Step-by-Step Detailed Solution
1. Define the point and the general point on the line:
Let the given point be \( P(5, 4, 2) \).
The equation of the line is \(\vec{r} = (-1 + 2\lambda)\hat{i} + (3 + 3\lambda)\hat{j} + (1 - \lambda)\hat{k}\).
Any general point \( F \) (Foot of Perpendicular) on this line is:
\( F(\alpha, \beta, \gamma) = (2\lambda - 1, 3\lambda + 3, 1 - \lambda) \).
2. Find the Direction Ratios of PF:
Direction ratios of \( \vec{PF} \) are \( (x_F - x_P, y_F - y_P, z_F - z_P) \):
\( \text{D.R.s} = (2\lambda - 1 - 5, 3\lambda + 3 - 4, 1 - \lambda - 2) \)
\( \text{D.R.s} = (2\lambda - 6, 3\lambda - 1, -\lambda - 1) \).
3. Use the Perpendicularity Condition:
The line is parallel to vector \( \vec{b} = 2\hat{i} + 3\hat{j} - \hat{k} \).
Since \( \vec{PF} \perp \vec{b} \), their dot product must be zero:
\( 2(2\lambda - 6) + 3(3\lambda - 1) - 1(-\lambda - 1) = 0 \)
\( 4\lambda - 12 + 9\lambda - 3 + \lambda + 1 = 0 \)
\( 14\lambda - 14 = 0 \implies \lambda = 1 \).
4. Find the coordinates of F \((\alpha, \beta, \gamma)\):
Substitute \( \lambda = 1 \) into the general point \( F \):
\( \alpha = 2(1) - 1 = 1 \)
\( \beta = 3(1) + 3 = 6 \)
\( \gamma = 1 - 1 = 0 \)
So, the vector \( \vec{v} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k} = \hat{i} + 6\hat{j} + 0\hat{k} \).
5. Calculate the Projection:
We need the projection of \( \vec{v} = \hat{i} + 6\hat{j} \) on \( \vec{w} = 6\hat{i} + 2\hat{j} + 3\hat{k} \).
Formula for projection of \( \vec{A} \) on \( \vec{B} \) is \( \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|} \).
\( \vec{v} \cdot \vec{w} = (1)(6) + (6)(2) + (0)(3) = 6 + 12 = 18 \).
\( |\vec{w}| = \sqrt{6^2 + 2^2 + 3^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7 \).
Length of Projection = \( \frac{18}{7} \).
Related Theory: 3D Geometry and Vector Projections
Three-Dimensional (3D) Geometry is a significant portion of the JEE Main and Advanced mathematics syllabus. It frequently intersects with Vector Algebra to solve problems involving lines, planes, and points.
1. Foot of the Perpendicular from a Point to a Line
Finding the foot of the perpendicular is a classic problem. Given a point \(P(x_1, y_1, z_1)\) and a line \(L: \frac{x-x_2}{a} = \frac{y-y_2}{b} = \frac{z-z_2}{c}\):
- Express a general point \(F\) on the line in terms of a parameter \(\lambda\).
- Determine the direction ratios of the segment \(PF\).
- Since \(PF\) is perpendicular to the line, the dot product of \(PF\)'s direction ratios and the line's direction ratios \((a, b, c)\) must be zero.
- Solve for \(\lambda\) to find the exact coordinates of \(F\).
2. Vector Projection: Geometric Meaning
The projection of vector \(\vec{a}\) on \(\vec{b}\) represents the "shadow" or the component of \(\vec{a}\) along the direction of \(\vec{b}\). It is a scalar value given by:
$$\text{Projection} = |\vec{a}| \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$$
Note: The "Vector Projection" would be this scalar multiplied by the unit vector \(\hat{b}\). JEE questions often specify "length of projection" to ask for the scalar magnitude.
