I have a question. Given are the points $a(1,3)$ and $b(-2,1)$. I have to compute $\|a\|$ and $\|a+b\|$, but I don't know how to do it. Can someone help me?
Thank you
Points in a plane
-1
$\begingroup$
vector-analysis
2 Answers
1
The magnitude of a vector $v=(x,y,z)$, also called its norm is given by $$\mid \mid v \mid \mid =\sqrt {x^2+y^2+z^2} $$ In three dimensional space.
Here $a =(1,3,0) $, so $\mid \mid a \mid \mid =\sqrt {1^2+3^2+0} = \sqrt {10} $. Hope you can take it from here and proceed likewise for the second case.
-
0Can I for the second case, just count a+b to get (-1,4) and do ||a+b|| = sqrt ((-1)^2+4^2)? – 2017-01-27
-
1@user410573 Yes, that is correct. – 2017-01-27
0
The magnitude (length) of a vector $\vec t=(t_1,t_2,\cdots,t_n)$ is given by $$||\vec t||=\sqrt{t_1^2+t_2^2+\ldots+t^2_n}$$
In this context, the vectors only have 2 component, so
$$||\vec A||=\sqrt{A_1^2+A_2^2}=\sqrt{1^2+3^2}=10$$
For the second part of question, we sum both vectors separately for every component. For instance
$$A=(a_1,a_2,a_3),B=(b_1,b_2,b_3)\implies A+B=(a_1+b_1,a_2+b_2,a_3+b_3)$$