Question
If the lines and are co-planar, then the distance of the plane containing these two lines from the point (, 0, 0) is :
Options
Solution
Key Concepts and Formulas
- Coplanarity of Two Lines: Two lines in 3D space, given by
$$\overrightarrow r = \overrightarrow{a_1} + \lambda \overrightarrow{b_1}$$and$$\overrightarrow r = \overrightarrow{a_2} + \mu \overrightarrow{b_2}$$, are coplanar if and only if the scalar triple product of the vector joining a point on the first line to a point on the second line ($$\overrightarrow{a_2} - \overrightarrow{a_1}$$) and their direction vectors ($$\overrightarrow{b_1}$$and$$\overrightarrow{b_2}$$) is zero. Mathematically:$$(\overrightarrow{a_2} - \overrightarrow{a_1}) \cdot (\overrightarrow{b_1} \times \overrightarrow{b_2}) = 0$$In Cartesian coordinates, this translates to:$$\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ b_{1x} & b_{1y} & b_{1z} \\ b_{2x} & b_{2y} & b_{2z} \end{vmatrix} = 0$$ - Equation of a Plane Containing Two Coplanar Lines: If two lines are coplanar, the equation of the plane containing them can be found using a point from one of the lines (e.g.,
$$\overrightarrow{a_1}$$) and their direction vectors ($$\overrightarrow{b_1}$$and$$\overrightarrow{b_2}$$). For any point$$\overrightarrow r = (x,y,z)$$on the plane:$$(\overrightarrow r - \overrightarrow{a_1}) \cdot (\overrightarrow{b_1} \times \overrightarrow{b_2}) = 0$$In Cartesian form:$$\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ b_{1x} & b_{1y} & b_{1z} \\ b_{2x} & b_{2y} & b_{2z} \end{vmatrix} = 0$$ - Distance of a Point from a Plane: The perpendicular distance of a point
$$P(x_0, y_0, z_0)$$from a plane$$Ax + By + Cz + D = 0$$is given by the formula:$$d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$$
Step-by-Step Solution
Step 1: Extract Parameters from the Given Lines
We begin by identifying the position vectors ($$\overrightarrow{a_1}, \overrightarrow{a_2}$$) and direction vectors ($$\overrightarrow{b_1}, \overrightarrow{b_2}$$) for each line, as these are crucial for applying the coplanarity condition and finding the plane equation.
The first line is $$\overrightarrow r = \left( {\widehat i - \widehat j + \widehat k} \right) + \lambda \left( {3\widehat j - \widehat k} \right)$$.
Comparing with $$\overrightarrow r = \overrightarrow{a_1} + \lambda \overrightarrow{b_1}$$:
$$\overrightarrow{a_1} = \widehat i - \widehat j + \widehat k \implies (x_1, y_1, z_1) = (1, -1, 1)$$
$$\overrightarrow{b_1} = 0\widehat i + 3\widehat j - \widehat k \implies (b_{1x}, b_{1y}, b_{1z}) = (0, 3, -1)$$
The second line is $$\overrightarrow r = \left( {\alpha \widehat i - \widehat j} \right) + \mu \left( {2\widehat i - 3\widehat k} \right)$$.
Comparing with $$\overrightarrow r = \overrightarrow{a_2} + \mu \overrightarrow{b_2}$$:
$$\overrightarrow{a_2} = \alpha \widehat i - \widehat j + 0\widehat k \implies (x_2, y_2, z_2) = (\alpha, -1, 0)$$
$$\overrightarrow{b_2} = 2\widehat i + 0\widehat j - 3\widehat k \implies (b_{2x}, b_{2y}, b_{2z}) = (2, 0, -3)$$
Next, we calculate the vector $$\overrightarrow{a_2} - \overrightarrow{a_1}$$, which is needed for the coplanarity condition:
$$\overrightarrow{a_2} - \overrightarrow{a_1} = (\alpha \widehat i - \widehat j + 0\widehat k) - (\widehat i - \widehat j + \widehat k)$$
$$\overrightarrow{a_2} - \overrightarrow{a_1} = (\alpha - 1)\widehat i + (-1 - (-1))\widehat j + (0 - 1)\widehat k$$
$$\overrightarrow{a_2} - \overrightarrow{a_1} = (\alpha - 1)\widehat i + 0\widehat j - \widehat k \implies (x_2-x_1, y_2-y_1, z_2-z_1) = (\alpha - 1, 0, -1)$$
Step 2: Apply Coplanarity Condition to Find $$\alpha$$
Since the lines are given to be coplanar, we set the determinant of the coefficients to zero:
$$\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ b_{1x} & b_{1y} & b_{1z} \\ b_{2x} & b_{2y} & b_{2z} \end{vmatrix} = 0$$
Substitute the values obtained in Step 1:
$$\begin{vmatrix} \alpha - 1 & 0 & -1 \\ 0 & 3 & -1 \\ 2 & 0 & -3 \end{vmatrix} = 0$$
Expand the determinant. It's easiest to expand along the second column due to the zeros:
$$0 \cdot C_{12} + 3 \cdot (-1)^{2+2} \begin{vmatrix} \alpha - 1 & -1 \\ 2 & -3 \end{vmatrix} + 0 \cdot C_{32} = 0$$
$$3 \cdot [(\alpha - 1)(-3) - (-1)(2)] = 0$$
$$3 \cdot [-3\alpha + 3 + 2] = 0$$
$$3 \cdot [-3\alpha + 5] = 0$$
$$-3\alpha + 5 = 0$$
$$3\alpha = 5$$
$$\alpha = \frac{5}{3}$$
Thus, the value of $$\alpha$$ is $$\frac{5}{3}$$.
