If $A = < 2, 6 >$ $A= <2 , 6 >$ and $B = < - 7, 4 >$ $B= <-7, 4 >$ , what is $| | A + B | | - | | A | | - | | B | |$ $||A+B|| -||A|| -||B||$ ?

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If $A = < 2, 6 >$ $A= <2 , 6 >$ and $B = < - 7, 4 >$ $B= <-7, 4 >$ , what is $| | A + B | | - | | A | | - | | B | |$ $||A+B|| -||A|| -||B||$ ?

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Answer 1 · 2016-06-01T21:22:59

It is $- 3.2$ $-3.2$ .

Explanation:

The norm of a vector with coordinates $v = < x, y >$ $v=< x, y >$ is defined as

$| | v | | = \sqrt{x^{2} + y^{2}}$ $||v||=sqrt(x^2+y^2)$ .

Then we have

$| | A + B | | = | | < 2, 6 > + < - 7, 4 > | |$ $||A+B||=||<2,6>+<-7,4>||$
$= | | < - 5, 10 > | | = \sqrt{{(- 5)}^{2} + 10^{2}} = \sqrt{125} \approx 11.18$ $=||<-5,10>||=sqrt((-5)^2+10^2)=sqrt(125)\approx11.18$

$| | A | | = \sqrt{2^{2} + 6^{2}} = \sqrt{4 + 36} = \sqrt{40} \approx 6.32$ $||A||=sqrt(2^2+6^2)=sqrt(4+36)=sqrt(40)\approx6.32$

$| | B | | = \sqrt{{(- 7)}^{2} + 4^{2}} = \sqrt{49 + 16} = \sqrt{65} \approx 8.06$ $||B||=sqrt((-7)^2+4^2)=sqrt(49+16)=sqrt(65)\approx8.06$

Finally

$| | A + B | | - | | A | | - | | B | | = 11.18 - 6.32 - 8.06 = - 3.2$ $||A+B||-||A||-||B||=11.18-6.32-8.06=-3.2$ .

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If $A = < 2, 6 >$ $A= <2 , 6 >$ and $B = < - 7, 4 >$ $B= <-7, 4 >$ , what is $| | A + B | | - | | A | | - | | B | |$ $||A+B|| -||A|| -||B||$ ?

1 Answer

Explanation:

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If A=<2,6>A= <2 , 6 > and B=<−7,4>B= <-7, 4 >, what is ||A+B||−||A||−||B||||A+B|| -||A|| -||B||?

1 Answer

Explanation:

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If $A = < 2, 6 >$ $A= <2 , 6 >$ and $B = < - 7, 4 >$ $B= <-7, 4 >$ , what is $| | A + B | | - | | A | | - | | B | |$ $||A+B|| -||A|| -||B||$ ?