Question

Find the angle between a and b. Given that a= 6i - 2j -3k and b= i+j+k

Answer 1

3d hvn't learnt yet

Answer 2

arccos( a dot b) / ||a|| * ||b||

arcos(6*1 - 2*1 + 3*1/sqrt(36+4+9)*sqrt(3))
= arccos(7/(7*sqrt(3)))
= arccos(1/sqrt(3))
= 54.7

Answer 3

\[\cos(\theta)=\frac{a \cdot b}{|a| |b|} => \theta=\cos^{-1}(\frac{a \cdot b}{|a||b|})\]

Answer 4

so we need to find a dot b
\[a \cdot b=6(1)+(-2)(1)+(-3)(1)=-7\]

Answer 5

is equal to or greater than theta?

Answer 6

we need to find |a|
\[|a|=\sqrt{6^2+(-2)^2+(-3)^2}=\sqrt{36+4+9}=\sqrt{49}=7\]
\[|b|=\sqrt{1^2+1^2+1^2}=\sqrt{3}\]

Answer 7

\[\theta=\cos^{-1}( \frac{-7}{7 \sqrt{3}})=\cos^{-1}(\frac{-1}{\sqrt{3}})=\cos^{-1}(-\frac{\sqrt{3}}{3})\]

Answer 8

this is the angle created by the vectors a and b
so =
unless you use an approximation
and depending on your approximation theta could be greater than or less than your approximation

Answer 9

curvey vector lol

Answer 10

lol

Answer 11

i guess i could have use that draw a straight line tool