Question

A line passes through the points \( (6, -7, -1) \) and \( (2, -3, 1) \). Find the direction ratios and the direction cosines of the line. Show that the line does not pass through the origin.

Hint:

To find direction ratios, subtract the coordinates of the two points. For direction cosines, divide each direction ratio by the magnitude of the direction vector.

Answer 1

Step 1: Find the direction ratios.
The direction ratios of the line are obtained by subtracting the coordinates of the two given points: \[ \text{Direction ratios} = (2 - 6, -3 - (-7), 1 - (-1)) = (-4, 4, 2) \]

Step 2: Find the direction cosines.
The direction cosines are the ratios of the direction ratios to the magnitude of the direction vector. The magnitude of the direction vector is: \[ |\text{Direction vector}| = \sqrt{(-4)^2 + 4^2 + 2^2} = \sqrt{16 + 16 + 4} = \sqrt{36} = 6 \] Thus, the direction cosines are: \[ \cos \alpha = \frac{-4}{6} = -\frac{2}{3}, \cos \beta = \frac{4}{6} = \frac{2}{3}, \cos \gamma = \frac{2}{6} = \frac{1}{3} \]

Step 3: Show that the line does not pass through the origin.
For the line to pass through the origin, we must check if the point \( (0, 0, 0) \) satisfies the equation of the line. Using the parametric form of the line: \[ x = 6 - 4t, y = -7 + 4t, z = -1 + 2t \] Setting \( x = 0, y = 0, z = 0 \) and solving for \( t \): \[ 6 - 4t = 0 $\Rightarrow$ t = \frac{3}{2} \] Substitute \( t = \frac{3}{2} \) into the equations for \( y \) and \( z \): \[ y = -7 + 4 \times \frac{3}{2} = -7 + 6 = -1 \text{(not zero)} \] Thus, the line does not pass through the origin.

Final Answer: The direction ratios are \( (-4, 4, 2) \), and the direction cosines are \( \left( -\frac{2}{3}, \frac{2}{3}, \frac{1}{3} \right) \). The line does not pass through the origin.