# Question Video: Finding the Measure of the Angle between Two Straight Lines in Three Dimensions given the Coordinates of Four Points Lying on Them Mathematics

A straight line 𝐿₁ passes through the two points 𝐴(−2, 2, −3) and 𝐵(−6, −4, −5), and a straight line 𝐿₂ passes through the two points 𝐶(1, 4, 1) and 𝐷(−9, −6, −9). Find the measure of the angle between the two lines, giving your answer to two decimal places if necessary.

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### Video Transcript

A straight line 𝐿 one passes through the two points 𝐴 negative two, two, negative three and 𝐵 negative six, negative four, negative five, and a straight line 𝐿 two passes through the two points 𝐶 one, four, one and 𝐷 negative nine, negative six, negative nine. Find the measure of the angle between the two lines, giving your answer to two decimal places if necessary.

To calculate the angle between two vectors, we use something called the scalar product. For two vectors 𝐚 and 𝐛, we know that cos of 𝜃 equals the dot product of 𝐚 and 𝐛 over the product of the magnitudes of 𝐚 and 𝐛. So, we’re going to need to find a way to represent 𝐴𝐵 and 𝐶𝐷 in vector form. Well, the vector 𝐀𝐁 is found by subtracting the vector 𝐎𝐀 from the vector 𝐎𝐁. That’s the vector that describes the movement from the origin to either point 𝐴 or 𝐵.

Now, it doesn’t really matter how we represent these vectors, I quite like using column vectors, so I’m going to represent 𝐎𝐁 as negative six, negative four, negative five. Of course, you could use 𝑖, 𝑗, and 𝑘 interchangeably. We’re going to subtract 𝐎𝐀, so that’s negative two, two, negative three. Negative six minus negative two is negative six plus two, which is negative four. Negative four minus two is negative six. And negative five minus negative three is negative two.

Now that we found vector 𝐀𝐁, let’s find the vector 𝐂𝐃. This time, that’s 𝐎𝐃 minus 𝐎𝐂. 𝐎𝐃 is the vector that describes movement from the origin zero, zero, zero to 𝐷. That’s negative nine, negative six, negative nine. And we subtract 𝐎𝐂, which is one, four, one. Then, negative nine minus one is negative 10. Negative six minus four is negative 10. And the third element is also negative 10. So, we found vector 𝐀𝐁, and we found vector 𝐂𝐃. To use the scalar product, we’re going to need to find the dot product and the individual magnitudes of each.

Let’s begin by calculating the dot product. This is the sum of the product of each individual element. So, it’s the sum of negative four times negative 10, negative six times negative 10, and negative two times negative 10, which is 120. Next, let’s find the magnitude of each of our vectors.

Now, we recall that the magnitude of a vector is the square root of the sum of the squares of each of its components. So, the magnitude of 𝐀𝐁 is the square root of negative four squared plus negative six squared plus negative two squared, which is the square root of 56. And the magnitude of 𝐂𝐃 is the square root of negative 10 squared plus negative 10 squared plus negative 10 squared, which is the square root of 300. And of course, we could simplify these, but there’s really no need since we’re going to be typing all of this into our calculator in a moment.

We pop everything into our formula for the scalar product. And we find that cos of 𝜃 equals 120 over the square root of 56 times the square root of 300. To solve for 𝜃, we’re going to find the inverse cos of both sides. The inverse cos of 120 over the square root of 56 times the square root of 300 is 22.2076 and so one, which, correct to two decimal places, is 22.21 degrees. And so, the angle between 𝐿 one and 𝐿 two, which we say we can represent as vectors 𝐀𝐁 and 𝐂𝐃, is 22.21 degrees.