+105
Watching videos on khan academy..hard to follow his length of a vector video. so need a bit of help here.
u= (1, 2, 3) v= (-2, 0, 1)
Find the legnth of u & v
i got ||u||= √1²+2²+3² =3.741
||v|| = -√2²+0²+1² = -2.236
3.741+ -2.236 = ||u v|| = 1.51
am i doing this right?
+33661 Length of v is √( (-2)2 + 02 + 12) = √(4+0+1) = √5
$${\sqrt{{\mathtt{5}}}} = {\mathtt{2.236\: \!067\: \!977\: \!499\: \!789\: \!7}}$$
If the last part is meant to be the length of the vector sum of u and v, then you have to sum the components first, before calculating the length, not add the lengths of the individual vectors together.
u+v = (-1, 2, 4)
length of u+v = √( (-1)2 + 22 + 42) = √21
$${\sqrt{{\mathtt{21}}}} = {\mathtt{4.582\: \!575\: \!694\: \!955\: \!84}}$$
+26393 u= (1, 2, 3)
v= (-2, 0, 1)
$$\vec{w}=\vec{u}+\vec{v}
=
\left(
\begin{array}{c}
1 & 2 & 3
\end{array}
\right)
+\left(
\begin{array}{c}
-2 & 0 & 1
\end{array}
\right)
=
\left(
\begin{array}{c}
1-2 & 2+0 & 3+1
\end{array}
\right)
=
\left(
\begin{array}{c}
-1 & 2 & 4
\end{array}
\right)$$
$$\begin{array}{rcl}
\|w\| &=& \sqrt{(-1)^2+2^2+4^2} \\\\
&=&\sqrt{1+4+16}\\\\
&=&\sqrt{21}\\\\
& \approx & 4.58257569496
\end{array}$$
