Given that x = \(\begin{pmatrix} -4 \\ 3 \end{pmatrix}\) and y= \(\begin{pmatrix} -9 \\ 15 \end{pmatrix}\) calculate, correct to the nearest degree, the angle between the vectors
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Correct Answer: Option
Explanation:
x = \(\begin{pmatrix} -4 \\ 3 \end{pmatrix}\) and y= \(\begin{pmatrix} -9 \\ 15 \end{pmatrix}\)
Changing x and y to the form xi+yj
we have x= -4i + 3j and y = -9i - 15j
using cosØ = \(\frac{xy}{|x|ly|}\)
where Ø is the angle between x and y
xy = (-4i+3j) (-9i - 15j)
-36 + 60 x 0 - 27 x 0- 45 = -81
|x| = √(4\(^2\) +3\(^2\)) =√(16 +9) = √25 = 5
|y| = √(-9)\(^2\) + (-15)\(^2\) = √(81 +225) = √306
cosØ =\(\frac{-81}{5√306}\)
= \(\frac{-9√306}{170}\)
\(\frac{17.49 * 9}{170}\)
Ø = cos\(^{-1}\)(-0.1029);
Ø = 95.91°
Ø = 96°
x = \(\begin{pmatrix} -4 \\ 3 \end{pmatrix}\) and y= \(\begin{pmatrix} -9 \\ 15 \end{pmatrix}\)
Changing x and y to the form xi+yj
we have x= -4i + 3j and y = -9i - 15j
using cosØ = \(\frac{xy}{|x|ly|}\)
where Ø is the angle between x and y
xy = (-4i+3j) (-9i - 15j)
-36 + 60 x 0 - 27 x 0- 45 = -81
|x| = √(4\(^2\) +3\(^2\)) =√(16 +9) = √25 = 5
|y| = √(-9)\(^2\) + (-15)\(^2\) = √(81 +225) = √306
cosØ =\(\frac{-81}{5√306}\)
= \(\frac{-9√306}{170}\)
\(\frac{17.49 * 9}{170}\)
Ø = cos\(^{-1}\)(-0.1029);
Ø = 95.91°
Ø = 96°