Question

The projections of a line segment on X, Y and Z axes are 12, 4 and 3 respectively. The length and direction cosines of the line segment are
(a) $13;\frac{12}{13},\frac{4}{13},\frac{3}{13}$

(b) $19;\frac{12}{19},\frac{4}{19},\frac{3}{19}$

(c) $11;\frac{12}{11},\frac{14}{11},\frac{3}{11}$

(d) none of these

Answer 1

(a) $13;\frac{12}{13},\frac{4}{13},\frac{3}{13}$

If a line makes angles α, β and γ with the axes, then
${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1...\left(1\right)$

Let r be the length of the line segment. Then,

$$\begin{gathered} r\cos \alpha =12,r\cos \beta =4,r\cos \gamma =3...\left(2\right) \\ \Rightarrow {\left(r\cos \alpha \right)}^2+{\left(r\cos \beta \right)}^2+{\left(r\cos \gamma \right)}^2={12}^2+4^2+3^2 \\ \Rightarrow r^2\left({\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma \right)=169 \\ \Rightarrow r^2\left(1\right)=169\left[\mathrm{From}\left(1\right)\right] \\ \Rightarrow r=\sqrt{169} \\ \Rightarrow r=\pm 13 \\ \Rightarrow r=13(\mathrm{Since}\mathrm{length}\mathrm{cannot}\mathrm{be}\mathrm{negative}) \end{gathered}$$


Substituting r = 13 in (2), we get

$\cos \alpha =\frac{12}{13},\cos \beta =\frac{4}{13},\cos \gamma =\frac{1}{13}$

Thus, the direction cosines of the line are $\frac{12}{13},\frac{4}{13},\frac{1}{13}$.