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In this video, we’ll see how we can find the equation of a straight line in space.
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So that means we’re looking at coordinates in three dimensions rather than just in two dimensions.
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We’ll see how we can write this equation in Cartesian form, which is sometimes called the general form.
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And we’ll also see how we can write it in vector form.
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Let’s begin by having a look at the vector form.
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The vector form of a line can be described as 𝐫 equals 𝐫 sub zero plus 𝑡𝐯, where 𝐫, 𝐫 sub zero, and 𝐯 are all vectors.
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𝐫 is the position vector of any general point on the line.
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𝐫 sub zero is the position vector of a given point on the line.
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𝐯 is the direction vector of or along the line.
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And 𝑡 is a scalar multiple.
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Vector form can be used in both two dimensions and three dimensions.
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The difference is that in three dimensions, all of our vectors will have 𝑥-, 𝑦-, and 𝑧-components.
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When we’re describing a line in vector form, remember that we’re thinking about how we navigate from the origin to a specific point.
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And then we’re moving along that line in scalar multiples of the vector 𝐯.
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We’ll now have a look at a question where we need to write the vector equation of a line given a point on the line and a direction vector.
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Give the vector equation of the line through the point three, seven, negative seven with direction vector zero, negative five, seven.
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We should remember that when we need to write an equation in vector form, it will be in the form 𝐫 equals 𝐫 sub zero plus 𝑡𝐯, where 𝐫 is the position vector of a general point on the line, 𝐫 sub zero is a position vector of a given point on the line, and 𝐯 is the direction vector.
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𝑡 is a scalar multiple.
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If we look at the information that we’re given in the question, we can see that we have a direction vector.
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And we’ve got a point on the line which can be written as a position vector.
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As we navigate from the origin to the point three, seven, negative seven, then we can write this as the position vector three, seven, negative seven.
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We can then simply plug in these two vectors into the vector form.
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𝐫 equals the position vector three, seven, negative seven plus 𝑡 times the direction vector zero, negative five, seven.
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And so that’s the answer for the vector equation of the line.
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In the following question, we’ll see how we can calculate a direction vector given two points.
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Find the direction vector of the straight line passing through 𝐴 one, negative two, seven and 𝐵 four, negative one, three.
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In this question, we’re given the position vectors of two points in space, 𝐴 and 𝐵, and we’re asked to find the direction vector.
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When we want to find a direction vector 𝐀𝐁, 𝐴 is the starting point and 𝐵 is the terminal point, we subtract the starting point from the terminal point.
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In order to find the direction vector, we can subtract each of the 𝑥-, 𝑦-, and 𝑧-components in 𝐴 from those in 𝐵.
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To begin then, we’ll have four subtract one, giving us three.
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Then we’ll have negative one subtract negative two, which is equivalent to negative one plus two, which is one.
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And finally, we’ll have three subtract seven, giving us negative four.
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And so we’ve got the answer for the direction vector of 𝐝 as three, one, negative four.
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In this question, however, we didn’t necessarily need to find the direction vector 𝐀𝐁.
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We could also have found the direction vector 𝐁𝐀.
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In this case, we would’ve got the vector inverse of 𝐝 equals negative three, negative one, four, which would also have been a valid answer.
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In the next question, we’ll see a slightly more complex example where we need to find the vector equation of a median of a triangle drawn in three-dimensional space.
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The points 𝐴 negative eight, negative nine, negative two; 𝐵 zero, negative seven, six; and 𝐶 negative eight, negative one, negative four form a triangle.
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Determine in vector form the equation of the median drawn from 𝐶.
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In this question, we have three points 𝐴, 𝐵, and 𝐶, which are given in three-dimensional space.
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These three points we’re told form a triangle.
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We’re told that there is a median drawn from 𝐶, and so it would be useful to recall that a median is a line segment joining a vertex to the midpoint of the opposite side.
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For example, if we drew this two-dimensional triangle 𝐴𝐵𝐶, the median from 𝐶 would look like this.
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Perhaps the best way to begin this question is to see if we can find the point that is the midpoint of 𝐴𝐵.
