\ } i + 2j - 3k + \phi(2i+2k) $$ is still the same line.

In fact you can get the same line from many different possible position vectors, but you must always use a direction vector that is some multiple of the same direction vector. The different choices of direction vector will always give the same answer from the angle formula (except that there are really usually two angles between two lines, an obtuse angle and an acute angle, and if we multiply one of the directions by a negative number we'll change which of those angles the formula gives us). The different choices of position vector will (except in special cases) give us many different answers.

So if you want the angle between two lines, use the angle between their direction vectors. Intuitively, it is the direction of a line that determines the angle with another line, not the position of the line.

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  • $\begingroup$ thank you, but aren't lines also classified as vectors or am i mistaken? $\endgroup$ – Andrew1024 Oct 22 '17 at 12:28
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    $\begingroup$ No, a line is not a vector. A line has different properties than a vector. A vector has a finite length, but a line does not. You can have two parallel lines that are different from each other, but two parallel vectors are either the same vector or one is a multiple of the other. $\endgroup$ – David K Oct 22 '17 at 12:34
  • $\begingroup$ these equations contain real parameters $\endgroup$ – Dr. Sonnhard Graubner Oct 22 '17 at 12:35
  • $\begingroup$ @Dr.SonnhardGraubner Yes, they do. That's how we can tell we're dealing with parameterized sets of something (which turn out to describe two lines) rather than just two constant vectors. $\endgroup$ – David K Oct 22 '17 at 12:40
  • $\begingroup$ ok i think the Problem is solved! $\endgroup$ – Dr. Sonnhard Graubner Oct 22 '17 at 12:41

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