Question

If a vector $\overrightarrow{r}$ has magnitude 14 and direction ratios 2, 3 and -6. Then, find the direction cosines and components of $\overrightarrow{r}$, given that $\overrightarrow{r}$ makes an acute angle with X- axis.

Answer 1

$\text{Here,}|\overrightarrow{r}|=14,\overrightarrow{a}=2k,\overrightarrow{3k}and\overrightarrow{c}=-6k\therefore \text{Direction cosines l, m and n are}l=\frac{\overrightarrow{a}}{|\overrightarrow{r}|}=\frac{2k}{14}=\frac{k}{7}m=\frac{\overrightarrow{b}}{|\overrightarrow{r}|}=\frac{3k}{14}\text{and}n=\frac{\overrightarrow{c}}{|\overrightarrow{r}|}=\frac{-6k}{14}=\frac{-3k}{7}$

Also, we know that

$l^2+m^2+n^2=1$

$\Rightarrow \frac{k^2}{49}+\frac{9k^2}{196}+\frac{9k^2}{49}=1$

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

$\Rightarrow k^2=\frac{196}{49}=4$

$\Rightarrow k=\pm 2$

So, the direction cosines (l, m, n) are $\frac{2}{7},\frac{3}{7}\text{and}\frac{-6}{7}.$

[since, $\overrightarrow{r}$ makes an acute angle with X-axis]

$\because \overrightarrow{r}=\hat{r}.|\overrightarrow{r}|$

$\therefore \overrightarrow{r}=(l\hat{i}+m\hat{j}+n\hat{k})|\overrightarrow{r}|$

$=(\frac{+2}{7}\hat{i}+\frac{3}{7}\hat{j}-\frac{6}{7}\hat{k}).14=+4\hat{i}+6\hat{j}-12\hat{k}$