Question

Plz help me
Maths
Topic : - Ratio ​

Answer 1

Given :

(a² + b² + c²)(x² + y² + z²) = (ax + by + cz)²

To prove :

• x : a = y : b = z : c

Proof :

→ (a² + b² + c²)(x² + y² + z²) = (ax + by + cz)²

• Apply identity
• (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac

→ x²(a² + b² + c²) + y²(a² + b² + c²) + z²(a² + b² + c²) = (ax)² + (by)² + (cz)² + 2*ax*by + 2*by*cz + 2*ax*cz

→ x²a² + x²b² + x²c² + y²a² + y²b² + y²c² + z²a² + z²b² + z²c² = a²x² + b²y² + c²z² + 2axby + 2bycz + 2axcz

• Cancel a²x², b²y² & c²z²

→ x²b² + x²c² + y²a² + y²c² + z²a² + z²b² = 2axby + 2bycz + 2axcz

→ x²b² + y²a² - 2axby + y²c² + z²b² - 2bycz + x²c² + a²z² - 2axcz = 0

• Apply identity
• (a - b)² = a² + b² - 2ab

→ (xb - ya)² + (yc - zb)² + (xc - az)² = 0

→ (xb - ya)² = 0, (yc - zb)² = 0, (xc - az)² = 0

→ xb - ya = 0, yc - zb = 0, xc - az = 0

→ xb = ya, yc = zb, xc = az

→ x/a = y/b, y/b = z/c, z/c = x/a

• Each one is equal to each other

→ x/a = y/b = z/c - proved

━━━━━━━━━━━━━━━━━━━━━━━━

Answer 2

Question:-

• $\tt \: ( { a }^{ 2 } + { b }^{ 2 } + { c }^{ 2 } )( { x }^{ 2 } + { y }^{ 2 } + { z }^{ 2 } ) = ( { ax + by + cz }^{ 2 } )$ then show that x : a = y : b = z : c.

To Find:-

• Prove it.

Given:-

• $\tt \: ( { a }^{ 2 } + { b }^{ 2 } + { c ) }^{ 2 } ( { x }^{ 2 } + { y }^{ 2 } + { z }^{ 2 } ) = ( { ax + by + cz ) }^{ 2 }$

Solution:-

Here ,

$\tt \: \dfrac { x } { a } = \dfrac { y } { b } = \dfrac { z } { c } = k$

• x = ak ; y = bk ; z = ck

$\tt\implies \: ( { a }^{ 2 } + { b }^{ 2 } + { c }^{ 2 } ) ( { ( ak ) }^{ 2 } + { ( bk ) }^{ 2 } + { ( ck ) }^{ 2 } ) = { ( { a }^{ 2 } k + { b }^{ 2 } k + { c }^{ 2 } k ) }^{ 2 }$

$\tt\implies \: ( { a }^{ 2 } + { b }^{ 2 } + { c }^{ 2 } ) ( { a }^{ 2 } + { b }^{ 2 } + { c }^{ 2 } ) { k }^{ 2 } = { k }^{ 2 } { ( { a }^{ 2 } + { b }^{ 2 } + { c }^{ 2 } ) }^{ 2 }$

$\tt\implies \: ( { a }^{ 2 } + { b }^{ 2 } + { c }^{ 2 } ) { k }^{ 2 } = ( { a }^{ 2 } + { b }^{ 2 } + { c }^{ 2 } ) { k }^{ 2 }$

$\tt\implies \: LHS = RHS$

Hence Proved