The line, that is coplanar to the line \(\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}\) is
- \(\frac{x-1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}\)
- \(\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}\)
- \(\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{4}\)
- \(\frac{x+1}{1}=\frac{y-2}{2}=\frac{z-5}{5}\)
The Correct Option is B
Solution and Explanation
The condition for two lines to be coplanar is that the scalar triple product of the direction ratios of the two lines, along with the vector joining any two points on the lines, must be zero.
We are given the first line in symmetric form: \[ \frac{x+3}{-3} = \frac{y-1}{1} = \frac{z-5}{5} \] The direction ratios of this line are \( \mathbf{a_1} = (-3, 1, 5) \). Let’s calculate the condition for coplanarity.
The second line is of the form: \[ \frac{x+1}{-1} = \frac{y-2}{2} = \frac{z-5}{5} \] The direction ratios of the second line are \( \mathbf{a_2} = (-1, 2, 5) \). Now, to check for coplanarity, we apply the condition of coplanarity for two lines.
The condition involves the scalar triple product of the direction ratios of the two lines and the vector joining any point on the first line to a point on the second line.
Let’s take points on the lines: - For the first line, take \( P_1(-3, 1, 5) \).
- For the second line, take \( P_2(-1, 2, 5) \). The vector joining these points is \( \overrightarrow{P_1P_2} = (-1 - (-3), 2 - 1, 5 - 5) = (2, 1, 0) \).
The scalar triple product condition for coplanarity is: \[ \left| \begin{matrix} x_2 - x_1 & a_1 & a_2 y_2 - y_1 & b_1 & b_2 z_2 - z_1 & c_1 & c_2 \end{matrix} \right| = 0 \]
Substitute the values into the determinant: \[ \begin{vmatrix} 2 & -3 & -1 1 & 1 & 2 0 & 5 & 5 \end{vmatrix} \]
Now, calculate the determinant: \[ = 2 \begin{vmatrix} 1 & 2 5 & 5 \end{vmatrix} - (-3) \begin{vmatrix} 1 & 2 0 & 5 \end{vmatrix} + (-1) \begin{vmatrix} 1 & 1 0 & 5 \end{vmatrix} \] \[ = 2 \left( (1 \times 5) - (2 \times 5) \right) + 3 \left( (1 \times 5) - (2 \times 0) \right) + (-1) \left( (1 \times 5) - (1 \times 0) \right) \] \[ = 2 \left( 5 - 10 \right) + 3 \left( 5 - 0 \right) + (-1) \left( 5 - 0 \right) \] \[ = 2(-5) + 3(5) - 1(5) \] \[ = -10 + 15 - 5 = 0 \] Since the determinant is zero, the two lines are coplanar.
Thus, the correct option is (2).
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