Problem 64 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If the angle between the line \(x=\frac{y-1}{2}=\frac{z-3}{\lambda}\) and the plane \(x+2 y+3 z=\) is \(\cos ^{-1}\left(\sqrt{\frac{5}{14}}\right)\), then \(\lambda\) equals [2011] (A) \(\frac{3}{2}\) (B) \(\frac{2}{5}\) (C) \(\frac{5}{3}\) (D) \(\frac{2}{3}\)

Step-by-Step Solution

Answer

\(\lambda = \frac{2}{3}\)

Step 1: Extract direction ratios of the line

The given equation of the line is parameterized as \(x=\frac{y-1}{2}=\frac{z-3}{\lambda}\). From this, we derive the direction ratios of the line to be (1, 2, \(\lambda\)).

Step 2: Identify normal direction ratios of the plane

The given plane equation is \( x + 2y + 3z = 0 \). Thus, the normal to the plane has direction ratios (1, 2, 3).

Step 3: Use the formula for the angle between a line and a plane

The cosine of the angle \( \theta \) between the line and the plane is given by \(\cos \theta = \frac{|\mathbf{a} \cdot \mathbf{n}|}{\|\mathbf{a}\| \|\mathbf{n}\|}\) where \(\mathbf{a} = (1, 2, \lambda)\) and \(\mathbf{n} = (1, 2, 3)\).

Step 4: Calculate dot product and magnitudes

Calculate \(\mathbf{a} \cdot \mathbf{n} = 1 \cdot 1 + 2 \cdot 2 + \lambda \cdot 3 = 1 + 4 + 3\lambda = 5 + 3\lambda\). Magnitudes: \(\|\mathbf{a}\| = \sqrt{1^2 + 2^2 + \lambda^2} = \sqrt{1 + 4 + \lambda^2}\) and \(\|\mathbf{n}\| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}\).

Step 5: Solve for \(\lambda\) using the angle condition

We know \(\cos \theta = \sqrt{\frac{5}{14}}\). Set the expression \(\cos \theta = \frac{|5 + 3\lambda|}{\sqrt{5 + \lambda^2} \cdot \sqrt{14}} = \sqrt{\frac{5}{14}}\). Squaring both sides, \(\frac{(5 + 3\lambda)^2}{(5 + \lambda^2) \cdot 14} = \frac{5}{14}\). Simplifying gives \((5 + 3\lambda)^2 = 5 \cdot (5 + \lambda^2)\).

Key Concepts

Direction Ratios

Direction ratios are fundamental when dealing with lines in 3D geometry. They give us a way to describe the direction of a line in space using three numbers. For any given line, such as the one from our exercise, direction ratios can be found from its parametric form.
For the line given by the equation \( x = \frac{y - 1}{2} = \frac{z - 3}{\lambda} \), the parameters help us identify the direction ratios as (1, 2, \( \lambda \)). Simply put, these numbers indicate how much the line moves in the x, y, and z directions, respectively, as the parameter increases.
This concept is especially important because it allows you to calculate other properties of the line, such as angles with respect to planes, which is precisely what we are solving in our exercise. The direction ratios also form the basis for vector calculations involving the line.

Dot Product

The dot product, also known as the scalar product, is a crucial operation when working with vectors. In simple terms, it takes two vectors and returns a single number (a scalar) that encapsulates information about the angle between them.
The dot product formula is given by \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \), where \( \mathbf{a} = (a_1, a_2, a_3) \) and \( \mathbf{b} = (b_1, b_2, b_3) \).
In our exercise, we use the dot product to find the cosine of the angle between the direction vector of the line \( \mathbf{a} = (1, 2, \lambda) \) and the normal vector of the plane \( \mathbf{n} = (1, 2, 3) \). The resulting dot product helps us calculate the angle between the line and the plane, which is central to solving for \( \lambda \) in the exercise.

Magnitude of Vectors

The magnitude of vectors (sometimes referred to as the length or norm) is a measure of a vector's extent in space. Calculating a vector's magnitude is like measuring the length of a line segment joining the origin to the vector point.
To find the magnitude of a vector \( \mathbf{a} = (a_1, a_2, a_3) \), use the formula \( \|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2} \). This formula is derived from the Pythagorean theorem in three dimensions.
In our exercise, we calculated the magnitudes of the direction vector of the line \( \mathbf{a} = (1, 2, \lambda) \) and the normal vector of the plane \( \mathbf{n} = (1, 2, 3) \). These magnitudes were used alongside the dot product to find the angle between the line and the plane via cosine. The magnitudes are essential for normalizing the vectors when comparing them, helping us translate the geometry of a problem into solvable algebraic expressions.