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Mrac [35]
4 years ago
8

C write function distance that calculates the distance between two points (x1, y1) and (x2, y2). all numbers and return values s

hould be of type double.
Mathematics
1 answer:
TEA [102]4 years ago
6 0
Your question is unclear but here is my try.

Given (0, 5) and (5, 0), find the distance between the two points.

Let (0, 5) and (5, 0) be the two points.

Let d = distance

d = sqrt{(5 - 0)^2 + (0 - 5)^2}

d = sqrt{(5)^2 + (-5)^2}

d = sqrt{25 + 25}

d = sqrt{50}

d = sqrt{25}•sqrt{2}

d = 5•sqrt{2}

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Find the distance of the point (4,4,−4) from the line r(t)=⟨−1+2t,1+2t,3−3t⟩.
Nadusha1986 [10]

Translate the given point and line together so that you get a new point and a new line that passes through the origin. This turns the problem into finding the distance between the new point,

<em>p</em> = (4, 4, -4) - (-1, 1, 3) = (5, 3, -7)

and the new line,

<em>r*</em><em>(t)</em> = <em>r</em><em>(t)</em> - ⟨-1, 1, 3⟩ = ⟨2<em>t</em>, 2<em>t</em>, -3<em>t</em>⟩

Let <em>p</em> = ⟨5, 3, -7⟩, the vector starting at the origin and pointing to <em>p</em>. Then the quantity ||<em>p</em> - <em>r*</em><em>(t)</em>|| is the distance from the point <em>p</em> to the line <em>r*</em><em>(t)</em>.

Let <em>u</em> be such that ||<em>p</em> - <em>r*</em><em>(t)</em>|| is minimized. At the value <em>t</em> = <em>u</em>, the vector <em>p</em> - <em>r*</em><em>(t)</em> is orthogonal to the line <em>r*</em><em>(t)</em>, so that

(<em>p</em> - <em>r*</em><em>(u)</em> ) • <em>r*</em><em>(u)</em> = 0

I've attached a sketch with all these elements in case this description is confusing. (The red dashed line is meant to be perpendicular to <em>r*</em><em>(t)</em>.)

Solve this equation for <em>u</em> :

<em>p</em> • <em>r*</em><em>(u)</em> - <em>r*</em><em>(u)</em> • <em>r*</em><em>(u)</em> = 0

<em>p</em> • <em>r*</em><em>(u)</em> = <em>r*</em><em>(u)</em> • <em>r*</em><em>(u)</em>

and <em>x</em> • <em>x</em> = ||<em>x</em>||² for any vector <em>x</em>, so

<em>p</em> • <em>r*</em><em>(u)</em> = ||<em>r*</em><em>(u)</em>||²

⟨5, 3, -7⟩ • ⟨2<em>u</em>, 2<em>u</em>, -3<em>u</em>⟩ = (2<em>u</em>)² + (2<em>u</em>)² + (-3<em>u</em>)²

10<em>u</em> + 6<em>u</em> + 21<em>u</em> = 4<em>u</em> ² + 4<em>u</em> ² + 9<em>u </em>²

17<em>u</em> ² - 37<em>u</em> = 0

<em>u</em> (17<em>u</em> - 37) = 0

==>   <em>u</em> = 0   or   <em>u</em> = 37/17

We ignore <em>u</em> = 0, since the dot product of any vector with the zero vector is 0.

Then the minimum distance distance between the given point and line is

||<em>p</em> - <em>r*</em><em>(u)</em>|| = ||⟨5, 3, -7⟩ - 37/17 ⟨2, 2, -3⟩|| = √(42/17)

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Step-by-step explanation:


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