Problem 44 - A Complete Resource Book in Mathematics for JEE Main

Question

A plane which passes through die point \((3,2,0)\) and the line \(\frac{x-4}{1}=\frac{y-7}{5}=\frac{z-4}{4}\) is: \(\quad\) (A) \(x-y+z=1\) (B) \(x+y+z=5\) (C) \(x+2 y-z=1\) (D) \(2 x-y+z=5\)

Step-by-Step Solution

Answer

The answer is (A) \(x - y + z = 1\).

Step 1: Identify the Direction Vector of the Line

First, recognize that the line is given in the symmetric form: \(\frac{x-4}{1}=\frac{y-7}{5}=\frac{z-4}{4}\). The direction vector of the line can be extracted as \(\mathbf{d} = (1,5,4)\).

Step 2: Find a Point on the Line

A point on the line can be found by setting the parameters equal, for example, setting them to zero in the symmetric equation: \(x=4, y=7, z=4\). So, a point on the line is \((4, 7, 4)\).

Step 3: Determine the Plane's Normal Vector

The plane passes through the point \((3,2,0)\) and contains the direction vector \((1,5,4)\). Determine its normal vector \(\mathbf{n}\) by two vectors: the vector connecting the point on the line \((4,7,4)\) and point on the plane \((3,2,0)\) is \((4-3, 7-2, 4-0) = (1,5,4)\).

Step 4: Calculate the Cross Product

To find the normal vector of the plane, calculate the cross product of \((1,5,4)\) and \((1,5,4)\). But this will result in zero because they are parallel. The correct approach is to verify using the given equations or options. Thus, comparing options gives \(\mathbf{n} = (1,-1,1)\) suits the data.

Step 5: Write the Equation of the Plane

Use the normal vector \(\mathbf{n} = (1,-1,1)\) and the point \((3,2,0)\) to write the equation of the plane: \(1(x-3) -1(y-2) +1(z-0) = 0\), which simplifies to \(x - y + z = 1\).

Step 6: Check the Plane Options

Comparing the simplified equation \(x - y + z = 1\) with the given options, identify that option (A), \(x - y + z = 1\), matches.

Key Concepts

Direction Vector

In geometry, understanding the direction vector of a line is crucial, especially in 3D space. A direction vector indicates the direction along which a line extends. It is derived from the symmetric form of a line equation. The symmetric form, such as \( \frac{x-a}{l} = \frac{y-b}{m} = \frac{z-c}{n} \), reveals the direction vector \( \mathbf{d} = (l, m, n) \). This vector tells us how much the line moves in the x, y, and z directions respectively.

• The direction vector is essential for determining line behavior.
• It can be used to calculate angles between lines and lengths of line segments.
• When combined with a point on the line, it can define the entire line in space.

Understanding this vector can help you solve problems regarding movement along a path or when defining geometric shapes like planes.

Symmetric Form

The symmetric form of a line equation in 3D geometry provides a concise representation that encompasses the line's path in space. It is written as \( \frac{x-a}{l} = \frac{y-b}{m} = \frac{z-c}{n} \). Here, \((a,b,c)\) is a known point on the line, and \((l,m,n)\) is the direction vector of the line.

• Useful for quickly recognizing the line's direction vector.
• Helps identify a point through which the line passes, by setting parameters to a common value.
• Ideal form for problems where comparison between different lines or geometrical objects is necessary.

By understanding the symmetric form, you can easily manipulate and interpret the properties of lines in spatial problems. It connects the abstract direction and tangible points of a line.

Normal Vector

A normal vector is a crucial element when dealing with planes in 3D geometry. It is a vector perpendicular to the plane's surface and is necessary for forming the plane's equation. To find a plane's normal vector, we typically cross two non-parallel vectors that lie on the plane. The resulting vector from this cross product is the normal vector.

• The normal vector \(\mathbf{n} =(a, b, c)\) helps form the plane equation \(ax + by + cz = d\).
• Provides a way to determine if two planes are parallel.
• Essential for calculating angles between planes.

In practice, deriving the correct normal vector is vital as it defines the orientation and positioning of the plane within space, influencing intersections and relationships with other geometric structures.

Cross Product

The cross product is an operation on two vectors in three-dimensional space that results in a third vector which is perpendicular to the plane containing the initial vectors. When you have vectors \(\mathbf{a} = (a_1, a_2, a_3)\) and \(\mathbf{b} = (b_1, b_2, b_3)\), the cross product \(\mathbf{a} \times \mathbf{b}\) is calculated as:\[\mathbf{a} \times \mathbf{b} = (a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1)\]

• The resultant vector is perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\).
• The magnitude (length) of the cross product vector corresponds to the area of the parallelogram spanned by \(\mathbf{a}\) and \(\mathbf{b}\).
• It's a vital tool in finding normal vectors to planes and calculating torque and rotational effects in physics.

Mastering the cross product empowers you to handle complex spatial problems, like determining the orientation of planes or defining the geometry of 3D structures.