<p>A right triangle in 3-dimensional space is formed by the points (2, -2, 1) and (5, 2, 1). What is the length of the hypotenuse of the triangle? Step by step please!</p>
<p>isn't it just 5?</p>
<p>they have the same z coordinate so just treat it as any other problem.</p>
<p>nope, the answer is 7.81</p>
<p>hmm. now i'm stuck</p>
<p>Use this formula to calculate distance:</p>
<p>square root of (x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2</p>
<p>It must be noted that only two sides are given, so we only know the length of one side. The distance between these two points is 5, as a few people have already explained. </p>
<p>7.81 is what you get if you round the square root of 61 to two decimal places (square that number to see that it equals 60.9961), and 61 is 36+25. So, they want you to realize that the second leg is 6 somehow. Based on the information given, this expectation is unrealistic. However, if we know that 5 is a leg, the hypotenuse must be greater than five. Still, there are an infinite number of possible choices.</p>
<p>Is there a diagram, and could you tell us what the other choices were?</p>
<p>haha, i was just doing some late night cramming. </p>
<p>@nickwasy, if you want me to tell you the other choices i could, but then it might take time to dig through papers. i'll comment again with the choices. and nope, i don't think there was a diagram.</p>
<p>This question is actually in the Sparknotes Math 2 Book. There's actually another point (2, -2, 7) that was provided in the question. I don't think its possible to solve your question without providing an angle or the final point.</p>
<p>If you had the last point, just use the distance formula 3 times to find the longest side, which is 7.81.</p>
<p>Yes, with the third point you can use the formula sak09 provided and find the answer...7.81, nice 'n easy ;)</p>