Question Video: Finding the Angle between Two Straight Lines in Two Dimensions Mathematics

Determine the measure of the acute angle between the two straight lines 𝐿₁: 𝐫 = βŸ¨βˆ’4, βˆ’3⟩ + 𝐾⟨4, βˆ’9⟩ and 𝐿₂: 7π‘₯ βˆ’ 3𝑦 + 17 = 0 to the nearest second.

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Video Transcript

Determine the measure of the acute angle between the two straight lines 𝐿 sub one 𝐫 is equal to the vector negative four, negative three plus 𝐾 times the vector four, negative nine and 𝐿 sub two seven π‘₯ minus three 𝑦 plus 17 is equal to zero to the nearest second.

In this question, we’re asked to determine the measure of the acute angle between two straight lines, and we’re given the equations of these straight lines. One’s given in vector form and one’s given in general form. We need to find this measure to the nearest second. To do this, we can start by recalling how we find the measure of the acute angle between two given straight lines. We know if two straight lines 𝐿 sub one and 𝐿 sub two have slopes of π‘š sub one and π‘š sub two, then the acute angle πœƒ between the two lines will satisfy the equation tan of πœƒ is equal to the absolute value of π‘š sub one minus π‘š sub two divided by one plus π‘š sub one times π‘š sub two.

And there’s a few things worth noting about this equation. For example, if our two lines are parallel and not vertical, then the slopes will be equal. So π‘š sub one will be equal to π‘š sub two. So the numerator of the right-hand side of this equation is zero. This then gives us tan of πœƒ is zero, so πœƒ is zero. Similarly, if the two lines are perpendicular, then we can have that π‘š sub one times π‘š sub two is negative one, once again, provided neither line is vertical. This then means tan πœƒ is undefined, so πœƒ is 90 degrees. Finally, we can only apply this formula if neither of our lines are vertical. So let’s start by finding the slopes of the two lines.

So we’ll start by finding the slope of line 𝐿 sub one. To do this, we recall we can find the slope of a vector equation of a line of the form 𝐫 is equal to 𝐫 sub zero plus 𝐾 times the vector π‘Ž, 𝑏 by just finding 𝑏 divided by π‘Ž. And this is provided that π‘Ž is nonzero. And applying this result to the equation of 𝐿 sub one, we get that π‘š sub one is equal to negative nine divided by four. And it’s worth noting we could’ve found this directly from the equation of the line. Its direction vector is four, negative nine. So for every four units we move to the right, the line moves nine units down. Its change in 𝑦 over its change in π‘₯ is negative nine over four.

To determine the equation of our second line, we’ll write it in slope–intercept form. We’ll start by adding three 𝑦 to both sides of the equation. This gives us that seven π‘₯ plus 17 is equal to three 𝑦. Now we’ll divide the equation through by three. This gives us that seven over three π‘₯ plus 17 over three is equal to 𝑦. And we know the coefficient of π‘₯ is the slope of the line. So our value of π‘š sub two is seven over three.

Now that we found the slopes of both of these lines, we can substitute these values into our equation. Doing this gives us the tan of πœƒ is equal to the absolute value of negative nine over four minus seven over three divided by one plus negative nine over four multiplied by seven over three. And if we evaluate the right-hand side of this equation, we get 55 over 51, which must be equal to the tan of πœƒ. We can then solve for the value of πœƒ in degrees by taking the inverse tangent to both sides of the equation. Well, we need to make sure our calculator is set to degrees mode.

This gives us that πœƒ is equal to 47.161 and this expansion continues degrees. And we could stop here, but the question wants us to give our answer to the nearest second. So we’re going to need to convert this into degrees, minutes, and seconds. To do this, we’ll start by recalling there’s 60 minutes in a degree and 60 seconds in a minute. And we can already see there are 47 degrees in our answer, so we’ll take these out. This then leaves us with 0.161 and this expansion continues degrees. And if we multiply this value by 60, we’ll get the remaining angle in minutes. Calculating this expression, where we need to be careful to use the exact value of our angle, we get 9.664 and this expansion continues minutes.

We can apply this process again. We can see that there are nine full minutes in this angle. This then leaves us with the remaining angle of 0.664 and this expansion continues minutes. If we multiply this value by 60, we’ll get the remaining angle in seconds. Calculating this, where once again we need to make sure we’re using the exact value, gives us 39.886 and this expansion continues seconds. Remember, the question wants us to give our answer to the nearest second. So we need to look at the first decimal digit, which is eight, which tells us we need to round this value up. So we round this up to 40 seconds, giving us our final answer of 47 degrees, nine minutes, and 40 seconds.

Therefore, we were able to show the measure of the acute angle to the nearest second between the two lines 𝐫 is equal to the vector negative four, negative three plus 𝐾 times the vector four, negative nine and the line seven π‘₯ minus three 𝑦 plus 17 is equal to zero is 47 degrees, nine minutes, and 40 seconds.

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