any help would be appreciated:)

H

hayood

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Feb 16, 2010
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Show that the plane whose intercepts with the coordinates axes are x =a, y=b , and z=c has equation

x/a + y/b +z/c =1

provided that a, b, and c are nonzero.


my answer:
i multiplied both sides by abc and i got:
bcx + acy + abz = abc

next i moved abc to the left side :
bcx + acy + abz - abc = 0

but then what should i do?

thank you :)
 
Hello, hayood![

You can't start with the equation . . . That's what you're trying to prove!

\(\displaystyle \text{Show that the plane whose intercepts are: }\:x =a,\:y=b,\: z=c\)

. . \(\displaystyle \text{has equation: }\;\frac{x}{a} + \frac{y}{b} +\frac{z}{c} \:=\:1\;\text{ provided that }a, b,c\text{ are nonzero.}\)
Click to expand...

I have a proof using vectors.


\(\displaystyle \text{We have three points: }\:A(a,0,0),\;B(0,b,0),\;C(0,0,c)\)

. . \(\displaystyle \text{Then: }\;\begin{array}{ccc}\overrightarrow{AB} &=& \langle a,\text{-}b,0\rangle \\ \overrightarrow{BC} &=& \langle 0,b,\text{-}c\rangle \end{array}\)


\(\displaystyle \text{The normal to the plane is: }\:\vec {n} \:=\:\overrightarrow{AB} \times \overrigyhtarrow{BC}\)

. . . \(\displaystyle \text{Hence: }\;\vec n \;=\;\left|\begin{array}{ccc} i&j&k \\ a& \text{-}b&0 \\ 0&b&\text{-}c \end{array}\right| \;=\; \langle bc,\:ac,\:ab\rangle\)


\(\displaystyle \text{The equation of the plane through }A(a,0,0)\,\text{ with normal vector }\:\vec n \:=\:\langle bc,\:ac,\:ab\rangle\:\text{ is:}\)

. . \(\displaystyle bc(x-a) + ac(y-0) + ab(z-0) \;=\;0 \quad\Rightarrow\quad bcx + acy + abz \:=\:abc\)


\(\displaystyle \text{Divide by }abc\!:\;\;\frac{bcx}{abc} + \frac{acy}{abc} + \frac{abx}{abc} \;=\;\frac{anc}{abc}\)

. . \(\displaystyle \text{Therefore: }\quad\frac{x}{a} + \frac{y}{b} + \frac{z}{c} \;=\;1\)

 
oh right:):)

thanks a lot for ur help!! do you have time to answer one more question?

if you can't its fine.. thanks again:)
 
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