Question

If \( L_1 \) and \( L_2 \) are two lines which pass through origin and have direction ratios \( (3, 1, -5) \) and \( (2, 3, -1) \) respectively, then the equation of the plane containing \( L_1 \) and \( L_2 \) is:

• A. \( 4x + 5y - 63 = 0 \)
• B. \( 5x - y + 3z = 0 \)
• C. \( 2x - y + z = 0 \)
• D. \( x - 5y + 3z = 0 \)

Hint:

In three dimensions, the normal vector to a plane can be obtained by the cross product of two non-collinear vectors lying on the plane. This vector provides the coefficients for the plane equation \( ax + by + cz = d \).

Answer 1

The Correct Option is C

To find the equation of the plane containing the lines \( L_1 \) and \( L_2 \) which pass through the origin, we use the cross product of their direction vectors. 
Step 1: The direction vectors of \( L_1 \) and \( L_2 \) are \( \mathbf{v}_1 = (3, 1, -5) \) and \( \mathbf{v}_2 = (2, 3, -1) \), respectively. 
Step 2: Compute the cross product to find a normal to the plane: \[ \mathbf{n} = \mathbf{v}_1 \times \mathbf{v}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 1 & -5 \\ 2 & 3 & -1 \end{vmatrix} = \mathbf{i}(1 \times (-1) - 3 \times (-5)) - \mathbf{j}(3 \times (-1) - 2 \times (-5)) + \mathbf{k}(3 \times 3 - 1 \times 2) \] \[ = \mathbf{i}( -1 + 15) - \mathbf{j}(-3 + 10) + \mathbf{k}(9 - 2) \] \[ = 16\mathbf{i} - 13\mathbf{j} + 7\mathbf{k} \] Step 3: The equation of the plane will be \( 16x - 13y + 7z = 0 \), since the normal vector to the plane is \( \mathbf{n} = (16, -13, 7) \). Now, simplify by dividing the entire equation by 8 to reduce the coefficients: \[ 2x - y + z = 0 \] Thus, the equation of the plane is \( 2x - y + z = 0 \).