A force f̂ is expressed with respect to the basis ê1 and ê2 by equation 2(3 ê1 +4 ê2) N. If the directions of ê1 makes an angle of 30 degree with f̂, find the vector ê2.
Given,
f=6e1^+8e2^f=6\hat{e_1}+8 \hat{e_2}f=6e1^+8e2^
The direction e1^\hat{e_1}e1^ makes angle with =30∘=30^\circ=30∘
⇒tan(30∘)=∣8e2^6e2^∣\Rightarrow \tan(30^\circ)=|\frac{8\hat{e_2}}{6\hat{e_2}}|⇒tan(30∘)=∣6e2^8e2^∣
⇒∣13∣=∣8e2^6e1^∣\Rightarrow |\frac{1}{\sqrt{3}}|=|\frac{8\hat{e_2}}{6\hat{e_1}}|⇒∣31∣=∣6e1^8e2^∣
squaring both side,
⇒13=64e22^36e^2\Rightarrow \frac{1}{3}=\frac{64\hat{e_2^2}}{36\hat{e}^2}⇒31=36e^264e22^
⇒36e12^=3×64e22^\Rightarrow 36 \hat{e_1^2}=3\times64\hat{e_2^2}⇒36e12^=3×64e22^
⇒e12^=163e22^\Rightarrow \hat{e_1^2}=\frac{16}{3}\hat{e_2^2}⇒e12^=316e22^
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