Question

find the angle between \\( \langle 3,-2\
angle \\) and \\( \langle 6,9\
angle \\).
the angle between \\( \langle 3,-2\
angle \\) and \\( \langle 6,9\
angle \\) is \\( \square \\).
(round to the nearest tenth as needed.)

Explanation:

Step1: Calculate dot product

$\langle 3, -2
angle \cdot \langle 6, 9
angle = 3(6) + (-2)(9) = 18 - 18 = 0$

Step2: Calculate magnitudes

$|\langle 3, -2
angle| = \sqrt{3^2 + (-2)^2} = \sqrt{13} \approx 3.6056$
$|\langle 6, 9
angle| = \sqrt{6^2 + 9^2} = \sqrt{117} \approx 10.8167$

Step3: Find angle using formula

$\cos\theta = \frac{0}{3.6056 \times 10.8167} = 0$
$\theta = \arccos(0) = 90.0^\circ$

Answer:

90.0