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

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 \([2005]\) (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 \(\frac{10}{3\sqrt{3}}\), corresponding to option (B).

Step 1: Identify Line Direction Vector and Point

The line is given in the form \(\vec{r} = (2, -2, 3) + \lambda(1, 1, 4)\). The direction vector of the line is \((1, 1, 4)\), and a point on the line is \((2, -2, 3)\).

Step 2: Determine Plane Normal Vector and Equation

The plane is defined by \(\vec{r} \cdot (1, 5, 1) = 5\), which gives the normal vector of the plane as \((1, 5, 1)\). This serves as the orientation of the plane.

Step 3: Find the Perpendicular Distance

To find the distance \(d\) from a point \(P_0(x_0, y_0, z_0)\) on the line to the plane, use the perpendicular distance formula: \[ d = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}} \], where \(a, b, c\) are components of the normal vector \((1, 5, 1)\), and \(d\) is the constant from the plane equation. The point is \((2, -2, 3)\).

Step 4: Substitute Values into the Distance Formula

The equation becomes:\[d = \frac{|1 \cdot 2 + 5 \cdot (-2) + 1 \cdot 3 - 5|}{\sqrt{1^2 + 5^2 + 1^2}} = \frac{|2 - 10 + 3 - 5|}{\sqrt{27}} = \frac{| -10 |}{3\sqrt{3}} = \frac{10}{3\sqrt{3}}.\]

Step 5: Conclusion

After calculating, we find that the distance from the line to the plane is \(\frac{10}{3\sqrt{3}}\). This corresponds to option (B).

Key Concepts

Direction vector

Understanding direction vectors is key for working with lines in three-dimensional space. A direction vector indicates the path along which the line extends. It is derived from the line's equation, usually in the parameterized form. For example, in the line equation \(\vec{r} = (2, -2, 3) + \lambda(1, 1, 4)\), the vector \((1, 1, 4)\) signifies the direction of the line.
This vector tells us how to move along the line starting from the point \((2, -2, 3)\).
If you imagine stepping along this line, you'd increment in the ratio of the direction vector's components; that is, one unit in the \(x\) direction, one unit in the \(y\) direction, and four units in the \(z\) direction. This forms the basis of understanding lines in vector algebra.

Normal vector

In contrast to the direction vector of a line, the normal vector is associated with a plane. It is a perpendicular vector to the plane's surface. For the plane equation \(\vec{r} \cdot (1, 5, 1) = 5\), the normal vector is \((1, 5, 1)\).
This vector is crucial because it provides the plane's orientation in space, showing which direction the plane is "facing."
If you visualize the plane as a flat surface, the normal vector sticks out perpendicular to that surface, akin to a pole. Understanding the normal vector helps determine how a line or another vector relates spatially to the plane.

Perpendicular distance formula

To measure the distance between a point on a line and a plane, we use the perpendicular distance formula. This formula accounts for the components of the normal vector, and the coordinates of the point from which distance is being measured.
The formula is:
\[d = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}\]
where \((a, b, c)\) are the components of the normal vector, \((x_0, y_0, z_0)\) are the coordinates of the point on the line, and \(d\) is the constant from the plane equation. In this example, the distance is calculated by plugging in these values. This formula is useful for finding the shortest path to the plane, which is always perpendicular.

Plane equation

The equation of a plane is a fundamental concept in geometry involving vectors. It is typically expressed in the form \(\vec{r} \cdot (a, b, c) = d\), where \((a, b, c)\) are the components of the normal vector, and \(d\) is a constant.
The plane equation \(\vec{r} \cdot (1, 5, 1) = 5\) describes a plane in three-dimensional space.
The normal vector \((1, 5, 1)\) provides the orientation of the plane, and the constant \(5\) determines its specific position. Any point \((x, y, z)\) on the plane will satisfy this equation, resulting in the same value as the constant. This equation is essential for calculating distances and understanding the relationship between different geometric constructs within a given space.