Let P be the foot of the perpendicular from the point (1, 2, 2) on the line $L:\frac{x-1}{1}=\frac{y+1}{-1}=\frac{z-2}{2}$ Let the line $\vec{r}=(-\hat{i}+\hat{j}-2\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})$, $\lambda\in R,$ intersect the line L at Q. Then $2(PQ)^{2}$ is equal to: Options: (A) 25 (B) 19 (C) 29 (D) 27

### Analyzing the Setup To find the foot of the perpendicular $P$ from point $A(1, 2, 2)$ to the line $L$ defined by $$\frac{x-1}{1} = \frac{y+1}{-1} = \frac{z-2}{2}$$ we first express the line in parametric form. By setting the equation equal to a parameter $t$, any point $P$ on the line can be represented as $(t+1, -t-1, 2t+2)$. The vector $\vec{AP}$ is calculated by subtracting the coordinates of $A$ from $P$: $$\vec{AP} = (t, -t-3, 2t)$$ ### The Master Equation Since $\vec{AP}$ is perpendicular to the line's direction vector $\vec{d_1} = (1, -1, 2)$, their dot product must be zero. This gives us the following equation: $$t(1) + (-t-3)(-1) + 2t(2) = 0$$ Simplifying this expression, we obtain: $$t + t + 3 + 4t = 0 \Rightarrow 6t + 3 = 0$$ Solving for $t$, we find $t = -\frac{1}{2}$. Substituting this value back into the parametric form, we determine the coordinates of the foot of the perpendicular: $$P = \left(\frac{1}{2}, -\frac{1}{2}, 1\right)$$ ### The Intersection: Finding Point $Q$ We now consider the second line $L_2$ given by $\vec{r} = (-1, 1, -2) + \lambda(1, -1, 1)$. Any point on $L_2$ can be expressed as $(\lambda-1, -\lambda+1, \lambda-2)$. For the lines to intersect at point $Q$, the coordinates must satisfy both parametric forms. Equating the components of $L$ and $L_2$ leads to a system of equations. Solving this system yields $t = -2$. Substituting $t = -2$ into the parametric form of $L$, we find the intersection point: $$Q = (-1, 1, -2)$$ ### The Final Calculation We now have the coordinates $P\left(\frac{1}{2}, -\frac{1}{2}, 1\right)$ and $Q(-1, 1, -2)$. The distance squared $PQ^2$ is calculated as follows: $$PQ^2 = \left(-1 - \frac{1}{2}\right)^2 + \left(1 - \left(-\frac{1}{2}\right)\right)^2 + (-2 - 1)^2$$ $$PQ^2 = \left(-\frac{3}{2}\right)^2 + \left(\frac{3}{2}\right)^2 + (-3)^2 = \frac{9}{4} + \frac{9}{4} + 9$$ $$PQ^2 = \frac{18}{4} + 9 = \frac{9}{2} + 9 = \frac{27}{2}$$ The problem asks for the value of $2(PQ)^2$. Therefore: $$2 \times \frac{27}{2} = \mathbf{27}$$

JEE Main 2025 (January)

LEVELJEE Main

Animated Solution for Mathematics - Three Dimensional Geometry: Let P be the foot of the perpendicular from the point (1, 2, 2) on the line $L : \frac{x - 1}{1} = \frac{y + 1}{- 1} = \frac{z - 2}{2}$ Let the line $r = (- \hat{i} + \hat{j} - 2 \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k})$ , $λ \in R,$ intersect the line L at Q. Then $2 (P Q)^{2}$ is equal to:

Q: Can I just use the projection formula instead of the dot product method?

Yes, you can! The projection of vector $\vec{AA'}$ (where $A'$ is a point on the line) onto the direction vector $\vec{d}$ gives you the distance along the line. Both methods are mathematically sound, but the parametric method is often more intuitive for finding the coordinates of the foot.

Select Answer:

Visualized Solution

Identify Point $A$ and Line $L$

Given point $A (1, 2, 2)$ and line $L : \frac{x - 1}{1} = \frac{y + 1}{- 1} = \frac{z - 2}{2}$ .
The direction vector of line $L$ is $d_{1} = (1, - 1, 2)$ .
We need to find point $P$ , the foot of the perpendicular from $A$ to $L$ .

General Point $P$ on Line $L$

Let $\frac{x - 1}{1} = \frac{y + 1}{- 1} = \frac{z - 2}{2} = t$ .
Any point $P$ on $L$ can be written as $(t + 1, - t - 1, 2 t + 2)$ .

Formulating Vector $A P$

Vector $A P = Position vector of P - Position vector of A$ .
$A P = (t + 1 - 1, - t - 1 - 2, 2 t + 2 - 2)$ .
$A P = (t, - t - 3, 2 t)$ .

