If $A = < 6, - 1 >$ $A= <6, -1 >$ and $B = < - 3, 6 >$ $B= <-3, 6>$ , what is $| | A + B | | - | | A | | - | | B | |$ $||A+B|| -||A|| -||B||$ ?

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Answer 1 · 2016-10-31T06:28:30

The answer is $\sqrt{34} - \sqrt{37} - \sqrt{45}$ $sqrt34-sqrt37-sqrt45$

The modulus of a vector $⟨ x, y ⟩$ $〈x,y〉$ is $= \sqrt{x^{2} + y^{2}}$ $=sqrt(x^2+y^2)$

So modulus of $\to A$ $vecA$ is $∥ ∥ ⟨ 6, - 1 ⟩ ∥ = \sqrt{36 + 1} = \sqrt{37}$ $∥〈6,-1〉∥=sqrt(36+1)=sqrt37$

So modulus of $\to B$ $vecB$ is $∥ ∥ ⟨ - 3, 6 ⟩ ∥ = \sqrt{9 + 36} = \sqrt{45}$ $∥〈-3,6〉∥=sqrt(9+36)=sqrt45$

$⟨ \to A ⟩ + ⟨ \to B ⟩ = ⟨ 6, - 1 ⟩ + ⟨ - 3, 6 ⟩ = ⟨ 3, 5 ⟩$ $〈vecA〉+〈vecB〉 =〈6,-1〉+〈-3,6〉=〈3,5〉$

So the modulus of $⟨ \to A ⟩ + ⟨ \to B ⟩$ $〈vecA〉+〈vecB〉$ is $∥ ∥ ∥ ⟨ \to A ⟩ + ⟨ \to B ⟩ ∥ ∥ ∥$ $∥〈vecA〉+〈vecB〉∥$

$= \sqrt{9 + 25} = \sqrt{34}$ $=sqrt(9+25)=sqrt34$

And finally
$∥ ∥ ∥ \to A + \to B ∥ - ∥ \to A ∥ - ∥ \to B ∥ = \sqrt{34} - \sqrt{37} - \sqrt{45}$ $∥vecA+vecB∥-∥vecA∥-∥vecB∥ =sqrt34-sqrt37-sqrt45$

If A= <6, -1 >A=<6,−1>A= <6, -1 > and B= <-3, 6>B=<−3,6>B= <-3, 6>, what is ||A+B|| -||A|| -||B||||A+B||−||A||−||B||||A+B|| -||A|| -||B||?