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

Question

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

Victoria Adebayo · Accepted Answer

#theta = arccos(94/sqrt{18685})approx 46.55^circ #

#A cdot C = |A||C| cos theta #

The dot product divided by the magnitudes gives the cosine of the angle.

#cos theta = { A cdot C}/{|A||C|} #

#C = A - B = (1,6,-8) -(-9,4,1) = (10, 2, -9)#

#cos theta = {(1,6,-8) cdot (10, 2, -9 ) } / {|((1,6,-8))||((10, 2, -9 )) |} #

#cos theta = {1(10) + 6(2) + -8(-9) }/{ sqrt{ (1^2+6^2+8^2 ) (10^2+2^2+9^2 ) } }#

#cos theta = 94/sqrt(18685)#

#theta = arccos(94/sqrt{18685})#

#theta approx 46.55^circ #

Madelyn Adams · Answer

undefined To find the angle between vectors $ A $ and $ C $, first calculate the dot product of vectors $ A $ and $ C $, then use the dot product formula to find the angle.

Given that $ A = \langle 1, 6, -8 \rangle $, $ B = \langle -9, 4, 1 \rangle $, and $ C = A - B $, calculate $ C $ by subtracting the corresponding components of $ B $ from $ A $:

$$ C = A - B = \langle 1, 6, -8 \rangle - \langle -9, 4, 1 \rangle = \langle 1 + 9, 6 - 4, -8 - 1 \rangle = \langle 10, 2, -9 \rangle $$

Next, calculate the dot product of $ A $ and $ C $:

$$ A \cdot C = (1)(10) + (6)(2) + (-8)(-9) = 10 + 12 + 72 = 94 $$

Then, calculate the magnitudes of vectors $ A $ and $ C $:

$$ |A| = \sqrt{1^2 + 6^2 + (-8)^2} = \sqrt{1 + 36 + 64} = \sqrt{101} $$
$$ |C| = \sqrt{10^2 + 2^2 + (-9)^2} = \sqrt{100 + 4 + 81} = \sqrt{185} $$

Now, use the dot product formula to find the angle $ \theta $ between $ A $ and $ C $:

$$ \cos(\theta) = \frac{A \cdot C}{|A| \cdot |C|} = \frac{94}{\sqrt{101} \cdot \sqrt{185}} $$

$$ \theta = \arccos\left(\frac{94}{\sqrt{101} \cdot \sqrt{185}}\right) $$

Finally, calculate the angle $ \theta $.

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

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