Question

The direction cosines of the line joining the points $\left(4,3,-5\right)$ and $\left(-2,1,-8\right)$ are

• A. $\left(\frac{6}{7}\right),\left(\frac{2}{7}\right),\left(\frac{3}{7}\right)$
• B. $\left(\frac{2}{7}\right),\left(\frac{3}{7}\right),\left(\frac{-6}{7}\right)$
• C. $\left(\frac{6}{7}\right),\left(\frac{3}{7}\right),\left(\frac{2}{7}\right)$
• D. None of these

Answer 1

The correct option is A

$\left(\frac{6}{7}\right),\left(\frac{2}{7}\right),\left(\frac{3}{7}\right)$

Explanation for correct option

Finding the direction cosines of the line joining the three points

Given, $P(4,3,-5),Q(-2,1,-8)$

$D{R}^{\text{'}}s=\left(x_2-x_1,y_2-y_1,z_2-z_1\right)$
$\mathrm{PQ}$ is a vector and direction ratio of $\mathrm{PQ}$ is,

$$\begin{gathered} =(-2-4,1-3,-8+5) \\ =(-6,-2,-8) \end{gathered}$$

The distance between the points $P$ and $Q$ will be,
$$\begin{gathered} \mathrm{PQ}=\sqrt{{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2+{\left({\mathrm{z}}_2-{\mathrm{z}}_1\right)}^2} \\ =\sqrt{(-6{)}^2+(-2{)}^2+(-8{)}^2} \\ \mathrm{PQ}=7 \end{gathered}$$
Therefore, the D's of the vector $\mathrm{PQ}$ is $\mathbf{=}\mathbf{}\left(\frac{x_2-x_1}{\left|PQ\right|},\frac{y_2-y_1}{\left|PQ\right|},\frac{z_2-z_1}{\left|PQ\right|}\right)$or$\left(\frac{6}{7},\frac{2}{7},\frac{3}{7}\right)$
Hence, the correct answer is Option (A).