Video: Finding the Angle between Vectors

Alex Cutbill

If |𝐀| = 9.5, |𝐁| = 15, and 𝐀 Γ— 𝐁 = βˆ’122𝐂, where 𝐂 is a unit vector, find the measure of the angle πœƒ between 𝐀 and 𝐁 rounded to the nearest minute given that 0Β°< πœƒ< 90Β°.

04:45

Video Transcript

If the magnitude of the vector 𝐀 is 9.5, the magnitude of the vector 𝐁 is 15, and 𝐀 cross 𝐁 is equal to negative 122𝐂, where 𝐂 is a unit vector, find the measure of the angle πœƒ between 𝐀 and 𝐁 rounded to the nearest minute given that zero degrees is less than πœƒ is less than 90 degrees.

So we’re given the magnitude of 𝐀, the magnitude of 𝐁, and we’re told something about the cross product of 𝐀 and 𝐁. And we want to know something about the angle πœƒ between 𝐀 and 𝐁. Can we think of something which links these quantities? Well, we know that the magnitude of the cross product of 𝐀 and 𝐁 is equal to the magnitude of 𝐀 times the magnitude of 𝐁 times the sine of the angle between these two vectors.

So we can substitute in what we have for the cross product of 𝐀 and 𝐁. So on the left-hand side of the equation, we have the magnitude of negative 122𝐂. And on the right-hand side, we can substitute in the values we have for the magnitude of the vector 𝐀 and the magnitude of the vector 𝐁. So we have 9.5 times 15 sin πœƒ. The magnitude of a scalar multiple of 𝐀 vector is equal to the absolute value of the scalar multiplier times the magnitude of that vector.

And so taking π‘˜ to be negative 102 [122] and the vector 𝐯 to be 𝐂, we can rewrite the left-hand side as the absolute value of negative 122 times the magnitude of 𝐂. We can simplify on the right-hand side as well: 9.5 times 15 is 142.5. And so on the right-hand side, we have 142.5 times sin πœƒ.

The absolute value of negative 122 is 122, and we are told in the question that 𝐂 is a unit vector; that means it’s a vector whose magnitude is one. So that means on the left-hand side, we have 122 times one. On the right, we still have 142.5 sin πœƒ. Cancelling the one and dividing both sides by 142.5, we get that 122 over 142.5 is equal to sin πœƒ. We can swap both sides and apply the inverse sine function or the arcsin function to get that πœƒ is equal to the inverse sine of 122 over 142.5.

Entering this into our calculators, we should get something like πœƒ is equal to 58.8859, and so on degrees. Of course, it’s required that your calculator to be in degree mode. Using your calculator, that should be a way of converting this angle, which is in decimal degrees, into an angle in degrees, minutes, and seconds. Doing this, you should get something like 58 degrees, 53 minutes, and 9.402 or so seconds.

The value yet for the number of seconds maybe more or less accurate, but you should be able to see it around 9.4. And as there is a 60 seconds to a minute and a fewer than thirty seconds here, we should be rounding down to 58 degrees and 53 minutes. This is the answer rounded to the nearest minute as we are asked. And we should just note that we have an angle in the correct range using the inverse sine function. Had we’ve been asked for the angle in the 90-degree to 180-degree range, then our answer would be 180 degrees minus our current answer.

So to be strictly correct, we should have added 180 degrees minus sine inverse of 122 over 142.5 as one of the possibilities for πœƒ. If your calculator doesn’t allow you to convert decimal degrees into degrees, minutes, and seconds or if you can’t work on how to make it do it, what you can do is you can write πœƒ is equal to 58 degrees plus 0.8859 dot dot dot degrees, where we got the 0.8859 dot dot dot degrees by subtracting 58 degrees from our answer. And using our calculator to multiply this fractional number of degrees by 60 to get the number of minutes, we get that 0.8859 dot dot dot degrees is equal to 53.1567 dot dot dot minutes, which again to the nearest minute is 58 degrees and 53 minutes.

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