If $A = < - 1, 2 >$ $A= <-1, 2 >$ and $B = < - 3, 4 >$ $B= <-3, 4>$ , what is $| | A + B | | - | | A | | - | | B | |$ $||A+B|| -||A|| -||B||$ ?

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If $A = < - 1, 2 >$ $A= <-1, 2 >$ and $B = < - 3, 4 >$ $B= <-3, 4>$ , what is $| | A + B | | - | | A | | - | | B | |$ $||A+B|| -||A|| -||B||$ ?

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Answer 1 · 2017-01-04T11:04:33

The answer is $= - 0.025$ $=-0.025$

Explanation:

$\to A + \to B = ⟨ - 1, 2 ⟩ + ⟨ - 3, 4 ⟩ = ⟨ - 4, 6 ⟩$ $vecA+vecB=〈-1,2〉+〈-3,4〉=〈-4,6〉$

The modulus of $\to A$ $vecA$ is $= ∥ \to A ∥ = ∥ ⟨ - 1, 2 ⟩ ∥ = \sqrt{1 + 4} = \sqrt{5}$ $=∥vecA∥=∥〈-1,2〉∥=sqrt(1+4)=sqrt5$

The modulus of $\to B$ $vecB$ is $= ∥ \to B ∥ = ∥ ⟨ - 3, 4 ⟩ ∥ = \sqrt{9 + 16} = \sqrt{25} = 5$ $=∥vecB∥=∥〈-3,4〉∥=sqrt(9+16)=sqrt25=5$

The modulus of $\to A + \to B$ $vecA+vecB$ is $= ∥ \to A + \to B ∥ = ∥ ⟨ - 4, 6 ⟩ ∥ = \sqrt{16 + 36} = \sqrt{52}$ $=∥vecA+vecB∥=∥〈-4,6〉∥=sqrt(16+36)=sqrt52$

Therefore,

$∥ ∥ ∥ \to A + \to B ∥ - ∥ \to A ∥ - ∥ \to B ∥ = \sqrt{52} - \sqrt{5} - 5 = - 0.025$ $∥vecA+vecB∥-∥vecA∥-∥vecB∥=sqrt52-sqrt5-5=-0.025$

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If $A = < - 1, 2 >$ $A= <-1, 2 >$ and $B = < - 3, 4 >$ $B= <-3, 4>$ , what is $| | A + B | | - | | A | | - | | B | |$ $||A+B|| -||A|| -||B||$ ?

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Explanation:

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If A=<−1,2>A= <-1, 2 > and B=<−3,4>B= <-3, 4>, what is ||A+B||−||A||−||B||||A+B|| -||A|| -||B||?

1 Answer

Explanation:

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If $A = < - 1, 2 >$ $A= <-1, 2 >$ and $B = < - 3, 4 >$ $B= <-3, 4>$ , what is $| | A + B | | - | | A | | - | | B | |$ $||A+B|| -||A|| -||B||$ ?