If $A = < - 2, - 5, 4 >$ $A= <-2 ,-5 ,4 >$ and $B = < - 7, 4, - 9 >$ $B= <-7 ,4 ,-9 >$ , what is $A \cdot B - | | A | | | | B | |$ $A*B -||A|| ||B||$ ?

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If $A = < - 2, - 5, 4 >$ $A= <-2 ,-5 ,4 >$ and $B = < - 7, 4, - 9 >$ $B= <-7 ,4 ,-9 >$ , what is $A \cdot B - | | A | | | | B | |$ $A*B -||A|| ||B||$ ?

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Answer 1 · 2016-10-23T15:10:24

The answer is $\to A . \to B - ∥ \to A ∥ ∥ \to B = - 123.1$ $vecA.vecB-∥vecA∥∥vecB=-123.1$

Explanation:

First we calculate the dot product
$\to A . \to B = ⟨ - 2, - 5, 4 ⟩ . ⟨ - 7, 4, - 9 ⟩ = 14 - 20 - 36 = - 42$ $vecA.vecB=〈-2,-5,4〉.〈-7,4,-9〉=14-20-36=-42$

Then we calculate the modulus of the vectors

$∥ ∥ ∥ \to A ∥ = \sqrt{4 + 25 + 16} = \sqrt{45}$ $∥vecA∥=sqrt (4+25+16)=sqrt45$

$∥ ∥ ∥ \to B ∥ = \sqrt{49 + 16 + 81} = \sqrt{146}$ $∥vecB∥=sqrt(49+16+81)=sqrt146$

So the result is $\to A . \to B - ∥ \to A ∥ ∥ \to B ∥ = - 42 - \sqrt{45 \cdot 146} = - 42 - 81.1 = - 123.1$ $vecA.vecB-∥vecA∥∥vecB∥=-42-sqrt(45*146)=-42-81.1=-123.1$

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If $A = < - 2, - 5, 4 >$ $A= <-2 ,-5 ,4 >$ and $B = < - 7, 4, - 9 >$ $B= <-7 ,4 ,-9 >$ , what is $A \cdot B - | | A | | | | B | |$ $A*B -||A|| ||B||$ ?

1 Answer

Explanation:

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If A=<−2,−5,4>A= <-2 ,-5 ,4 > and B=<−7,4,−9>B= <-7 ,4 ,-9 >, what is A⋅B−||A||||B||A*B -||A|| ||B||?

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Explanation:

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If $A = < - 2, - 5, 4 >$ $A= <-2 ,-5 ,4 >$ and $B = < - 7, 4, - 9 >$ $B= <-7 ,4 ,-9 >$ , what is $A \cdot B - | | A | | | | B | |$ $A*B -||A|| ||B||$ ?