If $A = < - 3, 6 >$ $A= <-3, 6 >$ and $B = < - 2, 5 >$ $B= <-2, 5 >$ , what is $| | A + B | | - | | A | | - | | B | |$ $||A+B|| -||A|| -||B||$ ?

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If $A = < - 3, 6 >$ $A= <-3, 6 >$ and $B = < - 2, 5 >$ $B= <-2, 5 >$ , what is $| | A + B | | - | | A | | - | | B | |$ $||A+B|| -||A|| -||B||$ ?

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Answer 1 · 2017-12-16T09:30:04

The answer is $= - 0.01$ $=-0.01$

Explanation:

The vectors are

$\to A = < - 3, 6 >$ $vecA= <-3,6>$

$\to B = < - 2, 5 >$ $vecB = <-2,5>$

The modulus of $\to A$ $vecA$ is $= ∣ ∣ ∣ ∣ ∣ ∣ \to A ∣ ∣ ∣ ∣ ∣ ∣ = | | < - 3, 6 > | | = \sqrt{{(- 3)}^{2} + {(6)}^{2}} = \sqrt{9 + 36} = \sqrt{45}$ $=||vecA||=||<-3,6>||=sqrt((-3)^2+(6)^2)=sqrt(9+36)=sqrt45$

The modulus of $\to B$ $vecB$ is $= ∣ ∣ ∣ ∣ ∣ ∣ \to B ∣ ∣ ∣ ∣ ∣ ∣ = | | < - 2, 5 > | | = \sqrt{{(- 2)}^{2} + {(5)}^{2}} = \sqrt{4 + 25} = \sqrt{29}$ $=||vecB||=||<-2,5>||=sqrt((-2)^2+(5)^2)=sqrt(4+25)=sqrt29$

The modulus of $(\to A + \to B)$ $(vecA+vecB)$ is

$∣ ∣ ∣ ∣ ∣ ∣ \to A + \to B ∣ ∣ ∣ ∣ ∣ ∣ = | | < - 3, 6, > + < - 2, 5 > | | = | | < - 5, 11 > | |$ $||vecA+vecB||= ||<-3,6,> +<-2,5>|| =||< -5,11>||$

$= \sqrt{{(- 5)}^{2} + 11^{2}}$ $=sqrt((-5)^2+11^2)$

$= \sqrt{25 + 121}$ $=sqrt(25+121)$

$= \sqrt{146}$ $=sqrt146$

Therefore,

$∣ ∣ ∣ ∣ ∣ ∣ \to A + \to B ∣ ∣ ∣ ∣ ∣ ∣ - ∣ ∣ ∣ ∣ ∣ ∣ \to A ∣ ∣ ∣ ∣ ∣ ∣ - ∣ ∣ ∣ ∣ ∣ ∣ \to B ∣ ∣ ∣ ∣ ∣ ∣ = \sqrt{146} - \sqrt{45} - \sqrt{29} = - 0.01$ $||vecA+vecB||-||vecA|| -||vecB||=sqrt(146)-sqrt(45)-sqrt(29)= -0.01$

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If $A = < - 3, 6 >$ $A= <-3, 6 >$ and $B = < - 2, 5 >$ $B= <-2, 5 >$ , what is $| | A + B | | - | | A | | - | | B | |$ $||A+B|| -||A|| -||B||$ ?

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Explanation:

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If A=<−3,6>A= <-3, 6 > and B=<−2,5>B= <-2, 5 >, what is ||A+B||−||A||−||B||||A+B|| -||A|| -||B||?

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Explanation:

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If $A = < - 3, 6 >$ $A= <-3, 6 >$ and $B = < - 2, 5 >$ $B= <-2, 5 >$ , what is $| | A + B | | - | | A | | - | | B | |$ $||A+B|| -||A|| -||B||$ ?