If #A = #, #B = # and #C=A-B#, what is the angle between A and C?

Question

If #A = <1 ,6 ,9 >#, #B = <-9 ,4 ,-8 ># and #C=A-B#, what is the angle between A and C?

Daniel Acosta · Accepted Answer

The angle is #=35.6#º

Let's start by calculating #vecC# #vecC=vecA-vecB=〈1,6,9〉-〈-9,4,-8〉=〈10,2,17〉# The angle #theta# is given by the dot product definition #vecA.vecC=∥vecA∥*∥vecC∥*costheta# #vecA.vecC=〈10,2,17〉.〈1,6,9〉=10+12+153=175# The modulus of #vecA=∥vecA∥=∥〈1,6,9〉∥=sqrt(1+36+81)=sqrt118# The modulus of #vecC=∥vecC∥=∥〈10,2,17〉∥=sqrt(100+4+289)=sqrt393# Therefore, #costheta=175/(sqrt118*sqrt393)=0.81# #theta=35.6#º

Brooklyn Anderson · Answer

undefined To find the angle between two vectors A and C, you can use the dot product formula:

$ \text{cos}(\theta) = \frac{{A \cdot C}}{{\|A\|\|C\|}} $

Where:
- $ A \cdot C $ is the dot product of vectors A and C.
- $ \|A\| $ and $ \|C\| $ are the magnitudes of vectors A and C respectively.

First, calculate the dot product of vectors A and C:

$ A \cdot C = (1)(1) + (6)(-9) + (9)(-8) $

Then, find the magnitudes of vectors A and C:

$ \|A\| = \sqrt{1^2 + 6^2 + 9^2} $
$ \|C\| = \sqrt{(1 - (-9))^2 + (6 - 4)^2 + (9 - (-8))^2} $

Finally, substitute the values into the formula to find the angle $ \theta $ between A and C:

$ \text{cos}(\theta) = \frac{{A \cdot C}}{{\|A\|\|C\|}} $

$ \theta = \text{cos}^{-1}\left(\frac{{A \cdot C}}{{\|A\|\|C\|}}\right) $

Calculate the value of $ \theta $ using the arccosine function.

If #A = &lt;1 ,6 ,9 &gt;#, #B = &lt;-9 ,4 ,-8 &gt;# and #C=A-B#, what is the angle between A and C?

If #A = <1 ,6 ,9 >#, #B = <-9 ,4 ,-8 ># and #C=A-B#, what is the angle between A and C?