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  4. A vector vecr has length 21 and its dr’s are < 2, -3, 6 >, the component of vector vecr are

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Q. A vector $\vec{r}$ has length $21$ and its dr’s are $<\,2, -3, 6\,>$, the component of vector $\vec{r}$ are

Three Dimensional Geometry

Solution:

Let $\vec{r} =a \hat{i}+b \hat{j}+c \hat{k} $
$\therefore \left|\vec{r}\right|=\sqrt{a^{2}+b^{2}+c^{2}}=21$ (given)
$\therefore \hat{r} =\frac{\vec{r}}{\left|\vec{r}\right|}= \frac{a}{21} \hat{i} +\frac{b}{21 } \hat{j} +\frac{c}{21} \hat{k} $
$\therefore $ dc’s are $ \frac{a}{21}, \frac{b}{21}, \frac{c}{21}$
Also the dr’s are $2, - 3, 6$
$\therefore \frac{a}{21}=2\lambda, \frac{b}{21}=-3\lambda, \frac{c}{21}=6\lambda $
$Rightarrow \left(2\lambda\right)^{2}+\left(-3\lambda\right)^{2}+\left(6\lambda\right)^{2}=1$
$=\lambda = \pm \frac{1}{7}$
$\therefore a=6, b=-9, c=18 $
$\therefore \vec{r} =6 \hat{i}-9 \hat{j} +18 \hat{k}$
Alternative Solution :
dr’s are $<\,2, -3, 6\,>$
$\therefore $ dc's are $l=\frac{2}{7}, m=\frac{-3}{7}, n=\frac{6}{7}$
$\therefore \vec{r}=\left|\vec{r}\right|\left(l \,\hat{i}+m \,\hat{j}+n\,\hat{k}\right)$
$=21\left(\frac{2}{7} \hat{i}-\frac{3}{7} \hat{j}+\frac{6}{7} \hat{k}\right)$
or $ \vec{r}=6 \hat{i}-9 \hat{j} +18 \hat{k}$