Question

A vector perpendicular to the vector ($\mathrm{i}+2\mathrm{j}$) and having magnitude $3\sqrt{5}$ units is _____

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Solution

Step 1: Given and assumeThe magnitude of r is $3\sqrt{5}$.Let,A vector, $\mathrm{r}=\mathrm{xi}+\mathrm{yj}$ perpendicular to $\mathrm{A}=\left(\mathrm{i}+2\mathrm{j}\right)$.Step 2: To findWe have to find a vector perpendicular to the vector ($\mathrm{i}+2\mathrm{j}$) and have magnitude $3\sqrt{5}$ units.Step 3: CalculationThe dot product of r and A must be zero because $\mathrm{Cos}90°=0$.Example., $\mathrm{r}.\mathrm{A}=\left(\mathrm{xi}+\mathrm{yj}\right).\left(\mathrm{i}+2\mathrm{j}\right)$then,$\mathrm{x}+2\mathrm{y}=0...\left(1\right)$According to the given information,$\left|\mathrm{r}\right|=\sqrt{\left({\mathrm{x}}^{2}+{\mathrm{y}}^{2}\right)}\phantom{\rule{0ex}{0ex}}=3\sqrt{5}$By taking square both sides, we get${\mathrm{x}}^{2}+{\mathrm{y}}^{2}=9×5\phantom{\rule{0ex}{0ex}}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}=45$From equation $\left(1\right)$,${\left(-2\mathrm{y}\right)}^{2}+{\mathrm{y}}^{2}=45\phantom{\rule{0ex}{0ex}}4{\mathrm{y}}^{2}+{\mathrm{y}}^{2}=45\phantom{\rule{0ex}{0ex}}5{\mathrm{y}}^{2}=45\phantom{\rule{0ex}{0ex}}{\mathrm{y}}^{2}=9\phantom{\rule{0ex}{0ex}}\mathrm{y}=3$By substituting the value of “y” in the equation $\left(1\right)$, we get$\mathrm{x}+2\left(3\right)=0\phantom{\rule{0ex}{0ex}}\mathrm{x}+6=0\phantom{\rule{0ex}{0ex}}\mathrm{x}=-6$Therefore, the vector perpendicular is $\mathrm{r}=\left(6\mathrm{i}-3\mathrm{j}\right)$.

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