If $A = < 2, 1, - 4 >$ $A= <2 ,1 ,-4 >$ and $B = < 3, 2, - 5 >$ $B= <3 ,2 ,-5 >$ , what is $A \cdot B - | | A | | | | B | |$ $A*B -||A|| ||B||$ ?

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Answer 1 · 2016-12-10T11:56:47

The answer is $= - 0.25$ $=-0.25$

Here we need the dot product and the modulus

The dot product is $= ⟨ a, b, c ⟩ . ⟨ d, e, f ⟩ = a d + b e + c f$ $=〈a,b,c〉.〈d,e,f〉=ad+be+cf$

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

Here we have $\to A = ⟨ 2, 1, - 4 ⟩$ $vecA=〈2,1,-4〉$ and $\to B = ⟨ 3, 2, - 5 ⟩$ $vecB=〈3,2,-5〉$

$\to A . \to B = ⟨ 2, 1, - 4 ⟩ . ⟨ 3, 2, - 5 ⟩ = 6 + 2 + 20 = 28$ $vecA.vecB=〈2,1,-4〉.〈3,2,-5〉=6+2+20=28$

The modulus of $\to A$ $vecA$

$= ∥ ⟨ 2, 1, - 4 ⟩ ∥ = \sqrt{4 + 1 + 16} = \sqrt{21}$ $=∥〈2,1,-4〉∥=sqrt(4+1+16)=sqrt21$

The modulus of $\to B$ $vecB$

$= ∥ ⟨ 3, 2, - 5 ⟩ ∥ = \sqrt{9 + 4 + 25} = \sqrt{38}$ $=∥〈3,2,-5〉∥=sqrt(9+4+25)=sqrt38$

Therefore,

$\to A . \to B - ∥ \to A ∥ ∥ \to B ∥ = 28 - \sqrt{21} \sqrt{38}$ $vecA.vecB-∥vecA∥∥vecB∥=28-sqrt21sqrt38$

$= 28 - \sqrt{798} = 28 - 28.25 = - 0.25$ $=28-sqrt798=28-28.25=-0.25$

If A=<2,1,−4>A= <2 ,1 ,-4 > and B=<3,2,−5>B= <3 ,2 ,-5 >, what is A⋅B−||A||||B||A*B -||A|| ||B||?