Problem 86 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The distance between the line \(\vec{r}=2 \hat{i}-2 \hat{j}+3 \hat{k}\) \(+\lambda(\hat{i}+\hat{j}+4 \hat{k})\) and the plane \(\vec{r} \cdot(\hat{i}+5 \hat{j}+\hat{k})=5\) is \([\mathbf{2 0 0 5}]\) (A) \(\frac{10}{9}\) (B) \(\frac{10}{3 \sqrt{3}}\) (C) \(\frac{3}{10}\) (D) \(\frac{10}{3}\)

Step-by-Step Solution

Answer

The distance is \(\frac{10}{3\sqrt{3}}\), which is option (B).

Step 1: Identify the Line and Plane Information

The line is given by the vector equation \( \vec{r} = 2\hat{i} - 2\hat{j} + 3\hat{k} + \lambda(\hat{i} + \hat{j} + 4\hat{k}) \). This can be interpreted such that the point on the line is \((2, -2, 3)\) and the direction vector is \((1, 1, 4)\). The plane is given by the equation \( \vec{r} \cdot (\hat{i} + 5\hat{j} + \hat{k}) = 5 \). Here, the normal vector to the plane is \((1, 5, 1)\).

Step 2: Check if the Line is Parallel to the Plane

We need to check if the direction vector of the line is parallel to the plane, which means it should be orthogonal to the normal of the plane. \((1, 1, 4) \cdot (1, 5, 1) = 1 + 5 + 4 = 10eq 0\). Since the dot product is non-zero, the line is not parallel and intersects the plane.

Step 3: Find Point of Intersection (Irrelevant Here)

Since the line is not parallel to the plane, they should intersect. However, we only require the perpendicular distance from the initial point on the line to the plane.

Step 4: Use Distance Formula for a Point to a Plane

The formula for the distance from a point \((x_0, y_0, z_0)\) to the plane \(Ax + By + Cz = D\) is given by: \[ \text{Distance} = \frac{|Ax_0 + By_0 + Cz_0 - D|}{\sqrt{A^2 + B^2 + C^2}} \]Substituting the given point \((2, -2, 3)\) and plane equation \(1x + 5y + 1z = 5\), we find the distance:\[ \text{Distance} = \frac{|1(2) + 5(-2) + 1(3) - 5|}{\sqrt{1^2 + 5^2 + 1^2}} \]

Step 5: Evaluate the Distance Expression

Calculate the numerator:\[ |2 + (-10) + 3 - 5| = |-10| = 10 \] Calculate the denominator:\[ \sqrt{1 + 25 + 1} = \sqrt{27} = 3\sqrt{3} \]Hence, the distance:\[ \text{Distance} = \frac{10}{3\sqrt{3}} \]

Step 6: Make a Selection Based on Calculation

The calculated distance \(\frac{10}{3\sqrt{3}}\) corresponds to option (B) as given in the problem.

Key Concepts

Vector Equation of Line

Understanding the vector equation of a line is a key aspect in vector algebra. A line in space can be described using a vector equation, which combines a point on the line and a direction vector.
The given vector equation \( \vec{r} = 2\hat{i} - 2\hat{j} + 3\hat{k} + \lambda(\hat{i} + \hat{j} + 4\hat{k}) \) tells us that:

• The point on the line is \((2, -2, 3)\).
• The direction vector of the line is \((1, 1, 4)\).
• The parameter \(\lambda\) varies along the line.

When \(\lambda = 0\), the vector \((2\hat{i} - 2\hat{j} + 3\hat{k})\) is a specific point on the line.
As \(\lambda\) changes, it describes the infinite set of points along the direction of the vector \((1, 1, 4)\). Thus, the line extends endlessly in both directions from the point \((2, -2, 3)\). By altering \(\lambda\), we move to different positions along this path. This concept is crucial for solving problems related to lines in 3D geometry.

Point to Plane Distance Formula

In geometry, calculating the perpendicular distance from a point to a plane is often necessary. For this, the point to plane distance formula comes in handy:
\[ \text{Distance} = \frac{|Ax_0 + By_0 + Cz_0 - D|}{\sqrt{A^2 + B^2 + C^2}} \]This formula measures how far a point \((x_0, y_0, z_0)\) is from the plane \(Ax + By + Cz = D\). The numerator captures the signed distance, while the denominator normalizes it with the plane's normal vector's magnitude.
Applying this to our exercise:

• The point is \((2, -2, 3)\).
• The plane equation is \(1x + 5y + 1z = 5\).

Plugging in these values results in:\[ \text{Distance} = \frac{|1(2) + 5(-2) + 1(3) - 5|}{\sqrt{1^2 + 5^2 + 1^2}} \]You calculate the numerator and denominator separately. The numerator becomes \(|-10| = 10\), and the denominator simplifies to \(\sqrt{27} = 3\sqrt{3}\).
Hence, the distance is \(\frac{10}{3\sqrt{3}}\), exemplifying how straightforward this formula can be when the components are correctly identified and calculated.

Dot Product

The dot product is a fundamental operation in vector algebra that calculates the scalar product of two vectors. It's useful in various contexts, especially in determining orthogonality (perpendicularity) of vectors. When two vectors are orthogonal, their dot product equals zero.
In the context of our line and plane, we used the dot product to check if the line's direction vector \((1, 1, 4)\) is perpendicular to the plane's normal vector \((1, 5, 1)\). If these were orthogonal, it would mean the line is parallel to the plane.

• The calculation is \((1, 1, 4) \cdot (1, 5, 1) = 1 \cdot 1 + 1 \cdot 5 + 4 \cdot 1 = 10\)
• Since the result is not zero, the vectors aren't orthogonal, indicating the line is not parallel to the plane.

The dot product doesn’t just help in assessing orthogonality; it also provides insights into angles between vectors when used with the cosine formula. Thus, being conversant with the dot product expands your toolset in tackling multidimensional geometry problems.