The perpendicular distance, of the line (x-1)/2 = (y+2)/-1 = (z+3)/2 from the point P(2,-10,1), is :

Q: Why can't we just find the projection of $P$ onto the line and use the distance formula?

You absolutely can! That is a valid method. However, the cross-product method is often faster because it avoids solving for the specific coordinates of the projection point. It gives you the magnitude directly, saving you precious time in the exam hall.

Q: Is the order of the cross product important?

The magnitude of the cross product is the same regardless of order ($|\vec{A} \times \vec{B}| = |\vec{B} \times \vec{A}|$), so you won't get the wrong distance. However, for consistency, we usually do $\vec{AP} \times \vec{d}$.

Q: What if the line was given in a different form?

Always convert it to the symmetric form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$ first. That is the universal key to unlocking any line problem in 3D geometry.

Question

The perpendicular distance, of the line $\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z+3}{2}$ from the point $P(2,-10,1)$, is :

Options:

(A) 6

(B) $5\sqrt{2}$

(C) $4\sqrt{3}$

(D) $3\sqrt{5}$

Sigma Percentile · Accepted Answer

### The Geometry of the Shortest Path

Imagine you are standing in a vast, three-dimensional space at point $P(2, -10, 1)$. Below you, stretching infinitely in both directions, lies a line defined by the equation:

$$\frac{x-1}{2} = \frac{y+2}{-1} = \frac{z+3}{2}$$

Your goal is to find the shortest distance from where you stand to that line. In the world of JEE Advanced, this is not just a calculation; it is a test of your spatial intuition. The shortest distance is always the perpendicular distance.

### Step 1: Extracting the DNA of the Line

Every line in 3D space has a unique signature. By looking at the equation, we can extract two vital pieces of information.

First, we identify a point $A$ that lies on the line. By setting the ratios to zero, we see that $A$ has coordinates $(1, -2, -3)$.

Second, we identify the direction vector $\vec{d}$, which tells us the orientation of the line. From the denominators, we get $\vec{d} = 2\hat{i} - \hat{j} + 2\hat{k}$. We now have the anchor point and the direction—the DNA of our line is fully revealed.

### Step 2: The Vector Bridge

To measure the distance from $P$ to the line, we need a vector that connects our point $P$ to the line. We use our known point $A$ to create this bridge.

We define the vector $\vec{AP}$ as the position vector of $P$ minus the position vector of $A$. Calculating this, we get:

$$\vec{AP} = (2-1)\hat{i} + (-10 - (-2))\hat{j} + (1 - (-3))\hat{k}$$

Simplifying this, we arrive at $\vec{AP} = \hat{i} - 8\hat{j} + 4\hat{k}$. This vector is the bridge between our point in space and the line below.

### Step 3: The Cross Product Magic

Here is where the elegance of vector algebra shines. The perpendicular distance $d$ is given by the formula:

$$d = \frac{|\vec{AP} 	imes \vec{d}|}{|\vec{d}|}$$

Geometrically, the magnitude of the cross product $|\vec{AP} 	imes \vec{d}|$ represents the area of a parallelogram formed by $\vec{AP}$ and $\vec{d}$. Since the area of a parallelogram is also base times height, and the base is the magnitude of the direction vector $|\vec{d}|$, dividing the area by the base gives us the height—which is exactly the perpendicular distance we seek.

### Step 4: The Calculation

Now, let us compute the cross product using a determinant:

$$\vec{AP} 	imes \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -8 & 4 \ 2 & -1 & 2 \end{vmatrix}$$

Expanding this, we get $\hat{i}((-8)(2) - (4)(-1)) - \hat{j}((1)(2) - (4)(2)) + \hat{k}((1)(-1) - (-8)(2))$. Simplifying the terms, we find:

$$\vec{AP} 	imes \vec{d} = -12\hat{i} + 6\hat{j} + 15\hat{k}$$

Next, we find the magnitudes. The magnitude of the cross product is:

$$|\vec{AP} 	imes \vec{d}| = \sqrt{(-12)^2 + 6^2 + 15^2} = \sqrt{144 + 36 + 225} = \sqrt{405}$$

The magnitude of the direction vector is:

$$|\vec{d}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$$

### Final Result

Finally, we substitute these values into our distance formula:

$$d = \frac{\sqrt{405}}{3}$$

We can simplify $\sqrt{405}$ as $\sqrt{81 	imes 5} = 9\sqrt{5}$. Thus:

$$d = \frac{9\sqrt{5}}{3} = **3\sqrt{5}**$$

We have successfully navigated the geometry and arrived at the solution. Remember, in JEE Advanced, it is not just about the final number; it is about understanding the beautiful relationship between vectors and space.

Animated Solution for Mathematics - Three Dimensional Geometry: The perpendicular distance, of the line $\frac{x - 1}{2} = \frac{y + 2}{- 1} = \frac{z + 3}{2}$ from the point $P (2, - 10, 1)$ , is :

Select Answer:

Visualized Solution

Visualizing the Setup

Extracting Line Parameters

Defining Vector $A P$

Calculating Vector $A P$

The Distance Formula

Setting up $A P \times d$

Expanding the Determinant

Simplifying the Cross Product

Magnitude of $A P \times d$

Magnitude of Direction Vector $d$

Substituting into the Formula

Final Simplification

The Sigma Insight: Equation of a Line in Space

The Geometry of the Shortest Path

Step 1

Step 2

Step 3

Step 4

Final Result

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Animated Solution for Mathematics - Three Dimensional Geometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The perpendicular distance, of the line 2x−1​=−1y+2​=2z+3​ from the point P(2,−10,1), is :