Orthogonality Condition

Since $A P$ is perpendicular to $L$ , $A P \cdot d_{1} = 0$ .
$A P \cdot (1, - 1, 2) = 0$ .
$t (1) + (- t - 3) (- 1) + 2 t (2) = 0$ .

Solving for Parameter $t$

Expanding the dot product: $t + t + 3 + 4 t = 0$ .
Combining like terms: $6 t + 3 = 0$ .
$6 t = - 3 \Rightarrow t = - \frac{1}{2}$ .

Coordinates of Point $P$

Substitute $t = - \frac{1}{2}$ into $P (t + 1, - t - 1, 2 t + 2)$ .
$P = (- \frac{1}{2} + 1, - (- \frac{1}{2}) - 1, 2 (- \frac{1}{2}) + 2)$ .
$P = (\frac{1}{2}, - \frac{1}{2}, 1)$ .

Define Line $L_{2}$ and Intersection $Q$

Line $L_{2} : r = (- \hat{i} + \hat{j} - 2 \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k})$ .
Any point on $L_{2}$ is $(λ - 1, - λ + 1, λ - 2)$ .
$L_{2}$ intersects $L$ at point $Q$ .

System of Equations for Intersection

At intersection $Q$ , a point on $L_{2}$ must match a point on $L$ .
Equating coordinates of $L (t + 1, - t - 1, 2 t + 2)$ and $L_{2} (λ - 1, - λ + 1, λ - 2)$ :
1) $λ - 1 = t + 1 \Rightarrow λ - t = 2$
2) $- λ + 1 = - t - 1 \Rightarrow λ - t = 2$
3) $λ - 2 = 2 t + 2 \Rightarrow λ - 2 t = 4$

Solving for Intersection Parameter

We have a system of equations:
$λ - t = 2$ (Equation A)
$λ - 2 t = 4$ (Equation B)
Subtracting Equation B from Equation A:
$(λ - t) - (λ - 2 t) = 2 - 4$
$t = - 2$

Coordinates of Point $Q$

Substitute $t = - 2$ into the general point of $L$ : $(t + 1, - t - 1, 2 t + 2)$ .
$Q = (- 2 + 1, - (- 2) - 1, 2 (- 2) + 2)$ .
$Q = (- 1, 1, - 2)$ .

Calculating $P Q^{2}$

We have $P (\frac{1}{2}, - \frac{1}{2}, 1)$ and $Q (- 1, 1, - 2)$ .
Using the distance formula squared: $P Q^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}$ .
$P Q^{2} = (- 1 - \frac{1}{2})^{2} + (1 - (- \frac{1}{2}))^{2} + (- 2 - 1)^{2}$ .
$P Q^{2} = (- \frac{3}{2})^{2} + (\frac{3}{2})^{2} + (- 3)^{2}$ .
$P Q^{2} = \frac{9}{4} + \frac{9}{4} + 9 = \frac{18}{4} + 9 = \frac{9}{2} + 9 = \frac{27}{2}$ .

Final Result: $2 (P Q)^{2}$

The question asks for the value of $2 (P Q)^{2}$ .
$2 (P Q)^{2} = 2 \times \frac{27}{2}$ .
$2 (P Q)^{2} = 27$ .
The correct answer is 27.

00:00 / 00:00

The Sigma Insight: Equation of a Line in Space

Solution Diagram

Analyzing the Setup

To find the foot of the perpendicular

P

from point

A (1, 2, 2)

to the line

L

defined by

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

we first express the line in parametric form. By setting the equation equal to a parameter

t

, any point

P

on the line can be represented as

(t + 1, - t - 1, 2 t + 2)

The vector

A P

is calculated by subtracting the coordinates of

A

from

P

A P = (t, - t - 3, 2 t)

The Master Equation

Since

A P

is perpendicular to the line's direction vector

d_{1} = (1, - 1, 2)

, their dot product must be zero. This gives us the following equation:

t (1) + (- t - 3) (- 1) + 2 t (2) = 0

Simplifying this expression, we obtain:

t + t + 3 + 4 t = 0 \Rightarrow 6 t + 3 = 0

Solving for

t

, we find

t = - \frac{1}{2}

. Substituting this value back into the parametric form, we determine the coordinates of the foot of the perpendicular:

P = (\frac{1}{2}, - \frac{1}{2}, 1)

The Intersection

Finding Point

Q

We now consider the second line

L_{2}

given by

r = (- 1, 1, - 2) + λ (1, - 1, 1)

. Any point on

L_{2}

can be expressed as

(λ - 1, - λ + 1, λ - 2)

For the lines to intersect at point

Q

, the coordinates must satisfy both parametric forms. Equating the components of

L

and

L_{2}

leads to a system of equations. Solving this system yields

t = - 2

Substituting

t = - 2

into the parametric form of

L

, we find the intersection point:

Q = (- 1, 1, - 2)

