Question Video: Finding the Distance between Two Parallel Lines | Nagwa Question Video: Finding the Distance between Two Parallel Lines | Nagwa

# Question Video: Finding the Distance between Two Parallel Lines Mathematics • Third Year of Secondary School

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Find, to the nearest hundredth, the distance between the parallel lines πΏβ: (π₯ + 7)/9 = (π¦ + 1)/5 = (π§ β 7)/β6, and πΏβ: (π₯ + 3)/9 = (π¦ + 10)/5 = (π§ + 10)/β6.

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### Video Transcript

Find, to the nearest hundredth, the distance between the parallel lines πΏ one, π₯ plus seven over nine equals π¦ plus one over five equals π§ minus seven over negative six, and πΏ two, π₯ plus three over nine equals π¦ plus 10 over five equals π§ plus 10 over negative six.

Okay, so here we have these two lines that are parallel, πΏ one and πΏ two. Our question asks us to solve for the distance between them, meaning the minimum or perpendicular distance between these lines. Letβs call that distance π, and we can recall the mathematical relationship for this distance. What weβll need to know are the components of a vector, weβve called it π¬, that runs parallel to our two lines. And weβll also need to know the components of a vector ππ that runs from a point on our second line to a point on our first line. We can say then that, to calculate π, the coordinates of points π and πΏ and the components of vector π¬ are what we need to know.

To solve for this information, letβs look at the given equations of lines one and two. Starting with line one, this is given in whatβs called symmetric form. We can write these three fractions as being equal to one another because we say that theyβre also equal to a scale factor. We can call it π‘ one. The fact that these three fractions are all equal to π‘ one means that we can write separate equations for the π₯-, π¦-, and π§-coordinates of line πΏ one. That is, since π₯ plus seven divided by nine equals π‘ one, we can write that π₯ equals nine times π‘ one minus seven. In a similar way, since π¦ plus one divided by five equals π‘ one, π¦ is equal to five times π‘ one minus one. And then, π§ minus seven over negative six equaling π‘ one implies that π§ equals negative six π‘ one plus seven.

Line πΏ one is now written in whatβs called parametric form. We can combine these three separate equations into one vector equation. The vector with components π₯, π¦, π§ is equal to the vector to negative seven, negative one, seven plus our scale factor π‘ one times the vector nine, five, negative six.

Now, if we take our π₯-, π¦-, and π§-components and write them as the vector π«, then this equals the vector negative seven, negative one, seven plus π‘ one times the vector nine, five, negative six. This first vector is a vector from the origin of a coordinate system to a point on line πΏ one. That is, the point with coordinates negative seven, negative one, seven lies on πΏ one. And then, regarding the second vector, this is a vector that lies along the axis of this line. That means itβs parallel to the line. And therefore, we can say that this is our vector π¬.

So far then, we have the components of a vector parallel to both of our lines. And we also have the coordinates of a point on our first line. To find the coordinates of a point on our second line, letβs look at that lineβs equation. Once again, this equation is given to us in symmetric form. That means we can say that each one of these fractions is equal to another scale factor weβll call π‘ two. If we once again write out separate equations for π₯, π¦, and π§, we find that π₯ equals nine π‘ two minus three, π¦ equals five π‘ two minus 10, and π§ equals negative six π‘ two minus 10.

Once again, we can write these parametric equations as a vector equation. A vector with components π₯, π¦, and π§ equals the vector negative three, negative 10, negative 10 plus π‘ two times the vector nine, five, negative six. Written this way, we know that this vector travels from the origin of a coordinate frame to a point along our line πΏ two. This tells us that the coordinates of that point are simply the components of the vector. We can write then that the coordinates of the point weβve called π are negative three, negative 10, negative 10. Our next step is to use the coordinates of points πΏ and π to solve for the vector ππ. This is the vector with components equal to the difference between the coordinates of point πΏ and those of point π. ππ, therefore, has components negative four, nine, 17.

We can now move ahead by calculating the cross product of vectors ππ and π¬. That equals the determinant of this matrix. And notice that, by column, we have the π₯-, π¦-, and π§-components of our two vectors. Calculating this cross product, we get π’ hat times negative 139 minus π£ hat times negative 129 plus π€ hat times negative 101. We can write this result then as a vector with components negative 139, positive 129, and negative 101. Now that we know ππ cross π¬, weβre ready to calculate the magnitude of this cross product and divide it by the magnitude of π¬.

The magnitude of ππ cross π¬ equals the square root of negative 139 squared plus 129 squared plus negative 101 squared. We divide this by the magnitude of π¬, which itself is the square root of nine squared plus five squared plus negative six squared. Calculating this entire fraction altogether, rounding to the nearest hundredth, we get an answer of 18.03. We say then that the distance between these two parallel lines is 18.03 length units.

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