If $A = < 4, 0, - 3 >$ $A= <4, 0 ,-3 >$ and $B = < - 2, 5, 7 >$ $B= <-2 ,5 ,7 >$ , what is $A \cdot B - | | A | | | | B | |$ $A*B -||A|| ||B||$ ?

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Answer 1 · 2016-11-27T08:35:13

The answer is $= - 29 - 5 \sqrt{78} = - 53.2$ $=-29-5sqrt78=-53.2$

Let $\to A = ⟨ a_{1}, a_{2}, a_{3} ⟩$ $vecA=〈a_1,a_2,a_3〉$

and $\to B = ⟨ b_{1}, b_{2}, b_{3} ⟩$ $vecB=〈b_1,b_2,b_3〉$

The dot product is

$\to A . \to B = ⟨ a_{1}, a_{2}, a_{3} ⟩ . ⟨ b_{1}, b_{2}, b_{3} ⟩$ $vecA.vecB=〈a_1,a_2,a_3〉.〈b_1,b_2,b_3〉$

$= a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}$ $=a_1b_1+a_2b_2+a_3b_3$

and the modulus is

$∥ ∥ ∥ \to A ∥ = \sqrt{a_{1}^{2} + a_{2}^{2} + a_{3}^{2}}$ $∥vecA∥=sqrt(a_1^2+a_2^2+a_3^2)$

$∥ ∥ ∥ \to B ∥ = \sqrt{b_{1}^{2} + b_{2}^{2} + b_{3}^{2}}$ $∥vecB∥=sqrt(b_1^2+b_2^2+b_3^2)$

Here, $\to A = ⟨ 4, 0, - 3 ⟩$ $vecA=〈4,0,-3〉$

and $\to B = ⟨ - 2, 5, 7 ⟩$ $vecB=〈-2,5,7〉$

$\to A . \to B = - 8 + 0 - 21 = - 29$ $vecA.vecB=-8+0-21=-29$

$∥ ∥ ∥ \to A ∥ = \sqrt{16 + 0 + 9} = \sqrt{25} = 5$ $∥vecA∥=sqrt(16+0+9)=sqrt25=5$

$∥ ∥ ∥ \to B ∥ = \sqrt{4 + 25 + 49} = \sqrt{78}$ $∥vecB∥=sqrt(4+25+49)=sqrt78$

Therefore,

$\to A . \to B - ∥ \to A ∥ ∥ \to B ∥ = - 29 - 5 \sqrt{78} = - 53.2$ $vecA.vecB-∥vecA∥∥vecB∥=-29-5sqrt78=-53.2$

If A=<4,0,−3>A= <4, 0 ,-3 > and B=<−2,5,7>B= <-2 ,5 ,7 >, what is A⋅B−||A||||B||A*B -||A|| ||B||?