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Let’s define this with the letter 𝑀.
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The formula to find the midpoint of two points in space is very similar to that which we might use for two coordinates in two-dimensional space.
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To find the midpoint 𝑀 of 𝑥 one, 𝑦 one, 𝑧 one and 𝑥 two, 𝑦 two, 𝑧 two, we have that 𝑀 is equal to 𝑥 one plus 𝑥 two over two, 𝑦 one plus 𝑦 two over two, 𝑧 one plus 𝑧 two over two.
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When we fill our values into this formula, we need to make sure we’re using the values for 𝐴 and 𝐵 as, after all, we need to find the midpoint of 𝐴𝐵.
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Note that when we’re plugging in our values, it doesn’t matter which point we use with our 𝑥 one, 𝑦 one, 𝑧 one values or the 𝑥 two, 𝑦 two, 𝑧 two values.
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So we have that the midpoint 𝑀 is equal to negative eight plus zero over two, negative nine plus negative seven over two, and negative two plus six over two.
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Simplifying this, we have that 𝑀 is equal to negative four, negative eight, two.
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We can now clear some space so we can begin to think about the vector form of the equation of this median.
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The vector form of an equation can be written in the form 𝐫 equals 𝐫 sub zero plus 𝑡𝐯, where 𝐫 is the position vector of a general point on the line, 𝐫 sub zero is the position vector of a given point on the line, and 𝐯 is the direction vector.
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𝑡 is a scalar multiple.
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Let’s think about what would happen if we model these three points in three-dimensional space.
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We’d have the triangle 𝐴𝐵𝐶 and the median, which would be the line segment of 𝐶𝑀.
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So when it comes to writing the median in vector form, the position vector can be the point 𝐶.
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But we still need to work out the direction vector of 𝐂𝐌.
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To find the vector 𝐂𝐌, we subtract the starting point 𝐶 from the terminal point 𝑀.
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So we have negative four subtract negative eight, negative eight subtract negative one, and two subtract negative four.
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Simplifying this, we have that vector 𝐂𝐌 is equal to four, negative seven, six.
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Now we have all the information that we need to plug in to the vector form of the line.
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𝐫 sub zero will be the position vector representing point 𝐶.
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Vector 𝐯 will be represented by the vector 𝐂𝐌.
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Therefore, the answer for the equation of the median from 𝐶 is 𝐫 equals negative eight, negative one, negative four plus 𝑡 four, negative seven, six.
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So far in this video, we’ve looked at equations of lines in vector form.
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Now we’ll think about how we change a line given in vector form to one in Cartesian form.
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You may be confused by the terminology of Cartesian form, but a line in two dimensions in Cartesian form can be written in the form 𝑦 equals 𝑚𝑥 plus 𝑏, where 𝑚 is the slope or gradient and 𝑏 is the 𝑦-intercept.
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But of course, it’s different in three-dimensional space as we need an equation that describes the 𝑥-, 𝑦-, and 𝑧-variables.
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So, to find the equation of a line in Cartesian form, we can say that given the equation of a line with direction vector 𝐯 equals 𝑙, 𝑚, 𝑛 which passes through the point 𝑥 sub one, 𝑦 sub one, 𝑧 sub one, then it’s given by 𝑥 minus 𝑥 sub one over 𝑙 equals 𝑦 minus 𝑦 sub one over 𝑚 equals 𝑧 minus 𝑧 sub one over 𝑛, where 𝑙, 𝑚, and 𝑛 are nonzero real numbers.
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So we can see how this information of an equation with a point and a direction, i.e., one in vector form, can be changed into one in Cartesian form.
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We’ll now have a look at two questions where we could apply this formula.
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Give the Cartesian equation of the line 𝐫 equals negative three, negative two, negative two plus 𝑡 four, two, four.
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In this question, we’re given this equation in vector form.
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Negative three, negative two, negative two is the position vector of a given point, and four, two, four is the direction vector.
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In order to change the equation in vector form into an equation in Cartesian form, there is a formula we can apply.
