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  • A vector r is inclined at equal angles to the three axes. If the magnitude of r is 2 root 3 units, find .vector r

A vector \overrightarrow{\mathrm{r}}  is inclined at equal angles to the three axes. If the magnitude of \overrightarrow{\mathrm{r}}  is  2\sqrt3 units, find \overrightarrow{\mathrm{r}} .

Answers (1)

Given that,

Magnitude of \overrightarrow{\mathrm{r}} = 2\sqrt3

\Rightarrow|\vec{r}|=2 \sqrt{3}

Also, given that
Vector \overrightarrow{\mathrm{r}}  is equally inclined to the three axes.

This means, direction cosines of the unit vector \overrightarrow{\mathrm{r}}  will be same. The direction cosines are (l, m, n).

\mathrm{l}=\mathrm{m}=\mathrm{n}

The direction cosines of a vector are simply the cosines of the angles between the vector and the three coordinate axes.

We know the relationship between direction cosines is,

\\ l^{2}+\mathrm{m}^{2}+\mathrm{n}^{2}=1 \\ \Rightarrow l^{2}+l^{2}+l^{2}=1[\because \mathrm{l}=\mathrm{m}=\mathrm{n}] \\ \Rightarrow 3.l^{2}=1 \\ \Rightarrow l=\pm \frac{1}{\sqrt{3}}

Also, we know that \overrightarrow{\mathrm{r}}  is represented in terms of direction cosines as,
\\ \hat{\mathrm{r}}=l \hat{\mathrm{u}}+\mathrm{m} \hat{\mathrm{j}}+\mathrm{n} \hat{\mathrm{k}}\\ \Rightarrow \hat{\mathrm{r}}=\pm \frac{1}{\sqrt{3}} \hat{\imath} \pm \frac{1}{\sqrt{3}} \hat{\jmath} \pm \frac{1}{\sqrt{3}} \hat{\mathrm{k}}

\\ \text{We are familiar with the formula,} $$ \hat{\mathrm{r}}=\frac{\overrightarrow{\mathrm{r}}}{|\overrightarrow{\mathrm{r}}|} $$

\\ \hat{\mathrm{r}}=\frac{\overrightarrow{\mathrm{r}}}{|\overrightarrow{\mathrm{r}}|} \\\text{To find } \overrightarrow{\mathrm{r}} \\ \Rightarrow \overrightarrow{\mathrm{r}}=\hat{\mathrm{r}}|\overrightarrow{\mathrm{r}}| \\ $Substituting values of$ |\overrightarrow{\mathrm{r}}| and \hat{\mathrm{r}}

\\ \begin{aligned} &\overrightarrow{\mathrm{r}}=\left(\pm \frac{1}{\sqrt{3}} \hat{\imath} \pm \frac{1}{\sqrt{3}} \hat{\jmath} \pm \frac{1}{\sqrt{3}} \hat{\mathrm{k}}\right)(2 \sqrt{3})\\ &\Rightarrow \overrightarrow{\mathrm{r}}=\pm \frac{1}{\sqrt{3}}(\hat{\imath}+\hat{\jmath}+\hat{\mathrm{k}})(2 \sqrt{3})\\ &\Rightarrow \overrightarrow{\mathrm{r}}=\pm 2(\hat{\imath}+\hat{\jmath}+\hat{\mathrm{k}})\\ &\text { Thus, the value of } \overrightarrow{\mathrm{r}}_{\text {is }} \pm 2(\hat{\imath}+\hat{\jmath}+\hat{\mathrm{k}}) \end{aligned}

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