Question

The direction cosines of a line passing through two points $P\left(x_1,y_1,z_1\right)$ and $Q\left(x_2,y_2,z_2\right)$ are

• A. $\left(x_2-x_1\right),\left(y_2-y_1\right),\left(z_2-z_1\right)$
• B. $\left(x_2+x_1\right),\left(y_2+y_1\right),\left(z_2+z_1\right)$
• C. $\frac{x_2-x_1}{PQ},\frac{y_2-y_1}{PQ},\frac{z_2-z_1}{PQ}$
• D. $\frac{x_2+x_1}{PQ},\frac{y_2+y_1}{PQ},\frac{z_2+z_1}{PQ}$

Answer 1

Answer: (C)

Since, one and only one line passes through two given points, we can determine the direction cosines of a line passing through the given points $P\left(x_1,y_1,z_1\right)$ and $Q\left(x_2,y_2,z_2\right)$ as follows

Let $l,m$ and $n$ be the direction cosines of the line $PQ$ and let it makes angles $\alpha ,\beta$ and $\gamma$ with the $X,Y$ and Z-axes, respectively.
Draw perpendiculars from $P$ and $Q$ to $xy$-plane to meet at $R$ and $S$. Draw a perpendicular from $P$ to $QS$ to meet at $N$. Now, in right angled $△PNQ,\angle PQN=\gamma$ from (figure).
Therefore, $\cos \gamma =\frac{NQ}{PQ}=\frac{z_2-z_1}{PQ}$
Similarly, $\cos \alpha =\frac{x_2-x_1}{PQ}$
and $\cos \beta =\frac{y_2-y_1}{PQ}$
Hence, the direction cosines of the line segment joining the points $P\left(x_1,y_1,z_1\right)$ and $Q\left(x_2,y_2,z_2\right)$ are
$\frac{x_2-x_1}{PQ},\frac{y_2-y_1}{PQ},\frac{z_2-z_1}{PQ}$
$\text{ where, }PQ=\sqrt{\frac{x_2-x_1}{PQ},\frac{y_2-y_1}{PQ},\frac{z_2-z_1}{PQ}}$