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  • A vector has magnitude 14 and direction ratios 2, 3, –6. Find the direction cosines and components of , given that makes an acute angle with x-axis.

A vector \overrightarrow{\mathrm{r}}  has magnitude 14 and direction ratios 2, 3, –6. Find the direction cosines and components of \overrightarrow{\mathrm{r}} , given that \overrightarrow{\mathrm{r}}  makes an acute angle with x-axis.

Answers (1)

Given that,

Magnitude of vector \overrightarrow{\mathrm{r}}  = 14

\Rightarrow|\overrightarrow{\mathrm{r}}|=14

Also, direction ratios = 2 : 3 : -6

\\ \begin{array}{l} \overrightarrow{\mathrm{a}}=2 \mathrm{k} \\ \overrightarrow{\mathrm{b}}=3 \mathrm{k} \end{array} \\ \overrightarrow{\mathrm{c}}=-6 \mathrm{k}\\

Also \overrightarrow{\mathrm{r}} can be defined as,

\overrightarrow{\mathrm{r}}=a \hat{\imath}+\mathrm{b} \hat{\jmath}+c \hat{\mathrm{k}}

Know that, the direction cosines of a vector are the cosines of the angles between the vector and the three coordinate axes.


 

∴, the direction cosines l, m and n are

\\ \mathrm{l}=\frac{\overrightarrow{\mathrm{a}}}{|\overrightarrow{\mathrm{r}}|} \\ \Rightarrow l=\frac{2 \mathrm{k}}{14}[\because]=2 \mathrm{k} \text { and }|\overrightarrow{\mathrm{r}}|=14_{\mathrm{J}} \\ \Rightarrow l=\frac{\mathrm{k}}{7} \\ \mathrm{~m}=\frac{\overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{r}}|}

\\ \Rightarrow \mathrm{m}=\frac{3 \mathrm{k}}{14}[\because \overrightarrow{\mathrm{b}}=3 \mathrm{k} \text { and }|\overrightarrow{\mathrm{r}}|=14 \mathrm{~g} \\ \mathrm{n}=\frac{\overrightarrow{\mathrm{c}}}{|\overrightarrow{\mathrm{r}}|} \\ \Rightarrow \mathrm{n}=-\frac{6 \mathrm{k}}{14}[\because]{\mathrm{c}}=-6 \mathrm{k} \text { and }|\overrightarrow{\mathrm{r}}|=14]

\\ \begin{aligned} &\Rightarrow \mathrm{n}=-\frac{3 \mathrm{k}}{7}\\ &\text { And we know that, }\\ &l^{2}+m^{2}+n^{2}=1\\ &\Rightarrow\left(\frac{\mathrm{k}}{7}\right)^{2}+\left(\frac{3 \mathrm{k}}{14}\right)^{2}+\left(-\frac{3 \mathrm{k}}{7}\right)^{2}=1 \end{aligned}

\\ \Rightarrow \frac{\mathrm{k}^{2}}{49}+\frac{9 \mathrm{k}^{2}}{196}+\frac{9 \mathrm{k}^{2}}{49}=1 \\ \Rightarrow \frac{4 \mathrm{k}^{2}+9 \mathrm{k}^{2}+36 \mathrm{k}^{2}}{196}=1 \\ \Rightarrow \frac{49 \mathrm{k}^{2}}{196}=1 \\ \Rightarrow 49 \mathrm{k}^{2}=196

\\ \Rightarrow \mathrm{k}^{2}=\frac{196}{49} \\ \Rightarrow \mathrm{k}^{2}=4 \\ \Rightarrow \mathrm{k}=\pm \sqrt{4} \\ \Rightarrow \mathrm{k}=\pm 2

Since, \overrightarrow{\mathrm{r}}  makes an acute angle with x-axis, then k will be positive.

\\ \begin{aligned} &\Rightarrow \mathrm{k}=2\\ &\text { The direction cosines are }\\ &l=\frac{\mathrm{k}}{7}=\frac{2}{7} \end{aligned}

\\ \begin{aligned} &\mathrm{m}=\frac{3 \mathrm{k}}{14}=\frac{3 \times 2}{14}=\frac{3}{7}\\ &\mathrm{n}=-\frac{3 \mathrm{k}}{7}=-\frac{3 \times 2}{7}=-\frac{6}{7}\\ &\text { The components of } \overrightarrow{\mathrm{r}} \text { can be found out by, }\\ &\overrightarrow{\mathrm{r}}=\hat{\mathrm{r}} \cdot|\overrightarrow{\mathrm{r}}| \end{aligned}

\\ \Rightarrow \overrightarrow{\mathrm{r}}=(l \hat{\imath}+\mathrm{m} \hat{\mathrm{j}}+\mathrm{n} \hat{\mathrm{k}})(14) \\ \Rightarrow \overrightarrow{\mathrm{r}}=\left(\frac{2}{7} \hat{\mathrm{r}}+\frac{3}{7} \hat{\mathrm{j}}-\frac{6}{7} \hat{\mathrm{k}}\right)(14) \\ \Rightarrow \overrightarrow{\mathrm{r}}=14 \times \frac{2}{7} \hat{\mathrm{i}}+14 \times \frac{3}{7} \hat{\mathrm{j}}-14 \times \frac{6}{7} \hat{\mathrm{k}} \\ \Rightarrow \overrightarrow{\mathrm{r}}=4 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-12 \hat{\mathrm{k}}

Thus, the direction cosines (l, m, n) are \left(\frac{2}{7}, \frac{3}{7},-\frac{6}{7}\right) ; and the components of \overrightarrow{\mathrm{r}}  are  (4,6,-12)

 

 

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