A force f is expressed with respect to the basis e1 and e2 by equation 2(3e1+4e2) N. If the directions of e1 makes an angle of 30 degree with f, find the vector e2.
Solution
Given force magnitude
"f=6e \n1\n\u200b\t\n +8e \n2\n\u200b"
Angle between e1 and f is =30°
Using basis vector concept
"f. e_1=(6e_1+8e_2). e_1\\\\\n\u2223(6e \n1\n\u200b\t\n +8e \n2\n\u200b\t\n )\u2223|e \n1\n\u200b\t\n \u2223cos30 \n\u00b0\n =6"
Magnitude of e1 is unit.
So vector e2 can be written as
"e_2=\\frac{4\\sqrt{3}-6e_1}{8}=\\frac{\\sqrt{3}}{2}-\\frac{3e_1}{4}"
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