Question

The direction cosines of the line of intersection of the planes \( x + 2y + z - 4 = 0 \) and \( 2x - y + z - 3 = 0 \) are:

• A. \( \left( \frac{3}{\sqrt{26}}, \frac{1}{\sqrt{26}}, \frac{-4}{\sqrt{26}} \right) \)
• B. \( \left( \frac{3}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{-1}{\sqrt{14}} \right) \)
• C. \( \left( \frac{3}{\sqrt{35}}, \frac{1}{\sqrt{35}}, \frac{-5}{\sqrt{35}} \right) \)
• D. \( \left( \frac{3}{\sqrt{22}}, \frac{-2}{\sqrt{22}}, \frac{3}{\sqrt{22}} \right) \)

Hint:

The direction cosines of a line are the normalized components of any non-zero vector parallel to the line. They are useful in determining the orientation of the line in three-dimensional space.

Answer 1

The Correct Option is C

To find the direction cosines of the line of intersection of the given planes, we compute the cross product of their normal vectors. 
Step 1: The normals to the planes \( x + 2y + z - 4 = 0 \) and \( 2x - y + z - 3 = 0 \) are \( \mathbf{n}_1 = (1, 2, 1) \) and \( \mathbf{n}_2 = (2, -1, 1) \), respectively. 
Step 2: Compute the cross product \( \mathbf{n}_1 \times \mathbf{n}_2 \) using the determinant of the matrix formed by the components of the normal vectors: \[ \mathbf{n}_1 \times \mathbf{n}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 1 \\ 2 & -1 & 1 \end{vmatrix} = \mathbf{i}(2 \cdot 1 - 1 \cdot (-1)) - \mathbf{j}(1 \cdot 1 - 1 \cdot 2) + \mathbf{k}(1 \cdot (-1) - 2 \cdot 2) \] \[ = \mathbf{i}(2 + 1) - \mathbf{j}(1 - 2) + \mathbf{k}(-1 - 4) \] \[ = 3\mathbf{i} + 1\mathbf{j} - 5\mathbf{k} \] Thus, \( \mathbf{n}_1 \times \mathbf{n}_2 = (3, 1, -5) \). 
Step 3: Normalize the vector \( \mathbf{n}_1 \times \mathbf{n}_2 = (3, 1, -5) \) to obtain the direction cosines: \[ \text{Magnitude of the vector} = \sqrt{3^2 + 1^2 + (-5)^2} = \sqrt{9 + 1 + 25} = \sqrt{35}. \] Thus, the direction cosines are: \[ \left( \frac{3}{\sqrt{35}}, \frac{1}{\sqrt{35}}, \frac{-5}{\sqrt{35}} \right). \]