Question Video: Finding the Measure of the Angle between Two Straight Lines in Three Dimensions given Their Equations | Nagwa Question Video: Finding the Measure of the Angle between Two Straight Lines in Three Dimensions given Their Equations | Nagwa

Question Video: Finding the Measure of the Angle between Two Straight Lines in Three Dimensions given Their Equations Mathematics • Third Year of Secondary School

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Find the measure of the angle between the two straight lines 𝐫₁ = 〈2/7, −2/3, −1〉 + 𝐭₁ 〈−2/7, −4/3, 9/3〉 and (−6𝑥 − 2)/7 = (4𝑦 − 3)/−6 = (3 − 8𝑧)/−5.

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Video Transcript

Find the measure of the angle between the two straight lines vector 𝐫 one is equal to two-sevenths, negative two-thirds, negative one plus 𝐭 one multiplied by negative two-sevenths, negative four-thirds, nine-fifths and negative six 𝑥 minus two over seven is equal to four 𝑦 minus three over negative six, which is equal to three minus eight 𝑧 over negative five.

As the angle between two straight lines in space is the angle between their direction vectors, we begin by finding the direction vector of our two lines. We can then use the formula shown to find the measure of the angle between the two lines. Cos 𝜃 is equal to the magnitude of the dot product of 𝐝 one and 𝐝 two divided by the magnitude of 𝐝 one multiplied by the magnitude of 𝐝 two, where 𝐝 one and 𝐝 two are the direction vectors. The first one of our straight lines is given in vector form, where 𝐫 is equal to 𝑥 one, 𝑦 one, 𝑧 one plus 𝜆 multiplied by 𝑎, 𝑏, 𝑐, where the straight line passes through a point with coordinates 𝑥 one, 𝑦 one, 𝑧 one and has direction vector 𝑎, 𝑏, 𝑐. If we let this first line be 𝑙 one, it has direction vector 𝐝 one equal to negative two-sevenths, negative four-thirds, nine-fifths.

Next, we recall that the Cartesian form of a straight line is 𝑥 minus 𝑥 one over 𝑎 is equal to 𝑦 minus 𝑦 one over 𝑏, which is equal to 𝑧 minus 𝑧 one, over 𝑐, where once again the line passes through the point with coordinates 𝑥 one, 𝑦 one, 𝑧 one and the direction vector has components 𝑎, 𝑏, and 𝑐. Our second line is given in a similar format to this. However, the coefficients of 𝑥, 𝑦, and 𝑧 are not equal to one. To write this equation in Cartesian form, we divide each term of the first expression by negative six, each term of the second expression by four, and each term of the third expression by negative eight.

Dividing each term of negative six 𝑥 minus two over seven by negative six gives us 𝑥 plus one-third over negative seven-sixths. This is equal to 𝑦 minus three-quarters over negative three over two, which is equal to 𝑧 minus three-eighths over five-eighths. Comparing this to the general form, we know that the denominators of our three expressions are the components of the direction vector. The direction vector 𝐝 two is equal to negative seven-sixths, negative three-halves, five-eights.

We are now in a position where we can calculate the dot product of 𝐝 one and 𝐝 two together with the magnitude of each of the direction vectors. To calculate the dot product of two vectors, we find the sum of the products of their corresponding components. In this case, we have negative two-sevenths multiplied by negative seven-sixths plus negative four-thirds multiplied by negative three-halves plus nine-fifths multiplied by five-eighths. This simplifies to one-third plus two plus nine-eighths, which is equal to 83 over 24. Since this value is positive, the magnitude of the dot product of the direction vectors 𝐝 one and 𝐝 two is 83 over 24.

Next, we can find the magnitude of the direction vector 𝐝 one. This is equal to the square root of the sum of the squares of the individual components. We have the square root of four over 49 plus 16 over nine plus 81 over 25. Adding the three fractions gives us 56221 over 11025 and the magnitude of the direction vector 𝐝 one is equal to the square root of this. We can now repeat this process to find the magnitude of the direction vector 𝐝 two. This is equal to the square root of 49 over 36 plus nine over four plus 25 over 64, which in turn is equal to the square root of 2305 over 576.

Whilst it is not required in this question, we could simplify these further using our knowledge of radicals. When square rooting a fraction, we can square root the numerators and denominators separately. And since the square root of 11025 is 105 and the square root of 576 is 24, the magnitudes of 𝐝 one and 𝐝 two can be simplified as shown.

We are now in a position where we can substitute in our three values to find an expression for cos 𝜃. This is equal to 83 over 24 divided by the square root of 56221 over 105 multiplied by the square root of 2305 over 24. We can then take the inverse cosine of both sides of this equation, giving us 𝜃 is equal to 40.042626 and so on degrees. Whilst we could round this to a given number of decimal places, we can also convert it into degrees, minutes, and seconds. This can be done directly on our calculator by using the degrees, minutes, and seconds button.

Alternatively, since there are 60 minutes in a degree, we can multiply the decimal part of our answer by 60. And since there are 60 seconds in a minute, we then repeat this process to give our final answer in degrees, minutes, and seconds. The measure of the angle between the two straight lines given is 40 degrees, two minutes, and 33 seconds to the nearest second.

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