Answer to Question #139092 in Electricity and Magnetism for Arsalan

Question #139092
If A=10ax-4ay+6az and B=2ax+ay, find a the component of A along ay b the magnitude of 3A-B, c a unit vector along A+2B
1
Expert's answer
2020-10-19T13:24:43-0400

A=10ax4ay+6azA=10a_x-4a_y+6a_z and B=2ax+ayB=2a_x+a_y

a) If A=Axax+Ayay+AzazA=A_xa_x+A_ya_y+A_za_z , then the component of AA along aya_y is AyA_y .

For A=10ax4ay+6azA=10a_x-4a_y+6a_z we have that Ay=4A_y=-4 is the component of AA along aya_y.


b) 3AB=3(10ax4ay+6az)(2ax+ay)=(3×102)ax+(3×(4)1)ay+3×6az3A-B=3 (10a_x-4a_y+6a_z )-(2a_x+a_y)=(3\times 10-2)a_x+(3\times(-4)-1)a_y+3\times6a_z

3AB=28ax13ay+18az3A-B=28a_x-13a_y+18a_z

3AB=282+(13)2+182=1277=35.74|3A-B|=\sqrt{28^2+(-13)^2+18^2}=\sqrt{1277}=35.74


c) A+2B=10ax4ay+6az+2(2ax+ay)=(10+2×2)ax+(4+2)ay+6azA+2B= 10a_x-4a_y+6a_z+2(2a_x+a_y)=(10+2\times2)a_x+(-4+2)a_y+6a_z

A+2B=14ax2ay+6azA+2B=14a_x-2a_y+6a_z

A unit vector along A+2BA+2B is vector 14ax2ay+6az142+(2)2+62=14ax2ay+6az2360.91ax0.13ay+0.39az\frac{14a_x-2a_y+6a_z}{\sqrt{14^2+(-2)^2+6^2}}= \frac{14a_x-2a_y+6a_z}{\sqrt{236}}\approx0.91a_x-0.13a_y+0.39a_z


Answer: a) 4-4 ; b) 35.7435.74 ; c) 0.91ax0.13ay+0.39az0.91a_x-0.13a_y+0.39a_z


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