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Question Number 144166 by liberty last updated on 22/Jun/21

$$\mathrm{Given}\overrightarrow{\mathrm{p}}=\sqrt{2}\hat{\mathrm{i}}+2\sqrt{3}\hat{\mathrm{j}}+\sqrt{3}\hat{\mathrm{k}}\mathrm{\&}$$ $$\overrightarrow{\mathrm{q}}=\mathrm{a}\hat{\mathrm{i}}+\hat{\mathrm{j}}+2\hat{\mathrm{k}}.\mathrm{If}{\mathrm{proj}}_{\overrightarrow{\mathrm{q}}}\overrightarrow{\mathrm{p}}=\frac{2\sqrt{2}}{9}\overrightarrow{\mathrm{q}}$$ $$\mathrm{then}\mid \overrightarrow{\mathrm{q}}\mid =?$$

Answered by benjo_mathlover last updated on 23/Jun/21

$$\left[\frac{\overrightarrow{\mathrm{p}}.\overrightarrow{\mathrm{q}}}{\mid \overrightarrow{\mathrm{q}}{\mid }^2}\right]\overrightarrow{\mathrm{q}}=\frac{2\sqrt{2}}{9}\overrightarrow{\mathrm{q}}$$ $$\left[\frac{\mathrm{a}\sqrt{2}+2\sqrt{3}+2\sqrt{3}}{5+{\mathrm{a}}^2}\right]=\frac{2\sqrt{2}}{9}$$ $$9\mathrm{a}\sqrt{2}+36\sqrt{3}=10\sqrt{2}+2\sqrt{2}{\mathrm{a}}^2$$ $$2\sqrt{2}{\mathrm{a}}^2-9\sqrt{2}\mathrm{a}+10\sqrt{2}-36\sqrt{3}=0$$ $$\mathrm{a}=\frac{9\sqrt{2}\pm \sqrt{162-4(40-73\sqrt{6})}}{4\sqrt{2}}$$ $$\mathrm{a}=\frac{9\sqrt{2}\pm \sqrt{292\sqrt{6}+2}}{4\sqrt{2}}$$

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