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
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°