Foot of Perpendicular and Distance from Point to Line

Foot of Perpendicular and Distance from Point to Line

Hello JEE aspirants! In this lesson, we'll explore a crucial concept in 3D geometry: finding the foot of the perpendicular from a point to a line, the perpendicular distance, and the image of a point with respect to a line. These concepts are frequently tested in JEE Main, and mastering them will significantly boost your score. Let's dive in!

Conceptual Understanding

Imagine a point $P$ floating in space and a straight line $L$. Now, picture dropping a perpendicular from $P$ onto the line $L$. The point where this perpendicular intersects the line is called the foot of the perpendicular. The length of this perpendicular is the perpendicular distance from $P$ to $L$. Finally, if you extend the perpendicular from $P$ to an equal distance on the other side of the line, you'll reach the image of $P$ with respect to the line (like a mirror image!).

Why is this important? Many problems in 3D geometry and vector algebra involve finding the shortest distance from a point to a line, which is essentially the perpendicular distance. Also, understanding reflections and symmetry is key in various geometrical problems, making the concept of the image of a point important.

Finding the Foot of the Perpendicular

Let's say we have a point $P(x_1, y_1, z_1)$ and a line given in vector form as $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\vec{a}$ is a known point on the line, $\vec{b}$ is the direction vector of the line, and $\lambda$ is a scalar parameter. Our goal is to find the coordinates of the foot of the perpendicular, which we'll call $Q$.

Here's the step-by-step process:

1. General point on the line: Any point $Q$ on the line can be represented as $\vec{q} = \vec{a} + \lambda \vec{b}$. Let's say $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$. Then, the coordinates of $Q$ are $(a_1 + \lambda b_1, a_2 + \lambda b_2, a_3 + \lambda b_3)$. This is a general point on the line.
$$Q = \vec{a} + \lambda \vec{b}$$
2. PQ must be perpendicular to b: Since $PQ$ is perpendicular to the line, the dot product of the vector $\vec{PQ}$ and the direction vector $\vec{b}$ must be zero. The vector $\vec{PQ} = \vec{q} - \vec{p} = (a_1 + \lambda b_1 - x_1)\hat{i} + (a_2 + \lambda b_2 - y_1)\hat{j} + (a_3 + \lambda b_3 - z_1)\hat{k}$. Therefore, $\vec{PQ} \cdot \vec{b} = 0$.
$$\vec{PQ} \cdot \vec{b} = 0$$
3. Solve for λ: Expanding the dot product, we get: $$(a_1 + \lambda b_1 - x_1)b_1 + (a_2 + \lambda b_2 - y_1)b_2 + (a_3 + \lambda b_3 - z_1)b_3 = 0$$ This equation can be solved for $\lambda$. Once you find the value of $\lambda$, substitute it back into the equation $\vec{q} = \vec{a} + \lambda \vec{b}$ to get the coordinates of the foot of the perpendicular $Q$.

Perpendicular Distance

Once you have the coordinates of $Q$, finding the perpendicular distance is straightforward. It's simply the magnitude of the vector $\vec{PQ}$.

Image of a Point

The image of point $P$ (let's call it $P'$) is such that $Q$ is the midpoint of $PP'$. If $P'$ has coordinates $(x', y', z')$, then:

$$x_Q = \frac{x_P + x'}{2}, \quad y_Q = \frac{y_P + y'}{2}, \quad z_Q = \frac{z_P + z'}{2}$$

Solving for $x', y', z'$, we get the coordinates of the image $P'$.

In component form:

$$x' = 2x_Q - x_P, \quad y' = 2y_Q - y_P, \quad z' = 2z_Q - z_P$$

Tips for Solving Problems

• Visualize: Always try to visualize the problem in 3D space. This will help you understand the relationships between the point, the line, and the perpendicular.
• Vector vs. Cartesian: You can solve these problems using vector algebra or Cartesian coordinates. Choose the method that you are more comfortable with and that seems easier for the specific problem.
• Check Perpendicularity: After finding the foot of the perpendicular, double-check that $\vec{PQ} \cdot \vec{b} = 0$ to ensure your calculations are correct.

Common Mistakes to Avoid