Question

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

• A. $(2,-3,9)$
• B. $(6,-9,18)$
• C. $4(-2,-3,9)$
• D. $(-2,-3,-9)$

Answer 1

Answer: (B)

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)
$\therefore \hat{r}=\frac{\vec{r}}{|\vec{r}|}=\frac{a}{21}\hat{i}+\frac{b}{21}\hat{j}+\frac{c}{21}\hat{k}$
$\therefore$ 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 $\vec{r}=6\hat{i}-9\hat{j}+18\hat{k}$