the vectors are orthogonal if the result is 0 after using the dot product. \[u \cdot v =
If vectors are parallel, then there exists a scalar \(c\) such that\[ \mathbf u = c \mathbf v \]
\[u \cdot v =
for dot product
but if it's neither than the dot product can't be 0 and a scalar c doesn't exist.
do i just plug in the numbers to solve?
you could.. since we have \[u=
, v= \]
and we can use dot product
\[u \cdot v = \]
u= <6,-2> and v = <8,24> \[u_1=6,u_2=-2,v_1=8,v_2=24 \] now we plug those values into the dot product formula \[u \cdot v = u_1v_1+u_2v_2\]
\[u \cdot v = (6)(8)+(-2)(24) \]
so what is 6 x 8 and what is -2 x 24 ?
6x8=48 -2x24= -48
mhm \[u \cdot v = 48-48 \] so now what's 48-48?
0. so is it neither?
no.. since our dot product is 0, our vectors are _______________
orthogonal since it equal 0