Question

Find the angle between v= 4i-3j and w=-12i-5j. Round to the nearest thenth of a degree.​

Answer 1

120.5

Explanation:

`\vec{v}=4\vec{i}-3\vec{j},\vec{w}=-12\vec{i}-5\vec{j}\\\vec{v}\cdot\vec{w}=|\vec{v}||\vec{w}|\cos\angle(\vec{v},\vec{w})\quad\Rightarrow\quad \cos\angle(\vec{v},\vec{w})=\frac{\vec{v}\cdot\vec{w}}{|\vec{v}||\vec{w}|}\\\cos\angle(\vec{v},\vec{w})=\frac{(4\vec{i}-3\vec{j})\cdot(-12\vec{i}-5\vec{j})}{√(4^2+(-3)^2)√((-12)^2+(-5^2))}=(4\cdot(-12)+(-3)\cdot(-5))/(5\cdot 13)=-(33)/(65)\\\angle(\vec{v},\vec{w})=\arccos(-{(33)/(65)})=120.5^o`

Answer 2

Explanation:

To find the angle between two vectors v and w, we can use the dot product formula:

θ = arccos((v · w) / (|v| |w|))

Where v · w is the dot product of vectors v and w, and |v| and |w| are the magnitudes of v and w, respectively.

Let's calculate the dot product first:

v · w = (4)(-12) + (-3)(-5)

= -48 + 15

= -33

Next, let's calculate the magnitudes of v and w:

|v| = sqrt((4)^2 + (-3)^2)

= sqrt(16 + 9)

= sqrt(25)

= 5

|w| = sqrt((-12)^2 + (-5)^2)

= sqrt(144 + 25)

= sqrt(169)

= 13

Now, we can substitute these values into the formula to find the angle:

θ = arccos((-33) / (5 * 13))

≈ arccos(-0.5077)

≈ 119.6 degrees

Rounded to the nearest tenth of a degree, the angle between v = 4i - 3j and w = -12i - 5j is approximately 119.6 degrees.