If $A = < 6, - 1 >$ $A= <6, -1 >$ and $B = < 4, 5 >$ $B= <4, 5 >$ , what is $| | A + B | | - | | A | | - | | B | |$ $||A+B|| -||A|| -||B||$ ?

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Answer 1 · 2016-12-29T06:19:34

The answer is $= \sqrt{116} - \sqrt{37} - \sqrt{41} = - 1.72$ $=sqrt116-sqrt37-sqrt41=-1.72$

$\to A = ⟨ 6, - 1 ⟩$ $vecA=〈6,-1〉$

$\to B = ⟨ 4, 5 ⟩$ $vecB=〈4,5〉$

$\to A + \to B = ⟨ 6, - 1 ⟩ + ⟨ 4, 5 ⟩ = ⟨ 10, 4 ⟩$ $vecA+vecB=〈6,-1〉+〈4,5〉=〈10,4〉$

The modulus of $\to A + \to B = ∥ ⟨ 10, 4 ⟩ ∥ = \sqrt{100 + 16} = \sqrt{116}$ $vecA+vecB=∥〈10,4〉∥=sqrt(100+16)=sqrt116$

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

The modulus of $\to B = ∥ ⟨ 4, 5 ⟩ ∥ = \sqrt{16 + 25} = \sqrt{41}$ $vecB=∥〈4,5〉∥=sqrt(16+25)=sqrt41$

Therefore,

$∥ ∥ ∥ \to A + \to B ∥ - ∥ \to A ∥ - ∥ \to B ∥ = \sqrt{116} - \sqrt{37} - \sqrt{41}$ $∥vecA+vecB∥-∥vecA∥-∥vecB∥=sqrt116-sqrt37-sqrt41$

$= - 1.72$ $=-1.72$

If A=<6,−1>A= <6, -1 > and B=<4,5>B= <4, 5 >, what is ||A+B||−||A||−||B||||A+B|| -||A|| -||B||?