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The equation of a line with direction vector 𝐯 equals 𝑙, 𝑚, 𝑛 that passes through 𝑥 sub one, 𝑦 sub one, 𝑧 sub one is given by 𝑥 minus 𝑥 sub one over 𝑙 equals 𝑦 minus 𝑦 sub one over 𝑚 equals 𝑧 minus 𝑧 sub one over 𝑛, where 𝑙, 𝑚, and 𝑛 are nonzero real numbers.
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We now need to take the direction vector four, two, four to have the values of 𝑙, 𝑚, and 𝑛, respectively.
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We can do the same and designate the coordinate 𝑥 sub one, 𝑦 sub one, 𝑧 sub one with the values negative three, negative two, negative two.
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Plugging these values into the formula, we have 𝑥 minus negative three over four equals 𝑦 minus negative two over two equals 𝑧 minus negative two over four.
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Simplifying the numerators, we have 𝑥 plus three over four equals 𝑦 plus two over two equals 𝑧 plus two over four.
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And that’s the answer for the Cartesian equation of the given line.
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Let’s have a look at one final question.
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Find the Cartesian form of the equation of the straight line passing through the points negative seven, negative three, negative seven and negative three, negative 10, negative four.
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In this question, although we’re asked for the Cartesian form of the equation, it might be useful to think about what this line would be like in a vector form.
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If we considered our two points as 𝐴 and 𝐵 and we wanted to find the direction vector of 𝐀𝐁, then we would subtract all the points on our starting point 𝐴 from those on 𝐵.
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So we’d have 𝐀𝐁 is equal to negative three subtract negative seven, which is equivalent to negative three plus seven, giving us four.
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Negative 10 subtract negative three is equivalent to negative 10 plus three, which is negative seven.
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And then we’d have negative four subtract negative seven, which gives us three.
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Now that we have a direction vector and a point on a line, we can find the Cartesian form of the equation of the line joining these two points.
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We should remember that the equation of a line with direction vector 𝐯 equals 𝑙, 𝑚, 𝑛 that passes through the point 𝑥 sub one, 𝑦 sub one, 𝑧 sub one is given by 𝑥 minus 𝑥 sub one over 𝑙 equals 𝑦 minus 𝑦 sub one over 𝑚 equals 𝑧 minus 𝑧 sub one over 𝑛.
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Note that 𝑙, 𝑚, and 𝑛 are nonzero real numbers.
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We can now use the direction vector of 𝐀𝐁 for the values of 𝑙, 𝑚, and 𝑛 and the point negative seven, negative three, negative seven for the values 𝑥 sub one, 𝑦 sub one, and 𝑧 sub one.
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Filling these values into the formula, we have 𝑥 subtract negative seven over four equals 𝑦 subtract negative three over negative seven equals 𝑧 subtract negative seven over three.
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Simplifying the numerators gives us the answer in Cartesian form 𝑥 plus seven over four equals 𝑦 plus three over negative seven equals 𝑧 plus seven over three.
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We can now summarize the key points of this video.
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Firstly, we saw the equation of a line given in vector form as 𝐫 equals 𝐫 sub zero plus 𝑡𝐯.
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𝐫 is the position vector of a general point on the line, 𝐫 sub zero is the position vector of a given point on the line, and 𝐯 is the direction vector.
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𝑡 is a scalar multiple.
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To find a direction vector 𝐀𝐁, we subtract the starting point 𝐴 from the terminal point 𝐵.
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Finally, we saw that the equation of a line with direction vector 𝐯 equals 𝑙, 𝑚, 𝑛 that passes through 𝑥 sub one, 𝑦 sub one, 𝑧 sub one is given by 𝑥 minus 𝑥 sub one over 𝑙 equals 𝑦 minus 𝑦 sub one over 𝑚 equals 𝑧 minus 𝑧 sub one over 𝑛, where 𝑙, 𝑚, and 𝑛 are nonzero real numbers.
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This final formula is very useful for changing the equation of a line given in vector form to an equation given in Cartesian form.