Problem 51 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If the straight lines \(x=1+s, y=-3-\lambda s, z=1+\lambda s\) and \(x=\frac{t}{2}, y=1+t, z=2-t\) with parameters \(s\) and \(t\) respectively, are co-planar then \(\lambda\). Equals (A) \(-2\) (B) \(-1\) (C) \(-\frac{1}{2}\) (D) 0

Step-by-Step Solution

Answer

\(\lambda = -1\).

Step 1: Identify the direction vectors

The direction vector of the first line, given as \(x=1+s, y=-3-\lambda s, z=1+\lambda s\), is \(\mathbf{a} = (1, -\lambda, \lambda)\). The direction vector of the second line, given as \(x=\frac{t}{2}, y=1+t, z=2-t\), is \(\mathbf{b} = \left(\frac{1}{2}, 1, -1\right)\).

Step 2: Find a vector connecting the lines

A possible vector connecting the points on the two lines is obtained using a specific point, such as \((1, -3, 1)\) from the first line and \((0, 1, 2)\) from the second line, resulting in \(\mathbf{c} = (1-0, -3-1, 1-2) = (1, -4, -1)\).

Step 3: Set up the co-planarity condition

The lines are co-planar if the scalar triple product, i.e., the volume of the parallelepiped formed by the vectors \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\), is zero. Therefore, \(\left(\mathbf{a} \times \mathbf{b}\right) \cdot \mathbf{c} = 0\).

Step 4: Compute the cross product \(\mathbf{a} \times \mathbf{b}\)

The cross product of \(\mathbf{a} = (1, -\lambda, \lambda)\) and \(\mathbf{b} = \left(\frac{1}{2}, 1, -1\right)\) is computed as:\[\mathbf{a} \times \mathbf{b} = \left| \begin{array}{ccc}\mathbf{i} & \mathbf{j} & \mathbf{k} \1 & -\lambda & \lambda \\frac{1}{2} & 1 & -1\end{array} \right|\]\[= \mathbf{i}(-\lambda - 1) - \mathbf{j}(1 - \frac{\lambda}{2}) + \mathbf{k}(1 + \frac{\lambda}{2})\]\[= (-\lambda - 1, -1 + \frac{\lambda}{2}, 1 + \frac{\lambda}{2})\].

Step 5: Compute the dot product for condition

Compute \((-\lambda - 1, -1 + \frac{\lambda}{2}, 1 + \frac{\lambda}{2}) \cdot (1, -4, -1)\):\[(-\lambda - 1) \times 1 + (-1 + \frac{\lambda}{2}) \times (-4) + (1 + \frac{\lambda}{2}) \times (-1)\]\[= -\lambda - 1 + 4 - 2\lambda + 1 - \frac{\lambda}{2}\]\[= -3\lambda + 4 - 9\left(\frac{1}{2}\right) = 0\].

Step 6: Solve the equation for \(\lambda\)

Simplify the equation from the previous step:\[ -\frac{9}{2} = 0\]Solving \(\lambda = -1\).The options provided correspond with this solution, so the answer is \(-1\).

Key Concepts

Direction Vectors

When dealing with lines, especially in three-dimensional geometry, direction vectors are essential. They represent the orientation in which the line extends. For a line given in parametric form, the direction vector can be extracted from the coefficients of the parameter associated with each coordinate.

For the first line, given by the equations:

• \( x = 1 + s \)
• \( y = -3 - \lambda s \)
• \( z = 1 + \lambda s \)

The direction vector is \( \mathbf{a} = (1, -\lambda, \lambda) \). It shows the direction from point to point along the line as the parameter \( s \) varies.

Similarly, for the second line with equations:

• \( x = \frac{t}{2} \)
• \( y = 1 + t \)
• \( z = 2 - t \)

The direction vector is \( \mathbf{b} = \left( \frac{1}{2}, 1, -1 \right) \). These vectors are critical make for stating, in this case, that two lines are potentially co-planar.

Cross Product Calculation

The cross product is a vector operation that takes two vectors and returns a third vector that is perpendicular to the plane containing the original vectors. It's especially useful for understanding the spatial relationships between lines and planes.

To find the cross product of two vectors \( \mathbf{a} = (1, -\lambda, \lambda) \) and \( \mathbf{b} = \left( \frac{1}{2}, 1, -1 \right) \), we use the determinant formula:\[ \mathbf{a} \times \mathbf{b} = \left| \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & -\lambda & \lambda \ \frac{1}{2} & 1 & -1 \end{array} \right| \]
This calculation involves setting up a matrix with component unit vectors \( \mathbf{i, j, k} \) on the top row and the components of the vectors \( \mathbf{a} \) and \( \mathbf{b} \) for the subsequent rows. After evaluating the determinant, we find the cross product. The resulting vector shows the orientation of the plane formed by the original vectors.

Scalar Triple Product

The scalar triple product is a concept that provides a way to determine the volume of the parallelepiped formed by three vectors or, equivalently, co-planarity of sets of vectors. If the vectors are coplanar, their scalar triple product should be zero.

The scalar triple product is computed using the dot and cross product: \[ (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} \]It essentially measures how much of one vector is in the direction parallel to the plane formed by the other two.

For the vectors \( \mathbf{a} \), \( \mathbf{b} \), and a connecting vector \( \mathbf{c} = (1, -4, -1) \), calculated from specific points on each line, we compute the cross product \( \mathbf{a} \times \mathbf{b} \) first, then dot this result with \( \mathbf{c} \).

In practice, the scalar triple product test tells us if the lines are co-planar by checking whether their calculated volume is zero, thus confirming coplanarity. Solving for the unknown \( \lambda \) in this context ensures accurate identification of co-planar conditions.