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Question-126859




Question Number 126859 by sdfg last updated on 24/Dec/20
Answered by mr W last updated on 24/Dec/20
G=10×sin 60°×cos 45° i+10×sin 60°×cos 45° j+10×cos 60° k  ⇒G=((5(√6))/2) i+((5(√6))/2) j+5 k    H=−15×sin 45°×sin 30° i+15×sin 45°×cos 30° j+15×cos 45° k  ⇒H=−((15(√2))/4) i+((15(√6))/4) j+((15(√2))/2) k
$${G}=\mathrm{10}×\mathrm{sin}\:\mathrm{60}°×\mathrm{cos}\:\mathrm{45}°\:{i}+\mathrm{10}×\mathrm{sin}\:\mathrm{60}°×\mathrm{cos}\:\mathrm{45}°\:{j}+\mathrm{10}×\mathrm{cos}\:\mathrm{60}°\:{k} \\ $$$$\Rightarrow\boldsymbol{{G}}=\frac{\mathrm{5}\sqrt{\mathrm{6}}}{\mathrm{2}}\:\boldsymbol{{i}}+\frac{\mathrm{5}\sqrt{\mathrm{6}}}{\mathrm{2}}\:\boldsymbol{{j}}+\mathrm{5}\:\boldsymbol{{k}} \\ $$$$ \\ $$$${H}=−\mathrm{15}×\mathrm{sin}\:\mathrm{45}°×\mathrm{sin}\:\mathrm{30}°\:{i}+\mathrm{15}×\mathrm{sin}\:\mathrm{45}°×\mathrm{cos}\:\mathrm{30}°\:{j}+\mathrm{15}×\mathrm{cos}\:\mathrm{45}°\:{k} \\ $$$$\Rightarrow\boldsymbol{{H}}=−\frac{\mathrm{15}\sqrt{\mathrm{2}}}{\mathrm{4}}\:\boldsymbol{{i}}+\frac{\mathrm{15}\sqrt{\mathrm{6}}}{\mathrm{4}}\:\boldsymbol{{j}}+\frac{\mathrm{15}\sqrt{\mathrm{2}}}{\mathrm{2}}\:\boldsymbol{{k}} \\ $$

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