1a) The two vectors are:
a = 2x + 3y, b = 4x + 2y
The angle of each vector with respect to the horizontal is:
[tex]\theta_a = arctan(\frac{3}{2})=56.3^{\circ}[/tex]
[tex]\theta_b = arctan(\frac{2}{4})=26.6^{\circ}[/tex]
So, the angle between the two vectors is [tex]\theta=\theta_a - \theta_b = 56.3-26.6=26.7^{\circ}[/tex]
1b) The two vectors are:
a = x + y, b = x - y
The angle of each vector with respect to the horizontal is:
[tex]\theta_a = arctan(\frac{1}{1})=45.0^{\circ}[/tex]
[tex]\theta_b = arctan(\frac{2}{4})=-45.0^{\circ}[/tex]
So, the angle between the two vectors is [tex]\theta=\theta_a - \theta_b = 45.0-(-45.0)=90.0^{\circ}[/tex]
1c) The two vectors are:
a = 2x - 3y, b = 5x + y
The angle of each vector with respect to the horizontal is:
[tex]\theta_a = arctan(\frac{-3}{2})=-56.3^{\circ}[/tex]
[tex]\theta_b = arctan(\frac{1}{5})=11.3^{\circ}[/tex]
So, the angle between the two vectors is [tex]\theta=\theta_a - \theta_b = 11.3-(-56.3)=67.6^{\circ}[/tex]
1d) The two vectors are:
a = x - 6y, b = 2x - 2y
The angle of each vector with respect to the horizontal is:
[tex]\theta_a = arctan(\frac{-6}{1})=-80.5^{\circ}[/tex]
[tex]\theta_b = arctan(\frac{-2}{2})=-45.0^{\circ}[/tex]
So, the angle between the two vectors is [tex]\theta=\theta_a - \theta_b = -45.0-(-80.5)=35.5^{\circ}[/tex]
2) The two vectors are:
a = 2x + 3y, b = 5x + 2y
The angle of each vector with respect to the horizontal is:
[tex]\theta_a = arctan(\frac{3}{2})=56.3^{\circ}[/tex]
[tex]\theta_b = arctan(\frac{2}{5})=21.8^{\circ}[/tex]
So, the angle between the two vectors is [tex]\theta=\theta_a - \theta_b = 56.3-21.8=34.5^{\circ}[/tex]
3a) The two vectors are
a = 2x + 2y – z, b = –3x – y + 3z
The dot vector is:
[tex]a\cdot b = 2 \cdot (-3) + 2 \cdot (-1) + (-1) \cdot 3 =-6 -2 -3=-11[/tex]
3b) the angle between two 3d vectors is given by
[tex]cos \theta = \frac{a \cdot b}{|a| |b|}[/tex]
Let's calculate the magnitude of each vector:
[tex]|a| = \sqrt{2^2 + 2^2 + (-1)^2}=\sqrt{9}=3[/tex]
[tex]|b|=\sqrt{(-3)^2+(-1)^2 +3^2}=\sqrt{19}=4.36[/tex]
So, if we apply the formula,
[tex]\cos \theta = \frac{-11}{3 \cdot 4.36}=-0.84[/tex]
and the angle is
[tex]\theta = arccos (-0.84)=147.1 ^{\circ}[/tex]
4) The two vectors are:
a = (250x + 350y – 50z), b = (100x – 500y + 50z)
the angle between two 3d vectors is given by
[tex]cos \theta = \frac{a \cdot b}{|a| |b|}[/tex]
The dot vector here is
[tex]a \cdot b = 250 \cdot 100 + 350 \cdot (-500) + (-50) \cdot 50 =-152500[/tex]
The magnitude of each vector is:
[tex]|a| = \sqrt{250^2 + 350^2 + (-50)^2}=\sqrt{187500}=433.0[/tex]
[tex]|b|=\sqrt{(100)^2+(500)^2 +50^2}=\sqrt{262500}=512.3[/tex]
So, if we apply the formula,
[tex]\cos \theta = \frac{-152500}{433.0 \cdot 512.3}=-0.687[/tex]
and the angle is
[tex]\theta = arccos (-0.687)=133.4 ^{\circ}[/tex]