Step 3: Find the Equation of the Plane Containing the Two Lines
The plane containing the two coplanar lines can be determined using a point from one line (e.g., $$\overrightarrow{a_1} = (1, -1, 1)$$) and the direction vectors of the lines ($$\overrightarrow{b_1} = (0, 3, -1)$$ and $$\overrightarrow{b_2} = (2, 0, -3)$$). The equation of the plane is given by:
$$\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ b_{1x} & b_{1y} & b_{1z} \\ b_{2x} & b_{2y} & b_{2z} \end{vmatrix} = 0$$
Substitute the values:
$$\begin{vmatrix} x-1 & y-(-1) & z-1 \\ 0 & 3 & -1 \\ 2 & 0 & -3 \end{vmatrix} = 0$$
$$\begin{vmatrix} x-1 & y+1 & z-1 \\ 0 & 3 & -1 \\ 2 & 0 & -3 \end{vmatrix} = 0$$
Expand the determinant along the first row:
$$(x-1) \begin{vmatrix} 3 & -1 \\ 0 & -3 \end{vmatrix} - (y+1) \begin{vmatrix} 0 & -1 \\ 2 & -3 \end{vmatrix} + (z-1) \begin{vmatrix} 0 & 3 \\ 2 & 0 \end{vmatrix} = 0$$
$$(x-1) ((3)(-3) - (-1)(0)) - (y+1) ((0)(-3) - (-1)(2)) + (z-1) ((0)(0) - (3)(2)) = 0$$
$$(x-1) (-9) - (y+1) (2) + (z-1) (-6) = 0$$
$$-9(x-1) - 2(y+1) - 6(z-1) = 0$$
$$-9x + 9 - 2y - 2 - 6z + 6 = 0$$
$$-9x - 2y - 6z + 13 = 0$$
Multiplying by -1 to express it in standard form ( with ):
$$9x + 2y + 6z - 13 = 0$$
This is the equation of the plane containing the two given lines.
Step 4: Calculate the Distance of the Point $$(\alpha, 0, 0)$$ from the Plane
The point is $$(\alpha, 0, 0) = \left(\frac{5}{3}, 0, 0\right)$$.
The equation of the plane is $$9x + 2y + 6z - 13 = 0$$.
Using the distance formula $$d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$$:
Here, $$x_0 = \frac{5}{3}, y_0 = 0, z_0 = 0$$ and $$A = 9, B = 2, C = 6, D = -13$$.
$$d = \frac{\left|9\left(\frac{5}{3}\right) + 2(0) + 6(0) - 13\right|}{\sqrt{9^2 + 2^2 + 6^2}}$$
$$d = \frac{|15 + 0 + 0 - 13|}{\sqrt{81 + 4 + 36}}$$
$$d = \frac{|2|}{\sqrt{121}}$$
$$d = \frac{2}{11}$$
Common Mistakes & Tips
- Sign Errors in Determinants: Be extremely careful with signs when expanding determinants, especially with negative terms. A small sign error can propagate through the entire calculation.
- Vector Components: Ensure correct identification of
$$\widehat i, \widehat j, \widehat k$$components, especially when a component is zero (e.g.,$$3\widehat j - \widehat k$$means$$0\widehat i + 3\widehat j - \widehat k$$). - Coplanarity vs. Plane Equation: Remember the slight difference in the determinant setup: for coplanarity, the first row is
$$(\overrightarrow{a_2} - \overrightarrow{a_1})$$, while for the plane equation, it's$$(\overrightarrow r - \overrightarrow{a_1})$$.
Summary
We first utilized the condition for two lines to be coplanar, expressed as a scalar triple product (determinant form), to solve for the unknown parameter $$\alpha$$. With $$\alpha = 5/3$$, we then determined the equation of the plane containing these two lines using a point from one of the lines and their direction vectors. Finally, we applied the formula for the perpendicular distance from a point to a plane to find the distance of the point $$(\alpha, 0, 0)$$ from the derived plane. Our calculations yielded a distance of $$2/11$$.
The final answer is which corresponds to option (B).