### Video Transcript

A straight line πΏ one passes through the two points π΄ negative two, two, negative three and π΅ negative six, negative four, negative five, and a straight line πΏ two passes through the two points πΆ one, four, one and π· 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 π and π, we know that cos of π equals the dot product of π and π over the product of the magnitudes of π and π. So, weβre going to need to find a way to represent π΄π΅ and πΆπ· in vector form. Well, the vector ππ is found by subtracting the vector ππ from the vector ππ. Thatβs the vector that describes the movement from the origin to either point π΄ or π΅.

Now, it doesnβt really matter how we represent these vectors, I quite like using column vectors, so Iβm going to represent ππ as negative six, negative four, negative five. Of course, you could use π, π, and π interchangeably. Weβre going to subtract ππ, so thatβs 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 ππ, letβs find the vector ππ. This time, thatβs ππ minus ππ. ππ is the vector that describes movement from the origin zero, zero, zero to π·. Thatβs negative nine, negative six, negative nine. And we subtract ππ, 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 ππ, and we found vector ππ. To use the scalar product, weβre going to need to find the dot product and the individual magnitudes of each.

Letβs begin by calculating the dot product. This is the sum of the product of each individual element. So, itβs the sum of negative four times negative 10, negative six times negative 10, and negative two times negative 10, which is 120. Next, letβs 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 ππ 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 ππ 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 thereβs really no need since weβre 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 π equals 120 over the square root of 56 times the square root of 300. To solve for π, weβre 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 πΏ one and πΏ two, which we say we can represent as vectors ππ and ππ, is 22.21 degrees.