Problem 88 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the image of the point \(\mathrm{P}(1,-2,3)\) in the plane, \(2 \mathrm{x}+3 \mathrm{y}-4 \mathrm{z}+22=0\) measured parallel to line, \(\frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{4}=\frac{\mathrm{z}}{5}\) is \(Q\), then \(P Q\) is equal to : [2017] (a) \(6 \sqrt{5}\) (b) \(3 \sqrt{5}\) (c) \(2 \sqrt{42}\) (d) \(\sqrt{42}\)

Step-by-Step Solution

Answer

The length of \(P Q\) is \(\sqrt{42}\).

Step 1: Determine the normal vector of the plane

The plane equation is given by: \(2x + 3y - 4z + 22 = 0\). The normal vector \(\mathbf{n}\) is composed of the coefficients of \(x\), \(y\), and \(z\): \(\mathbf{n} = \langle 2, 3, -4 \rangle\).

Step 2: Determine the direction vector of the line

The direction vector \(\mathbf{d}\) of the line given by the equation \(\frac{x}{1} = \frac{y}{4} = \frac{z}{5}\) is \(\langle 1, 4, 5 \rangle\).

Step 3: Find the projection of the line direction on the normal

We need to find the component of the direction vector \(\mathbf{d}\) that is along the normal vector \(\mathbf{n}\). Calculate this using the dot product: \(\text{proj}_{\mathbf{n}}(\mathbf{d}) = \frac{\mathbf{d} \cdot \mathbf{n}}{\|\mathbf{n}\|^2} \cdot \mathbf{n}\).\(\mathbf{d} \cdot \mathbf{n} = 1 \times 2 + 4 \times 3 + 5 \times (-4) = 6 - 20 = -14\). \\(\|\mathbf{n}\|^2 = 2^2 + 3^2 + (-4)^2 = 4 + 9 + 16 = 29\) . \\(\text{proj}_{\mathbf{n}}(\mathbf{d}) = \frac{-14}{29} \cdot \langle 2, 3, -4 \rangle = \frac{-28}{29}, \frac{-42}{29}, \frac{56}{29}\).

Step 4: Determine the parameter of the image point on the plane

The image point \(Q\) of the point \(P(1, -2, 3)\) is on the plane and lies along the vector direction. Let it have a parameter \(t\): \\[ (1 + t, -2 + 4t, 3 + 5t) \]. \It also satisfies the plane equation: \\[ 2(1 + t) + 3(-2 + 4t) - 4(3 + 5t) + 22 = 0.\]Simplifying, \\[ 2 + 2t - 6 + 12t - 12 - 20t + 22 = 0.\] \\[-6t + 6 = 0.\] \\[t = 1.\]

Step 5: Write expression for image point Q

Substituting \(t = 1\) into \( (1 + t, -2 + 4t, 3 + 5t)\): \\[Q(1+1, -2+4 \times 1, 3+5 \times 1) = (2, 2, 8).\]

Step 6: Calculate length of PQ

The distance \(PQ\) is calculated as: \\[ \|PQ\| = \sqrt{(2-1)^2 + (2+2)^2 + (8-3)^2} = \sqrt{1 + 16 + 25} = \sqrt{42}.\]

Key Concepts

Normal Vector

In geometry, a normal vector is a vector perpendicular to a surface or a plane. Understanding the normal vector of a plane is crucial because it gives you a direct way to identify the plane's orientation in space. For any plane equation like \( ax + by + cz + d = 0 \), the normal vector \( \mathbf{n} \) is formed using the coefficients of \( x \), \( y \), and \( z \). In this exercise, the plane equation is \( 2x + 3y - 4z + 22 = 0 \), so the normal vector becomes \( \langle 2, 3, -4 \rangle \).

• Normal vectors are not unique; they can be scaled by multiplying with a scalar.
• In context of reflecting points, the normal vector helps identify the shortest distance path perpendicular to the plane.

Recognizing the normal vector helps in solving projections and distances effectively.

Direction Vector

The direction vector is a fundamental concept in vector geometry, indicating the direction in which a line moves in space. It is particularly important when dealing with lines as this vector tells us about the line’s orientation. For the line described by \( \frac{x}{1} = \frac{y}{4} = \frac{z}{5} \), the direction vector is extracted from the denominators of these ratios, thus being \( \langle 1, 4, 5 \rangle \).

• The direction vector can be used to express any point on the line using a parameter, often denoted \( t \).
• Direction vectors are also critical in calculating projections and in determining angles between vectors.

Understanding how to derive and use the direction vector is key to analyzing vector movement and distance calculations.

Projection

Projection in vector mathematics is a technique used to "throw" one vector onto another, effectively measuring one vector along the direction of another. The projection of a vector \( \mathbf{d} \) onto another vector \( \mathbf{n} \) (a normal vector, in this context) is crucial in understanding the component of \( \mathbf{d} \) that lies along \( \mathbf{n} \). This can be visualized as the shadow or footprint of one vector upon another. The formula for projecting \( \mathbf{d} \) onto \( \mathbf{n} \) is \( \text{proj}_{\mathbf{n}}(\mathbf{d}) = \frac{\mathbf{d} \cdot \mathbf{n}}{\|\mathbf{n}\|^2} \cdot \mathbf{n} \).

• The projection tells us how one vector contributes along another specified direction.
• In the solution, the projection helps find the point where a line is parallel to the normal vector to reach the plane.

Effective use of projection allows for decomposition of vector operations into plane and line components, making complex spatial calculations more manageable.

Distance Formula

The distance formula is a key mathematical tool for finding the length between two points in space. When dealing with vectors and coordinate geometry, the distance formula \[ \|PQ\| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \] enables accurate calculation of how far apart two points are. In context, after determining the coordinates of point \( Q \), this formula helps us calculate \( PQ \), the distance between points \( P \) and \( Q \).

• The formula derives from the Pythagorean theorem, allowing for three-dimensional space calculations.
• Using the distance formula helps ensure the accuracy when calculating line segments in geometry problems.

Understanding and applying the distance formula is essential for most vector geometry problems, ensuring precise results in distance measurements.