# Question Video: Finding the Dot Product of Two Vectors Using the Properties of Isosceles Triangles Mathematics • 12th Grade

Given that 𝐴𝐵𝐶 is an isosceles triangle, where 𝐴𝐵 = 𝐴𝐶 = 6 cm and 𝑚∠𝐴 = 120°, determine 𝐂𝐀⋅𝐁𝐂.

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### Video Transcript

Given that 𝐴𝐵𝐶 is an isosceles triangle, where 𝐴𝐵 is equal to 𝐴𝐶 is equal to six centimeters and the measure of angle 𝐴 is 120 degrees, determine the scalar or dot product of vector 𝐂𝐀 with vector 𝐁𝐂.

Let’s begin by sketching the isosceles triangle. We’re given that the measure of angle 𝐴 is 120 degrees and that sides 𝐴𝐵 and 𝐴𝐶 both have a length of six centimeters. We’re asked to find the dot product of the vectors 𝐂𝐀 and 𝐁𝐂. And we know that the scalar or dot product of two vectors 𝐮 and 𝐯 is the product of their magnitudes with the cosine of the angle between them. So in our case this is the product of the magnitude of vector 𝐂𝐀 with the magnitude of vector 𝐁𝐂 and the cosine of the angle between them.

Now we know the magnitude of vector 𝐂𝐀, and that’s six centimeters. So we’ll need to find the magnitude of vector 𝐁𝐂 and the angle 𝜃 between the two vectors. To find the angle 𝜃, we recall that the angle between two vectors is defined uniquely as the angle between their directions when the lines representing them either both converge or both diverge. In the case of our vectors 𝐂𝐀 and 𝐁𝐂, we see that 𝐁𝐂 converges onto the vertex 𝐶, whereas 𝐂𝐀 diverges from the vertex 𝐶. And so to specify the angle 𝜃, we must extend 𝐁𝐂 so that both vectors diverge from the vertex 𝐶. Our angle 𝜃 then is the obtuse angle between the two vectors.

Now, to find the measure of the angle 𝜃, we refer back to the fact that our triangle is an isosceles triangle. And this means that the measures of angles 𝐶 and 𝐵 are the same. And since the angles in a triangle must sum to 180 degrees, we have the measures of angles 𝐴, 𝐵, and 𝐶 sum to 180. And since the measures of angles 𝐵 and 𝐶 are the same, we have 180 is equal to the measure of angle 𝐴 plus two times the measure of angle 𝐵. And this, of course, is the same as the measure of angle 𝐴 plus two times the measure of angle 𝐶.

And since the measure of angle 𝐴 is 120 degrees, we can subtract 120 from both sides. And we have 60 degrees is two times the measure of angle 𝐵. And finally, dividing through by two, we have the measure of angle 𝐵 is 30 degrees. And of course this is the same as the measure of angle 𝐶.

So now making some space and making a note of the measure of the angles at 𝐵 and 𝐶, we can use this to find the measure of the angle 𝜃 by noting that 𝜃 plus the measure of angle 𝐶 must equal 180 degrees. And this is because the vectors 𝐂𝐁 and 𝐮 lie on a straight line and the angle between them is 180 degrees. Since we just found that the measure of angle 𝐶 is 30 degrees, subtracting 30 from both sides, we have the measure of our angle 𝜃 is 150 degrees.

So now making a note of this, our next step is to find the magnitude of the vector 𝐁𝐂, that is, its length in centimeters, since the lengths of 𝐴𝐵 and 𝐴𝐶 are both in centimeters. And to find the length 𝐵𝐶, we’re gonna use the sine rule. This tells us that for triangle 𝐴𝐵𝐶, the length of side 𝑎 over the sin of the angle 𝐴 is the length of side 𝑏 over the sin of the angle 𝐵 is the length of side 𝑐 over the sin of the angle at 𝐶. And applying this to our triangle, we have six over the sin of 30 degrees is equal to 𝐵𝐶 over the sin of 120 degrees. And now multiplying through by the sin of 120 degrees, we have six times the sin of 120 degrees over the sin of 30 degrees is equal to 𝐵𝐶.

Now since the sin of 120 degrees is root three over two and the sin of 30 degrees is one-half, we have six times the square root of three over two divided by one-half. Dividing by two on the top leaves us with three root three. And dividing by a half is the same as multiplying by two. And so we have our length 𝐵𝐶 is equal to six root three centimeters.

Now making some space so we can calculate our dot product, we have everything we need to find the dot product of the vectors 𝐂𝐀 and 𝐁𝐂, that is, six, which is the magnitude of vector 𝐂𝐀, multiplied by six root three, that’s the magnitude of vector 𝐁𝐂, multiplied by the cos of 150 degrees, that’s the cos of our angle 𝜃. The cos of 150 degrees is negative root three over two.

We can then cancel the two in our denominator with one of the sixes to give three, which leaves us with 18 multiplied by root three multiplied by negative root three. And finally this evaluates to negative 54. The dot or scalar product of vectors 𝐂𝐀 and 𝐁𝐂 is therefore equal to negative 54.