The modulus of a vector A=5i+PJ+4√2k is 11 the value of p is?

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Answer 1 · 2018-04-28T14:19:48

#color(blue)(P=+-8)#

To find the modulus of vector we use:

#||A||=sqrt((5-0)^2+(P-0)^2+(4sqrt(2)-0)^2)=11#

#||A||=sqrt(25+P^2+32)=11#

Squaring:

#25+P^2+32=121#

#P^2=121-25-32=64#

#P=sqrt(64)=+-8#

Check:

#P=8#

#||A||=sqrt((5-0)^2+(8-0)^2+(4sqrt(2)-0)^2)#

#=sqrt(25+64+32)=sqrt(121)=11#

#P=-8#

#||A||=sqrt((5-0)^2+(-8-0)^2+(4sqrt(2)-0)^2)#

#=sqrt(25+64+32)=sqrt(121)=11#