Given #(ay-bx)/p=(cx-az)/q=(bz-cy)/r#. How to prove that #x/a=y/b=z/c#?

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Answer 1 · 2017-05-20T13:41:04

See below.

Calling

#M = ((-b, a, 0),(c, 0, -a),(0, -c, b))#

and

#P = ((x),(y),(z))#

#B=((lambda p),(lambda q),(lambda r))#

we have

#M.P=B#

now we have also

#P_0 = mu ((a),(b),(c))#

If #P = P_0# then

#M.P_0 = B#

#M cdot P_0 = ((0),(0),(0)) = B rArr lambda = 0# so the affirmation

If #M cdot P = B rArr M cdot P_0 = B#

is true when considering #lambda = 0# and #lambda in RR# so it is true

Answer 2 · 2017-05-20T14:37:22

Refer to the Proof given in the Explanation.

Given that,

#(ay-bx)/p=(cx-az)/q=(bz-cy)/r#

#rArr" Each Ratio="{c(ay-bx)+b(cx-az)+a(bz-cy)}/(cp+bq+ar),#

#=0/(cp+bq+ar),#

#=0.#

# rArr (ay-bx)/p=(cx-az)/q=(bz-cy)/r=0.#

# rArr ay=bx, cx=az, bz=cy," or, what is the same as to say, "#

# x/a=y/b, z/c=x/a, y/b=z/c, or, x/a=y/b=z/c.#

Hence, the Proof.

Enjoy Maths.!