Asked by Anonymous
Two vectors 𝑢⃗ and 𝑣 are acting from the point P. If |𝑢⃗ | = 8, |𝑣 | = 10 and |𝑢⃗ + 𝑣 | = 14 then determine |𝑢⃗ − 𝑣 |
Answers
Answered by
Anonymous
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
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
Answered by
oobleck
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
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
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