How to Prove an Inequality Using Vectors

[utkarshakash]

Homework Statement

If a,b,c and k are real constants and α,β,γ are variables subject to the condition that atanα+btanβ+ctanγ = k, then prove using vectors that tan^2 α+tan^2 β+ tan^2 γ ≥ k^2/(a^2+b^2+c^2)

Homework Equations

The Attempt at a Solution

$(ai+bj+ck).(tan \alpha i+ tan \beta j+ tan \gamma k) = k \\<br /> k^2 = (a^2+b^2+c^2)(tan^2 \alpha + tan^2 \beta + tan^2 \gamma)$

But what I arrive at is an equation instead of the inequality required to prove.

 

[ehild]

Homework Statement

If a,b,c and k are real constants and α,β,γ are variables subject to the condition that atanα+btanβ+ctanγ = k, then prove using vectors that tan^2 α+tan^2 β+ tan^2 γ ≥ k^2/(a^2+b^2+c^2)

Homework Equations

The Attempt at a Solution

$(ai+bj+ck).(tan \alpha i+ tan \beta j+ tan \gamma k) = k \\$

Correct so far...

$k^2 = (a^2+b^2+c^2)(tan^2 \alpha + tan^2 \beta + tan^2 \gamma)$

That is wrong. How is the dot product of two vectors related to their magnitudes?


ehild

 

[utkarshakash]

Correct so far...


That is wrong. How is the dot product of two vectors related to their magnitudes?


ehild

$\vec{c} ^2 = \vec{c} . \vec{c} = |\vec{c}|^2$

I've used this identity to simplify it further.

 

[ehild]

$\vec{c} ^2 = \vec{c} . \vec{c} = |\vec{c}|^2$

I've used this identity to simplify it further.

Yes, but you have the dot product of two different vectors to be squared. $(\vec a \cdot\vec b)^2≠|\vec a |^2 |\vec b|^2$.

ehild

 

[haruspex]

$\vec{c} ^2 = \vec{c} . \vec{c} = |\vec{c}|^2$

That's only the special case where the two vectors are the same. What is the more general relationship?

 

[Saitama]

Homework Statement

If a,b,c and k are real constants and α,β,γ are variables subject to the condition that atanα+btanβ+ctanγ = k, then prove using vectors that tan^2 α+tan^2 β+ tan^2 γ ≥ k^2/(a^2+b^2+c^2)

Homework Equations

The Attempt at a Solution

$(ai+bj+ck).(tan \alpha i+ tan \beta j+ tan \gamma k) = k \\<br /> k^2 = (a^2+b^2+c^2)(tan^2 \alpha + tan^2 \beta + tan^2 \gamma)$

But what I arrive at is an equation instead of the inequality required to prove.

Do you know about Cauchy-Schwarz inequality?

 

[utkarshakash]

Do you know about Cauchy-Schwarz inequality?

I encountered it in Calculus but never bothered to go through it as it is not in the JEE syllabus.

 

[Saitama]

I encountered it in Calculus but never bothered to go through it as it is not in the JEE syllabus.

Cauchy-Schwarz inequality uses dot product, it isn't too difficult.

Check out Wikipedia.

 

[utkarshakash]

Cauchy-Schwarz inequality uses dot product, it isn't too difficult.

Check out Wikipedia.

Thanks mate. This inequality helped me to solve some other problems as well.

 

[Saitama]

Thanks mate. This inequality helped me to solve some other problems as well.

Glad to help. :)