A vector $\vec{r}$ has magnitude 14 and direction ratios $2, 3, –6$. Find the direction cosines and components of $\vec{r}$ , given that $\vec{r}$ makes an acute angle with x-axis.

The solution given in my reference is: $(l,r,m)=\big(\frac{2}{7},\frac{3}{7},\frac{-6}{7}\big)$ and components are $4\hat{i}, 6\hat{j}, -12\hat{k}$

My Doubt $$ |\vec{r}|=14\neq \sqrt{4+9+36}=7 $$

So, are the terms direction ratios and scalar components not the same ?

My Understanding

$$ \vec{r}=\Big(x\hat{i}+y\hat{j}+z\hat{k}\Big)=|\vec{r}|\Big(l\hat{i}+m\hat{j}+n\hat{k}\Big)=|\vec{r}|\Big(\cos\alpha\hat{i}+\cos\beta\hat{j}+\cos\gamma\hat{k}\Big) $$ where $x=l|\vec{r}|, y=m|\vec{r}|, z=n|\vec{r}|$ are the direction ratios.


1 Answer 1


We have that

$$\vec r=(r_x,r_y,r_z)=k(2,3,-6)$$


$$|\vec r|=k\sqrt{4+9+36}=7k=14 \implies k=2$$


$$\vec r=(4,6,-12)\implies \frac{\vec r}{|\vec r|}=\left(\frac27,\frac37,-\frac67\right)$$

  • $\begingroup$ so the terms scalar components and direction ratios are not the same ? $\endgroup$
    – Sooraj S
    Commented Jul 4, 2018 at 20:31
  • $\begingroup$ @ss1729FRankly I never heard the term "direction ratios" but it seems that the term clearly refers to the components of a vector multiple of $\vec r$. $\endgroup$
    – user
    Commented Jul 4, 2018 at 20:39
  • $\begingroup$ I've seen "direction ratios" used exactly as gimusi described here (but separated by colons). For example, a vector might be specified as having direction ratios 1:-4:8 and magnitude 18, meaning the components are $(2, -8, 16)$, since $\sqrt{2^2 + (-8)^2 + 16^2} = \sqrt{324} = 18$. $\endgroup$ Commented Jul 5, 2018 at 7:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .