Do not round any Intermediate computations Round your answers to the nearest degree. Complete the following for each pair of vectors. 1.Let w=-3i+2j and r=-6i+4j (B) Find the angle between w and r. square ^circ (b) Determine whether w and r are parallel, orthogonal, or neither. parallel orthogonal neither 2. Let t=-5i-4j and v=-6i+5j (a) Find the angle between t and v. square ^circ (b) Determine whether t and V are parallel, orthogonal, or neither. parallel orthogonal neither

1. (B) To find the angle between vectors \( w \) and \( r \), we can use the dot product formula:<br /><br />\[ \cos(\theta) = \frac{w \cdot r}{\|w\| \cdot \|r\|} \]<br /><br />where \( \theta \) is the angle between the vectors, \( w \cdot r \) is the dot product of \( w \) and \( r \), and \( \|w\| \) and \( \|r\| \) are the magnitudes of \( w \) and \( r \) respectively.<br /><br />First, let's calculate the dot product of \( w \) and \( r \):<br /><br />\[ w \cdot r = (-3)(-6) + (2)(4) = 18 + 8 = 26 \]<br /><br />Next, let's calculate the magnitudes of \( w \) and \( r \):<br /><br />\[ \|w\| = \sqrt{(-3)^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13} \]<br /><br />\[ \|r\| = \sqrt{(-6)^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52} \]<br /><br />Now, we can substitute these values into the formula for \( \cos(\theta) \):<br /><br />\[ \cos(\theta) = \frac{26}{\sqrt{13} \cdot \sqrt{52}} = \frac{26}{\sqrt{676}} = \frac{26}{26} = 1 \]<br /><br />Since \( \cos(\theta) = 1 \), the angle \( \theta \) is 0 degrees.<br /><br />(B) To determine whether \( w \) and \( r \) are parallel, orthogonal, or neither, we can compare their direction ratios. If the direction ratios are proportional, then the vectors are parallel. If the direction ratios are perpendicular, then the vectors are orthogonal. Otherwise, they are neither.<br /><br />In this case, the direction ratios of \( w \) and \( r \) are proportional, as \( r \) is simply twice \( w \). Therefore, \( w \) and \( r \) are parallel.<br /><br />2. (A) To find the angle between vectors \( t \) and \( v \), we can use the same dot product formula as in the previous question:<br /><br />\[ \cos(\theta) = \frac{t \cdot v}{\|t\| \cdot \|v\|} \]<br /><br />First, let's calculate the dot product of \( t \) and \( v \):<br /><br />\[ t \cdot v = (-5)(-6) + (-4)(5) = 30 - 20 = 10 \]<br /><br />Next, let's calculate the magnitudes of \( t \) and \( v \):<br /><br />\[ \|t\| = \sqrt{(-5)^2 + (-4)^2} = \sqrt{25 + 16} = \sqrt{41} \]<br /><br />\[ \|v\| = \sqrt{(-6)^2 + 5^2} = \sqrt{36 + 25} = \sqrt{61} \]<br /><br />Now, we can substitute these values into the formula for \( \cos(\theta) \):<br /><br />\[ \cos(\theta) = \frac{10}{\sqrt{41} \cdot \sqrt{61}} = \frac{10}{\sqrt{2501}} \]<br /><br />To find the angle \( \theta \), we take the inverse cosine of \( \cos(\theta) \):<br /><br />\[ \theta = \cos^{-1}\left(\frac{10}{\sqrt{2501}}\right) \]<br /><br />Using a calculator, we find that \( \theta \approx 73.74^\circ \).<br /><br />(A) To determine whether \( t \) and \( v \) are parallel, orthogonal, or neither, we can again compare their direction ratios. In this case, the direction ratios of \( t \) and \( v \) are not proportional, and they are not perpendicular either. Therefore, \( t \) and \( v \) are neither parallel nor orthogonal.