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Problem 16 Easy Difficulty
Find an equation of the sphere that passes through the origin and whose center is $ (1, 2, 3) $.
Answer
$(x-1)^{2}+(y-2)^{2}+(z-3)^{2}=14$
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Watch More Solved Questions in Chapter 12
Video Transcript
find an equation spared passes through the region and whose center is 123 Okay, we have no the center. But we do know this. The radio's so we're first gonna write down the equation like this. X minus one plus Y minus two. Square one is two and square Z minus three. Because our square what is are the c. R s over radios. Really? Is, uh, this fair? Yeah. And, uh, okay, it passes through the region so we can plug in according to the our region. And the equation holds. What does it mean? So if you plug in 000 it's going to be one square plus two square, three square equals are square. So, uh, Oscar is going to be one plus four plus 9. 14. Good. So now we have known Oscars 14. We're gonna replace it here, so, uh, square equals because 14. Yeah, and final equation is this. That's just one square plus women's to square. Plus, the mystery square equals equals 14
University of Illinois at Urbana-Champaign
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Vectors
Top Calculus 3 Educators
Lily A.
Johns Hopkins University
Catherine R.
Missouri State University
Heather Z.
Oregon State University
Kayleah T.
Harvey Mudd College