If a vector ${r}$ has magnitude $14$ and direction ratios $2, 3$ and $-6$. Then, find the direction cosines and components of ${r}$, given that ${r}$ makes an acute angle with X-axis. |
Direction Cosines: $\left(\frac{2}{7}, \frac{3}{7}, -\frac{6}{7}\right)$; Components: $(4, 6, -12)$ Direction Cosines: $\left(-\frac{2}{7}, -\frac{3}{7}, \frac{6}{7}\right)$; Components: $(-4, -6, 12)$ Direction Cosines: $\left(\frac{2}{7}, \frac{3}{7}, \frac{6}{7}\right)$; Components: $(4, 6, 12)$ Direction Cosines: $\left(\frac{2}{49}, \frac{3}{49}, -\frac{6}{49}\right)$; Components: $(\frac{4}{7}, \frac{6}{7}, -\frac{12}{7})$ |
Direction Cosines: $\left(\frac{2}{7}, \frac{3}{7}, -\frac{6}{7}\right)$; Components: $(4, 6, -12)$ |
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 $k = 2$] $∵{r} = \hat{{r}} \cdot |{r}|$ $∴{r} = (l \hat{{i}} + m \hat{{j}} + n \hat{{k}}) |{r}|$ $= \left( \frac{2}{7}\hat{{i}} + \frac{3}{7}\hat{{j}} - \frac{6}{7}\hat{{k}} \right) \cdot 14$ $= 4\hat{{i}} + 6\hat{{j}} - 12\hat{{k}}$ |