The Final Calculation

We now have the coordinates

P (\frac{1}{2}, - \frac{1}{2}, 1)

and

Q (- 1, 1, - 2)

. The distance squared

P Q^{2}

is calculated as follows:

P Q^{2} = (- 1 - \frac{1}{2})^{2} + (1 - (- \frac{1}{2}))^{2} + (- 2 - 1)^{2}

P Q^{2} = (- \frac{3}{2})^{2} + (\frac{3}{2})^{2} + (- 3)^{2} = \frac{9}{4} + \frac{9}{4} + 9

P Q^{2} = \frac{18}{4} + 9 = \frac{9}{2} + 9 = \frac{27}{2}

The problem asks for the value of

2 (P Q)^{2}

. Therefore:

2 \times \frac{27}{2} = 27

Common Traps & FAQs

Why do we use the dot product to find the foot of the perpendicular?

The dot product is the most elegant tool for orthogonality. When two vectors are perpendicular, their dot product is zero. By setting

A P \cdot d = 0

, we are essentially forcing the vector

A P

to be perpendicular to the line

L

Can I just use the projection formula instead of the dot product method?

Yes, you can! The projection of vector

A A^{'}

(where

A^{'}

is a point on the line) onto the direction vector

d

gives you the distance along the line. Both methods are mathematically sound, but the parametric method is often more intuitive for finding the coordinates of the foot.

How do I know if the lines intersect?

For two lines to intersect, they must share a common point. By equating their parametric forms, we get a system of equations. If this system has a consistent solution for the parameters

t

and

λ

, the lines intersect at that point.

Why did we multiply by 2 at the end?

That is a classic JEE trap! Always read the final question carefully. The question asks for

2 (P Q)^{2}

, not just

(P Q)^{2}

. It is a test of your attention to detail under pressure.

Select Answer:

Visualized Solution

Identify Point $A$ and Line $L$

General Point $P$ on Line $L$

Formulating Vector $A P$

Orthogonality Condition

Solving for Parameter $t$

Coordinates of Point $P$

Define Line $L_{2}$ and Intersection $Q$

System of Equations for Intersection

Solving for Intersection Parameter

Coordinates of Point $Q$

Calculating $P Q^{2}$

Final Result: $2 (P Q)^{2}$

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Master Equation

The Intersection

The Final Calculation

Similar Questions

Let $A$ and $B$ be two distinct points on the line $L : \frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}$ . Both $A$ and $B$ are at a distance $217$ from the foot of perpendicular drawn from the point $(1, 2, 3)$ on the line $L$ . If $O$ is the origin, then $O A \cdot O B$ is equal to:

Let $L$ be the line of intersection of planes $r \cdot (\hat{i} - \hat{j} + 2 \hat{k}) = 2$ and $r \cdot (2 \hat{i} + \hat{j} - \hat{k}) = 2$ . If $P (α, β, γ)$ is the foot of perpendicular on $L$ from the point $(1, 2, 0)$ , then the value of $35 (α + β + γ)$ is equal to :

Let P be the foot of the perpendicular from the point $Q (10, - 3, - 1)$ on the line $\frac{x - 3}{7} = \frac{y - 2}{- 1} = \frac{z + 1}{- 2}$ Then the area of the right angled triangle PQR, where R is the point $(3, - 2, 1)$ , is

Let a line pass through two distinct points $P (- 2, - 1, 3)$ and Q, and be parallel to the vector $3 \hat{i} + 2 \hat{j} + 2 \hat{k}$ If the distance of the point Q from the point $R (1, 3, 3)$ is 5, then the square of the area of $△ P QR$ is equal to:

A line with direction ratios $2, 1, 2$ meets the lines $x = y + 2 = z$ and $x + 2 = 2 y = 2 z$ respectively at the point $P$ and $Q$ . if the length of the perpendicular from the point $(1, 2, 12)$ to the line $P Q$ is $l$ , then $l^{2}$ is

If the foot of the perpendicular drawn from $(1, 9, 7)$ to the line passing through the point $(3, 2, 1)$ and parallel to the planes $x + 2 y + z = 0$ and $3 y - z = 3$ is $(α, β, γ)$ , then $α + β + γ$ is equal to

Select Answer:

Visualized Solution

Identify Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L

General Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P on Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L

Formulating Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AP

Orthogonality Condition

Solving for Parameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

Coordinates of Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Define Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​ and Intersection .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

System of Equations for Intersection

Solving for Intersection Parameter

Coordinates of Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ2

Final Result: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2(PQ)2

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Master Equation

The Intersection

The Final Calculation

Similar Questions

Identify Point $A$ and Line $L$

General Point $P$ on Line $L$

Formulating Vector $A P$

Solving for Parameter $t$

Coordinates of Point $P$

Define Line $L_{2}$ and Intersection $Q$

Coordinates of Point $Q$

Calculating $P Q^{2}$

Final Result: $2 (P Q)^{2}$