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I gave two lines $q_1 = 2x - y + 2 = 0$ and $q_2 = x + 2y - 3 = 0$.

They have vectors $n_1 = (2,1)$ and $n_2 = (1,2)$.

When I have to find angle between them I must apply that formula:

$$\cos \theta = \frac{2\cdot 1 + (-1)\cdot2}{\sqrt{4 + 1}\sqrt{1 + 4}} = 0 \Rightarrow \theta = 90^\circ$$

Replacing $n_1$ and $n_2$ is clear to me, but can you please explain how I get $90^\circ$ from solving the formula. Thank you.

Where those $\sqrt{4 + 1}\sqrt{1 + 4}$ comes from?

If $ \theta $ is $90^ \circ$ at $0$, how much degrees it will be if it is $\dfrac{1}{5}$?

2 Answers 2

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Let $\theta$ be an angle.

$\cos \theta = \frac{2.1 + (-1).2}{\sqrt 5. \sqrt 5}$

$\cos \theta = \frac{2 - 2}{\sqrt 5. \sqrt 5}$

$\cos \theta = \frac{0}{5}$

$\cos \theta = 0$

$\cos \theta = \cos 90°$

$\theta = 90°$

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    Comments are not for extended discussion; this conversation has been [moved to chat](http://chat.stackexchange.com/rooms/52693/discussion-on-answer-by-kanwaljit-singh-angle-between-two-lines-exmpanation). – 2017-01-29
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    @Jyrki Lahtonen sorry sir I forget. But from next time I take care of that. – 2017-01-29
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Given two vectors in $\mathbb{R}^2$, say $(x_1, y_1)$ and $(x_2, y_2)$, we have the formula $$\cos \theta = \frac{(x_1, y_1) \cdot (x_2, y_2)}{\|(x_1, y_1)\| \|(x_2,y_2)\|}.$$ Some clarification about the used notation: $\cdot$ denotes the inproduct of two vectors. It is defined as follows: $$(x_1,y_1) \cdot (x_2,y_2) = x_1x_2 + y_1y_2.$$ The norm of the vector $(x_1,y_1)$ is defined as $$\|(x_1, y_1\| = \sqrt{(x_1,y_1) \cdot (x_1,y_1)} = \sqrt{x_1^2 + y_1^2}.$$ This is the length of the vector and if you would draw the vector in the plane and draw the projections to the x and y axis, you'll see that this just is Pythagoras' theorem.

You know that if $\cos \theta = 0$, then we have that $\theta = 90^{\circ}$. The square roots come from the formula I gave you (and you probably find in your coursenotes). If you want to find $\theta$ where $\cos \theta$ = 1/5, you just compute the arccosine of 1/5.

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    from $n1$$(3,1)$ and $n2$$(4,-2)$ I get $ \frac{10}{\sqrt10\sqrt20}$. How is that equals to $\frac{\pi}{4}$? – 2017-01-29
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    note that $\sqrt{20} = \sqrt{2} \sqrt{10}$, which makes the denominator equal to $\sqrt{10} \sqrt{10} \sqrt{2} = 10 \sqrt{2}$. Therefore the 10 in the numerator and the 10 in the denominator cancel out, so we find $\cos \theta = \frac{1}{\sqrt{2}}$. This implies that $\theta = \pi / 4$. – 2017-01-29