Two vectors π’β and π£ are acting from the point P. If |π’β | = 8, |π£ | = 10 and |π’β + π£ | = 14 then determine |π’β β π£ |
I draw u along x axis length 8
I draw v in quadrant 1, length 10
I finish the parallelogram with the long diagonal length = 14, call intersection point A so diagonal is 14 from origin O to A
call angle C between v and side opposite u
then
14^2 = 10^2 + 8^2 - 2 * 10 * 8 cos C
196 = 100 + 64 - 160 cos C
cos C = -0.2
C = 180 - 78.46 = 101.53 degrees (in quadrant 2)
2 C = 203
the angles in the parallogram add up to 360 so the other two add to 360 - 203 = 157
so the angles at origin and at A are 157/2 = 78.5 degrees
so call that short diagonal, which is the difference we want x
x^2 = 10^2 + 8^2 - 2*10*8 cos 78.5 = 100 + 64 - 160 *.2
= 164 - 32 = 132
so
x = 11.5
a^2+b^2 = 8^2
c^2+d^2 = 10^2
(a+c)^2 + (b+d)^2 = 14^2
a^2+2ac+c^2 + b^2+2bd+d^2 = 14^2
8^2 + 10^2 + 2ac+2bd = 14^2
2ac+2bd = 32
Now do (a-c)^2 + (b-d)^2
To determine |π’β β π£ |, we can use the formula for the magnitude of the difference of two vectors:
|π’β β π£ | = β((π’β β π£ )Β·(π’β β π£ ))
Here, (π’β β π£ )Β·(π’β β π£ ) represents the dot product of π’β β π£ with itself.
First, let's find (π’β + π£ ) using the given information:
|π’β + π£ | = 14
Given that |π’β | = 8 and |π£ | = 10, we can use the properties of the dot product to find |π’β + π£ |.
|π’β + π£ | = β((π’β + π£ )Β·(π’β + π£ ))
= β(π’β Β·π’β + 2(π’β Β·π£ ) + π£ Β·π£ )
= β(|π’β |^2 + 2(π’β Β·π£ ) + |π£ |^2 )
= β(8^2 + 2(π’β Β·π£ ) + 10^2 )
= β(64 + 2(π’β Β·π£ ) + 100 )
= β(164 + 2(π’β Β·π£ ))
Simplifying this equation, we have:
β(164 + 2(π’β Β·π£ )) = 14
Now, solve for (π’β Β·π£ ):
164 + 2(π’β Β·π£ ) = 14^2
164 + 2(π’β Β·π£ ) = 196
2(π’β Β·π£ ) = 196 - 164
2(π’β Β·π£ ) = 32
π’β Β·π£ = 32/2
π’β Β·π£ = 16
Now that we know π’β Β·π£ = 16, we can find |π’β β π£ |:
|π’β β π£ | = β((π’β β π£ )Β·(π’β β π£ ))
= β(π’β Β·π’β - 2(π’β Β·π£ ) + π£ Β·π£ )
= β(|π’β |^2 - 2(π’β Β·π£ ) + |π£ |^2 )
= β(8^2 - 2(π’β Β·π£ ) + 10^2 )
= β(64 - 2(16) + 100 )
= β(64 - 32 + 100 )
= β(132)
= β(4 * 33)
= 2β(33)
Therefore, |π’β β π£ | = 2β(33).