1
$\begingroup$

Let $v= -3 i-10j$. I found the dot product to be $109$ and the magnitude to be $\sqrt{11881}$. I divided it out and it came out to be one. I don't know what I'm doing wrong. Please help.

$\endgroup$
2
  • 2
    $\begingroup$ It seems that you computed the dot-product of $\vec{v}$ with itself, getting the square of the magnitude of $\vec{v}$. For the direction angle with the $x$-axis try the dot-product of $\vec{v}$ with $\vec{i}$, then divide it by the product of the magnitudes of $\vec{v}$ and $\vec{i}$. $\endgroup$
    – LeoTheKub
    Mar 6, 2014 at 21:53
  • $\begingroup$ There are various ways to find direction angles of vectors. What kind of tools do you have? What is your definition of direction angle? $\endgroup$
    – David K
    Sep 1, 2014 at 0:54

3 Answers 3

1
$\begingroup$

If $\vec{v}=a\vec{i}+b\vec{j}$, then changing to polar coordinates gives $r=\sqrt{a^2+b^2}$ for the magnitude and $\theta=\arctan(b/a)$ is the direction angle.

A quick internet search yields many documents instructing how to find the direction angle of a vector.

$\endgroup$
0
$\begingroup$

A couple of things: (a) Isn't the dot-product an operation defined between 2 different vectors? If so, it's not clear in your example what the 2nd vector is. (b) Have you graphed this vector with respect to the origin on an XY-graph? After doing that, it should be a matter of employing well-known trigonometric rules to find out the "angle of the vector".

$\endgroup$
0
$\begingroup$

Dot product of two vectors $\vec{X}= ai+bj $ and $\vec{Y}= ci+dj $ is precisely,

$\vec{X}\cdot\vec{Y} = ac+bd $

Dot product is not defined for one vector. You need to have 2 vectors (same or different, that doesn't matter ) to define their dot product.

And magnitude of a vector $\vec{X}= ai+bj$ is equal to $\sqrt{a^2+b^2}$

So, I think you've calculated dot product of $\vec{v}= -3i-10j $ with itself. Only that way you can get $109$ .

And $|\vec{v}| =\sqrt{(-3)^2+(-10)^2} = \sqrt{100+9}=\sqrt{109}$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .