If $A = < 2 - 5 >$ $A= <2 -5 >$ and $B = < 9, 4 >$ $B= <9, 4 >$ , what is $| | A + B | | - | | A | | - | | B | |$ $||A+B|| -||A|| -||B||$ ?

Question

1569 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-26T14:42:36

The answer is $= \sqrt{122} - \sqrt{29} - \sqrt{97} = - 4.2$ $=sqrt122-sqrt29-sqrt97=-4.2$

$\to A = ⟨ 2, - 5 ⟩$ $vecA=〈2,-5〉$

$\to B = ⟨ 9, 4 ⟩$ $vecB=〈9,4〉$

The modulus of the vector $∥ ∥ ⟨ a, b ⟩ ∥ = \sqrt{a^{2} + b^{2}}$ $∥〈a,b〉∥=sqrt(a^2+b^2)$

$\to A + \to B = ⟨ 2, - 5 ⟩ + ⟨ 9, 4 ⟩ = ⟨ 11, - 1 ⟩$ $vecA+vecB=〈2,-5〉+〈9,4〉=〈11,-1〉$

The modulus of $\to A = ∥ ⟨ 2, - 5 ⟩ ∥ = \sqrt{4 + 25} = \sqrt{29}$ $vecA=∥〈2,-5〉∥=sqrt(4+25)=sqrt29$

The modulus of $\to B = ∥ ⟨ 9, 4 ⟩ ∥ = \sqrt{81 + 16} = \sqrt{97}$ $vecB=∥〈9,4〉∥=sqrt(81+16)=sqrt97$

The modulus of $\to A + \to B = ∥ ⟨ 11, - 1 ⟩ ∥ = \sqrt{121 + 1} = \sqrt{122}$ $vecA+vecB=∥〈11,-1〉∥=sqrt(121+1)=sqrt122$

So, $∥ ∥ ∥ \to A + \to B ∥ - ∥ \to A ∥ - ∥ \to B ∥ = \sqrt{122} - \sqrt{29} - \sqrt{97}$ $∥vecA+vecB∥-∥vecA∥-∥vecB∥=sqrt122-sqrt29-sqrt97$

$= - 4.2$ $=-4.2$

If A= <2 -5 >A=<2−5>A= <2 -5 > and B= <9, 4 >B=<9,4>B= <9, 4 >, what is ||A+B|| -||A|| -||B||||A+B||−||A||−||B||||A+B|| -||A|| -||B||?