3. Essential Formulas for 3D Geometry
- Distance between two points: \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}\)
- Shortest distance between two skew lines: \(\frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}\)
- Angle between two lines: \(\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}\)
4. Core Concept: Direction Cosines and Ratios
Direction ratios (DRs) are proportional to direction cosines (DCs). If \((a, b, c)\) are DRs, then the DCs are \(\left(\frac{a}{\sqrt{\Sigma a^2}}, \frac{b}{\sqrt{\Sigma a^2}}, \frac{c}{\sqrt{\Sigma a^2}}\right)\). In perpendicularity problems, using DRs is sufficient because the constant of proportionality cancels out in the dot product equation \(a_1 a_2 + b_1 b_2 + c_1 c_2 = 0\).
5. Common Mistakes students make
- Sign Errors: Mistaking the sign of \(\lambda\) or the direction vector components. Always verify the line equation \(\vec{r} = \vec{a} + \lambda\vec{b}\).
- Projection Confusion: Using \(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|}\) instead of \(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\) when asked for the projection on \(\vec{b}\). Remember, the denominator is always the magnitude of the vector on which the shadow is cast.
- Foot vs Image: Calculating the image of the point instead of the foot. The image \(I\) is such that the foot \(F\) is the midpoint of \(PI\).
6. Shortcut Trick: Parametric Visualization
Instead of writing out full equations, mental mapping of \(\lambda\) components directly into the dot product equation can save up to 45 seconds. Practice identifying the "direction vector" of the line immediately.
7. Exam Relevance: JEE Main vs Advanced
In JEE Main, questions are usually direct like this one, focusing on calculation accuracy. In JEE Advanced, you might find this as a sub-part of a larger problem involving planes or spheres.
8. Dot Product and Orthogonality
The dot product is the primary tool for testing orthogonality in 3D space. If \(\vec{u} \cdot \vec{v} = 0\), the vectors are perpendicular. This principle is used for finding the foot of the perpendicular, the shortest distance between lines, and the normal to a plane.
9. Projection of a Point on a Plane
Similar to the line, the foot of the perpendicular from \(P(x_1, y_1, z_1)\) to a plane \(ax+by+cz+d=0\) is given by the formula:
$$\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = -\frac{(ax_1+by_1+cz_1+d)}{a^2+b^2+c^2}$$
10. Summary for Students
Mastering these two concepts—parameterizing a point on a line and the dot product for perpendicularity—will allow you to solve nearly 40% of all 3D geometry questions in the JEE Main syllabus.
Frequently Asked Questions (FAQs)
1. What is the foot of the perpendicular?
It is the unique point on a line or plane that is closest to a given external point, forming a 90-degree angle with the line/plane.
2. How do you find a general point on a line?
By setting the symmetric form of the line equal to a parameter \(\lambda\) and solving for x, y, and z in terms of \(\lambda\).
3. What is the formula for vector projection?
The projection of \(\vec{a}\) on \(\vec{b}\) is \( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \).
4. Why did we set the dot product to zero?
Because the line connecting the external point and its foot is perpendicular to the given line, and the dot product of perpendicular vectors is zero.
5. What is the magnitude of vector \(6\hat{i} + 2\hat{j} + 3\hat{k}\)?
It is \(\sqrt{6^2 + 2^2 + 3^2} = \sqrt{49} = 7\).
6. Can the length of a projection be negative?
"Length" is typically treated as a magnitude (absolute value), but the scalar projection itself can be negative if the angle between vectors is obtuse.
7. How is the foot related to the 'image' of a point?
The foot is the midpoint of the segment connecting the point and its mirror image.
8. What are direction ratios?
They are any three numbers proportional to the direction cosines of a line, indicating its orientation in space.
9. Is \(\alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}\) a position vector?
Yes, it is the position vector of the foot of the perpendicular \(F\).
10. Why is \(\lambda=1\) in this problem?
Solving the equation \(14\lambda - 14 = 0\) resulting from the dot product condition gives \(\lambda=1\).