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Find, to the nearest hundredth, the distance between the parallel lines ๐ฅ equals one plus three ๐ก, ๐ฆ equals seven plus two ๐ก, ๐ง equals four plus five ๐ก and ๐ฅ equals three minus three ๐ก, ๐ฆ equals six minus two ๐ก, ๐ง equals four minus five ๐ก.
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Okay, so here we have these two parallel lines, and weโll call the first line line one and the second line two.
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We want to solve for the perpendicular distance between these lines.
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Weโll call that ๐.
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To do this, there are three bits of information weโll need to find out.
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First, weโll need to know the coordinates of a point on the first line; weโll call that ๐ one.
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Weโll also need to know a point on line two; weโll call that ๐ two.
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And lastly, weโll need to know the components of a vector thatโs parallel to both these lines.
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Weโll call this vector ๐ฌ.
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Once we know all this, weโll be able to use this expression to compute the distance between our two parallel lines.
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We see this equation involves a vector ๐ฌ, thatโs the vector thatโs parallel to our lines, and a second vector weโve called ๐ one ๐ two.
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On our sketch, thatโs a vector that looks like this.
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It goes from point one on line one to point two on line two.
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Letโs start by figuring out a point on line one, in other words, the coordinates of a point ๐ one.
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Weโll do this by looking at the equation of line one, which we see is given to us in parametric form.
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Written this way, we have separate equations for the ๐ฅ-, ๐ฆ-, and ๐ง-coordinates of every point on this line, and itโs possible to convert this lineโs equation from parametric form to whatโs called vector form.
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This involves combining all three equations into one, where ๐ซ is a vector with components ๐ฅ, ๐ฆ, and ๐ง.
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This form of the lineโs equation begins with the vector from the origin of our coordinate frame to the point one, seven, four.
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This point lies on line one, and then it moves out along this vector three, two, five multiplied by the scale factor ๐ก.
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We can say then that the point with coordinates one, seven, four lies along line ๐ฟ one.
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So weโll call this point ๐ one.
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Furthermore, this vector three, two, five is parallel to our line, and therefore we can call this the vector ๐ฌ.
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Now that we know a point ๐ one and a vector ๐ฌ, letโs clear a bit of space and start looking at the given equation for line two.
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Our goal in doing this is to discover a point ๐ two that lies along this line.
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Just like for line one, line two is given to us in parametric form.
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That means we can write this line in vector form as a vector to the point three, six, four plus ๐ก times another vector parallel to line two.
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Since the point three, six, four is on ๐ฟ two, we can call this ๐ two.
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And so, now we have all the information we need to begin calculating the distance ๐.
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The first thing weโll do is solve for the components of this vector ๐ one ๐ two.
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We find these components by subtracting the coordinates of point ๐ one from those of ๐ two.
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With those values substituted in, we find that the vector ๐ one ๐ two has components three minus one or two, six minus seven or negative one, and four minus four, zero.
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Now that we know this vector, next, weโll take the cross product of this vector with ๐ฌ.
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That cross product is given by the determinant of this three-by-three matrix.
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In the top row, we have the ๐ข, ๐ฃ, and ๐ค unit vectors and then the ๐ฅ-, ๐ฆ-, and ๐ง-components of ๐ one, ๐ two, and ๐ฌ, respectively.
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The magnitude of the ๐ข-component is given by the determinant of this two-by-two matrix.
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Negative one times five minus zero times two is negative five.
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And then the ๐ฃ-component is negative the determinant of this matrix, which is two times five minus zero times three or 10.
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And lastly, thereโs the ๐ค-component of this cross product equal to the determinant of this matrix.
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Two times two minus negative one times three is positive seven.
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This, then, is our overall cross product, which we can write in vector form with components negative five, negative 10, seven.
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Okay, weโre ready now to calculate ๐ by computing the magnitude of ๐ one ๐ two cross ๐ฌ and dividing that by the magnitude of ๐ฌ.
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The magnitude of our cross product equals the square root of negative five squared plus negative 10 squared plus seven squared, while the magnitude of ๐ฌ equals the square root of three squared plus two squared plus five squared.
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Entering this whole expression on our calculator, to the nearest hundredth, our answer is 2.14.
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This is the minimum distance between these two parallel lines.