If a vector r has magnitude 14 and direction ratios 2, 3 and -6. Then…

The correct answer is Option (1) → Direction Cosines: $\left(\frac{2}{7}, \frac{3}{7}, -\frac{6}{7}\right)$; Components: $(4, 6, -12)$ ##

Here, $|{r}| = 14, a = 2k, b = 3k$ and $c = -6k$

$∴$ Direction cosines $l, m$ and $n$ are

$l = \frac{a}{|{r}|} = \frac{2k}{14} = \frac{k}{7}$

$m = \frac{b}{|{r}|} = \frac{3k}{14}$

and $n = \frac{c}{|{r}|} = \frac{-6k}{14} = \frac{-3k}{7}$

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 $\frac{2}{7}, \frac{3}{7}$ and $\frac{-6}{7}$.

since, ${r}$ makes an acute angle with X-axis and take $k = 2$]

$∵{r} = \hat{{r}} \cdot |{r}|$

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

$= \left( \frac{2}{7}\hat{{i}} + \frac{3}{7}\hat{{j}} - \frac{6}{7}\hat{{k}} \right) \cdot 14$

$= 4\hat{{i}} + 6\hat{{j}} - 12\hat{{k}}$