# What is open expression of a vector's magnitude?

I'm 1st grade Physics student and my school started this week at 16.09.2019. I have seen 4.5 hours of lesson about vectors this week. We have seen such thing called Products of Vectors. My teacher told to class about Properties of Scalar Product with 6 items.

Properties of Scalar Product

1. $$\vec{A}\cdot\vec{B}=\vec{A}\cdot\vec{B}$$
2. $$\vec{A}\cdot(\vec{B}+\vec{C})=\vec{A}\cdot\vec{B}+\vec{A}\cdot\vec{C}$$
3. $$\vec{A}\cdot\vec{A}=AA\cos(\hat{\vec{A}, \vec{A}})=A^2\Rightarrow A=\sqrt{AA}$$
4. $$\vec{A}\cdot\vec{B}\lt0\Rightarrow \epsilon\gt\frac{\pi}{2}$$
5. $$\vec{A}\cdot\vec{B}=0\Rightarrow AB\cos{\theta}, \theta=(\frac{\pi}{2}+2k\pi), (\frac{3\pi}{4}+2k\pi)$$

$$\vec{A}$$ and $$\vec{B}$$ are perpendicular or one of $$\vec{A}$$ or $$\vec{B}$$ is $$0$$. 6. $$\vec{A}\cdot\vec{X}=k, k\in\Re\Rightarrow$$ What is $$\vec{X}$$? This question doesn't have only one right answer. There are many $$\vec{X}$$ vectors that ensure $$\vec{A}\cdot\vec{X}=k$$ equation.

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My first question is about 3rd item. Is the item correct?

My second question is about 4th question. Is $$\theta\gt\frac{\pi}{2}$$ or $$\frac{\pi}{2}\lt\theta\lt\pi$$ true?

And lastly we have seen such concept called Open Expression of A Vector's Magnitude.

$$\vec{A}\cdot\vec{A}=(A_x\cdot\hat{x}+A_y\cdot\hat{y}+A_z\cdot\hat{z})\cdot(A_x\cdot\hat{x}+A_y\cdot\hat{y}+A_z\cdot\hat{z})$$

$$\vec{A}\cdot\vec{A}=A_x^2+A_y^2+A_z^2$$

$$\vec{A}\cdot\vec{A}=A^2$$

$$A=\left|\vec{A}\right|=\sqrt{A_x^2+A_y^2+A_z^2}$$

Is the derivation and additional information correct? And also, what does Open Expression of A Vector's Magnitude mean?

Thanks!

• Might Mathematics be better suited for this maths question? – Kyle Kanos Oct 10 '19 at 11:31

By 3rd item do you mean $$\vec A \cdot \vec A = A^2$$? If so, yes that is correct. The dot product of 2 vectors is a scaler. The 4th equation is correct also but note that $$\theta$$ varies by $$0 \rightarrow \pi$$. I've never seen the term "open expression" used but from the equations above I think it means the obvious expressions. All the expressions look fine but #5. Again $$0 \rightarrow \pi$$.
• Yeah, I mean that $\vec{A}\cdot\vec{A}=\left|\vec{A}\right|^2$ at 3rd item. I just translated what does mean the term "açık ifade" on the title of subject in Turkish. It showed me up "open expression" but, as I check it once more with "obvious expression" it shows same as "açık ifade". Your definition must be correct. Thank you for validating the properties. I can not write things that my teacher wrote on the black board while I listening to him at the same time. So that, I can miss something and I can't be sure that I wrote correctly. Thank you again, have a good day! – ICCQBE Sep 21 '19 at 20:04