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What is the angle between #<-1,8,2 ># and #< 0,-3,1>#?

Answer 1

Asher Friedman

Approximately
#2.563color(white)(l)"rad"#

or equivalently,
#147^sf("o")#.

Let #theta# be the angle between the two vectors . By the geometric definition of scalar products,

#"Angle between vector " bb(a)" and "bb(b)=arccos ((bb(a)*bb(b))/(abs(bb(a))*abs(bb(b))))#

Where #bb(a)*bb(b)# the dot product of #bb(a)# and #bb(b)#, and #abs(bb(a))*abs(bb(b))# the product of their magnitude.

Let #bb(a)=<-1,8,2># and #bb(b)=<0,-3,1>#,

#theta=arccos((<-1,8,2>*<0,-3,1>)/(sqrt(1+8^2+2^2)*sqrt(3^2+1^2)))# #color(white)(theta)=arccos((-24+2)/(sqrt(69)*sqrt(10)))# #color(white)(theta)=arccos(-22/sqrt(690))# #color(white)(theta)~~2.563color(white)(l)"rad"=147^sf("o")#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

What is the angle between #&lt;-1,8,2 &gt;# and #&lt; 0,-3,1&gt;#?

What is the angle between #<-1,8,2 ># and #< 0,-3,1>#?