Problem 68 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If the lines \(\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}\) and \(\frac{x-1}{k}=\frac{y-4}{2}\) \(=\frac{z-5}{1}\) are coplanar, then \(k\) can have [2013] (A) exactly one value (B) exactly two values (C) exactly three values (D) any value

Step-by-Step Solution

Answer

(D) any value

Step 1: Identify Direction Ratios

Each line is given in a symmetric form. For the first line, we have direction ratios as \((1, 1, -k)\). For the second line, the direction ratios are \((k, 2, 1)\).

Step 2: Identify a Condition for Coplanarity of Lines

Two lines are coplanar if the scalar triple product of their direction vectors and the vector formed by any point on one line and any point on the other line is zero. The condition is \((b_1 - a_1, b_2 - a_2, b_3 - a_3) \cdot [(d_1 \times e_1) + (d_2 \times e_2) + (d_3 \times e_3)] = 0\) where \((a_1, a_2, a_3)\) and \((b_1, b_2, b_3)\) are points on the lines and \((d_1, d_2, d_3)\), \((e_1, e_2, e_3)\) are direction ratios.

Step 3: Choose Points on the Lines

From the first line, we choose point \(P(2, 3, 4)\) by letting the parameter in symmetric form be 0. From the second line, choose point \(Q(1, 4, 5)\) by letting the parameter also be 0.

Step 4: Calculate Vector between Points

We need the vector formed by points on both lines: \(\vec{PQ} = (1 - 2, 4 - 3, 5 - 4) = (-1, 1, 1)\).

Step 5: Compute Cross Product of Direction Ratios

Find the cross product of the direction vectors: \[(1, 1, -k) \times (k, 2, 1) = |i \quad j \quad k| | 1 \quad 1 \quad -k| | k \quad 2 \quad 1|\].

Step 6: Evaluate the Determinant for the Cross Product

Calculate:\(\vec{r} = i ((1)(1) - (-k)(2)) - j ((-k)(k) - (1)) + k((1)(2) - (1)(k))\). This simplifies to \(\vec{r} = (1 + 2k)i + (k^2 + 1)j + (2-k)k\).

Step 7: Compute the Scalar Triple Product

The scalar triple product \(\vec{PQ} \cdot \vec{r}\) is:\((-1)(1 + 2k) + 1(k^2 + 1) + 1(2-k)k = 0\).This further simplifies to\(-1 - 2k + k^2 + 1 + 2k -k^2 = 0\).

Step 8: Simplify and Solve

Combine like terms to get:\(0 = 0\).The expression holds for any \(k\), indicating that the condition for coplanarity is satisfied for any \(k\).

Step 9: Determine the Solution

Since the scalar triple product results in an identity, \(k\) can be any value. Therefore, \(k\) can take an infinite number of values.

Key Concepts

Direction Ratios

When lines in three-dimensional space are represented, each of them is defined in part by a set of values known as direction ratios. These ratios help describe the direction of the line in space. Let's consider the line equation given in the symmetric form, such as \[ \frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{-k} \]The terms in the denominators, \(1, 1, -k\), serve as the direction ratios of the line. They tell us how much we should move in each coordinate direction to stay on the line. For example:

• The ratio \(1\) indicates a movement of one unit along the x-axis.
• Another \(1\) describes the y-axis movement.
• The \(-k\) suggests movement along the z-axis is negative, being scaled by an unknown \(k\).

Understanding direction ratios is crucial as they significantly influence properties such as parallelism and perpendicularity of lines, and come into play heavily when we discuss concepts like coplanarity.

Scalar Triple Product

The scalar triple product is a powerful tool in vector algebra, especially when determining the coplanarity of lines. Simply put, it involves calculating the volume of a parallelepiped defined by three vectors, and checking if it's zero can determine coplanarity:

• If the scalar triple product is zero, the vectors—and by extension, lines—are coplanar.

For two lines to be coplanar, we use their direction ratios and a vector between known points on each line. In this problem, the points \(P(2,3,4)\) and \(Q(1,4,5)\) are selected, creating a connecting vector:\[ \vec{PQ} = (-1, 1, 1) \]The direction ratios of the lines are:

• For the first line: \((1, 1, -k)\)
• For the second line: \((k, 2, 1)\)

We calculate the cross product of these direction vectors, then find the dot product with \(\vec{PQ}\). The result is checked for equality to zero, confirming when the lines are coplanar, thus allowing us to think about the value of \(k\).

Symmetric Form of Lines

The symmetric form of lines in vector algebra is an efficient way to express the equation of a line. This form highlights the direction ratios and makes it easier to analyze different line properties, especially in the context of coplanarity.For the line equation:\[ \frac{x-a_1}{d_1} = \frac{y-a_2}{d_2} = \frac{z-a_3}{d_3} \]

• \( (a_1, a_2, a_3) \) represents a point on the line.
• \( (d_1, d_2, d_3) \) are the direction ratios which define the direction of the line.

The beauty of the symmetric form lies in its ability to provide quick insights into a line's geometry:

• It allows for easy parallel and perpendicular checks by comparing direction ratios.
• Aids in detecting intersection points by equating symmetric forms of two lines.
• Facilitates calculations for coplanarity when combined with the scalar triple product.

In this exercise, understanding the symmetric form helps us swiftly identify direction ratios and employ them correctly in our computations.