# Video: Finding the Angle between a Given Vector and a Unit Directional Vector

Find the angle between the vector 𝐀 = ⟨9, 10, −4⟩ and the unit vector 𝐣. Round your answer to the nearest degree.

03:53

### Video Transcript

Find the angle between the vector 𝐀 nine, 10, negative four and the unit vector 𝐣. Round your answer to the nearest degree.

In this question, we’re given two vectors: the vector 𝐀 and the unit directional vector 𝐣. We need to determine the angle between these two vectors. We need to round our answer to the nearest degree. To answer this question, let’s start by recalling how we find the angle between two vectors. We know if 𝜃 is the angle between two vectors 𝐮 and 𝐯, then the cos of 𝜃 will be equal to the dot product between 𝐮 and 𝐯 divided by the modulus of vector 𝐮 times the modulus of vector 𝐯. We can use this to find the angle 𝜃. All we need to do is take the inverse cosine of both sides of the equation.

Well it’s worth pointing out here since the range of the inverse cosine function is between zero degrees and 180 degrees, this will always give us the smaller angle between vectors 𝐮 and 𝐯. Therefore, to find the angle between the two vectors given to us in the question, we’re going to need to find their dot products and the modulus of both of these vectors. Let’s start by finding the dot product between these two vectors. To find the dot product between these two vectors, we’ll start by writing our unit directional vector 𝐣 in component notation.

We’re told the size and direction of this vector. The size of this vector is one because it’s a unit directional vector and its direction is 𝐣. That’s the second component. So this vector is the vector zero, one, zero. Therefore, the dot product between 𝐀 and 𝐣 is equal to the dot product between the vector nine, 10, negative four and the vector zero, one, zero.

Now, we need to calculate the dot product between these two vectors. Remember, this means we need to find the sum of the product of the corresponding components of each of the two vectors. For our two vectors, that’s nine multiplied by zero plus 10 multiplied by one plus negative four multiplied by zero. And if we calculate this expression, the first and last terms are just equal to zero. So the dot product between vector 𝐀 and 𝐣 is just equal to 10.

Next, we’re going to need to find the modulus of these two vectors. Let’s start with the modulus of vector 𝐀. To do this, we recall the modulus of a vector is equal to the square root of the sum of the squares of its components. In other words, the modulus of the vector 𝐚, 𝐛, 𝐜 is equal to the square root of 𝑎 squared plus 𝑏 squared plus 𝑐 squared. We can use this to find the modulus of vector 𝐀. It’s equal to the square root of nine squared plus 10 squared plus negative four all squared. And if we calculate the expression inside of our square root symbol, we see the modulus of vector 𝐀 is the square root of 197.

We could then do exactly the same to find the modulus of vector 𝐣 hat. However, this is not necessary. We’re already told that this is a unit vector. This means that the modulus of this vector must be equal to one. In fact, this is also represented by the hat notation. We use this to represent unit vectors. We’re now ready to find an expression for 𝜃 since we know the dot product between these two vectors and the modulus of both of these two vectors.

First, we know if 𝜃 is the angle between these two vectors, the cos of 𝜃 will be equal to the dot product of 𝐀 and vector 𝐣 divided by the modulus of vector 𝐀 times the modulus of vector 𝐣. Next, we can substitute in the values for the dot product between vector 𝐀 and 𝐣 and the modulus of vector 𝐀 and vector 𝐣. We get the cos of 𝜃 will be equal to 10 divided by root 197 multiplied by one. Then, we can solve this for the value of 𝜃. First, dividing by one doesn’t change the value. And next, we can take the inverse cosine of both sides of the equation. Remember, since we’re finding this to the nearest degree, we need to make sure our calculator is set to degrees mode.

This gives us that 𝜃 is equal to the inverse cos of 10 divided by root 197, which we can calculate is equal to 44.56 and this expansion continues degrees. We want to give this to the nearest degree. We can see the first decimal place in our expansion is five. This means we’re going to need to round up, giving us our final answer that the angle between the vector 𝐀 nine, 10, negative four and the unit directional vector 𝐣 to the nearest degree is 45 degrees.