JEE Advanced 2015

LEVELJEE Advanced

Animated Solution for Mathematics - Three Dimensional Geometry: In $R^{3}$ , let $L$ be a straight line passing through the origin. Suppose that all the points on $L$ are at a constant distance from the two planes $P_{1} : x + 2 y - z + 1 = 0$ and $P_{2} : 2 x - y + z - 1 = 0$ . Let $M$ be the locus of the feet of the perpendiculars drawn from the points on $L$ to the plane $P_{1}$ . Which of the following points lie(s) on $M$ ?

Q: Why do we take the cross product of the normals to find the line's direction?

Think about the geometry: the line $L$ is parallel to plane $P_1$, so its direction vector $\vec{v}$ is perpendicular to the normal $\vec{n}_1$. Similarly, $L$ is parallel to $P_2$, so $\vec{v}$ is perpendicular to $\vec{n}_2$. If a vector is perpendicular to two other vectors, it must be parallel to their cross product, $\vec{n}_1 \times \vec{n}_2$. It is the most elegant way to lock down the direction of $L$ in space.

Select Answer:

* Multiple Correct

Progressive Hints

Hint 1

Hint 1: Visualize the geometric constraint. If a line maintains a constant distance from a plane, it cannot intersect it. What does this imply about the relationship between the line's direction vector and the plane's normal vector?

Hint 2

Hint 2: You need the direction vector of line

L

. Since

L

is parallel to both planes

P_{1}

and

P_{2}

, its direction vector

v

must be perpendicular to the normal vectors

n_{1}

and

n_{2}

of the planes. How do you find a vector perpendicular to two given vectors?

Hint 3

Hint 3: Once you have the parametric equation of line

L

Q (t) = (t, - 3 t, - 5 t)

, use the standard formula for the foot of the perpendicular from a point

(x_{1}, y_{1}, z_{1})

to a plane

a x + b y + cz + d = 0

\frac{x - x _{1}}{a} = \frac{y - y _{1}}{b} = \frac{z - z _{1}}{c} = - \frac{a x _{1} + b y _{1} + c z _{1} + d}{a ^{2} + b ^{2} + c ^{2}}

Visualized Solution

Understanding the Geometry

Line $L$ passes through the origin $(0, 0, 0)$ .
Points on $L$ are at a constant distance from planes $P_{1}$ and $P_{2}$ .
$P_{1} : x + 2 y - z + 1 = 0$
$P_{2} : 2 x - y + z - 1 = 0$

Implication of Constant Distance

If a line is at a constant distance from a plane, it must be parallel to that plane.
Therefore, $L ∥ P_{1}$ and $L ∥ P_{2}$ .
This means $L$ is perpendicular to the normal vectors of both planes.

Normal Vectors of the Planes

Normal to $P_{1}$ : $n_{1} = \hat{i} + 2 \hat{j} - \hat{k}$
Normal to $P_{2}$ : $n_{2} = 2 \hat{i} - \hat{j} + \hat{k}$
Direction of $L$ , say $v$ , is given by $n_{1} \times n_{2}$ .

Direction Vector of Line $L$

$v = \hat{i} 12 \hat{j} 2 - 1 \hat{k} - 1 1$
$v = \hat{i} (2 - 1) - \hat{j} (1 - (- 2)) + \hat{k} (- 1 - 4)$
$v = \hat{i} - 3 \hat{j} - 5 \hat{k}$

Parametric Equation of Line $L$

Line $L$ passes through $(0, 0, 0)$ with direction $(1, - 3, - 5)$ .
Equation of $L$ : $r = t (\hat{i} - 3 \hat{j} - 5 \hat{k})$
A general point $Q$ on $L$ is $(t, - 3 t, - 5 t)$ .

Defining the Locus $M$

$M$ is the locus of the feet of the perpendiculars from points on $L$ to $P_{1}$ .
Let the foot of the perpendicular from $Q (t, - 3 t, - 5 t)$ to $P_{1}$ be $(x, y, z)$ .

Foot of Perpendicular Formula

For a point $(x_{1}, y_{1}, z_{1})$ and plane $a x + b y + cz + d = 0$ , the foot $(x, y, z)$ satisfies:
$\frac{x - x _{1}}{a} = \frac{y - y _{1}}{b} = \frac{z - z _{1}}{c} = - \frac{a x _{1} + b y _{1} + c z _{1} + d}{a ^{2} + b ^{2} + c ^{2}}$

Applying the Formula for $M$

Substitute $Q (t, - 3 t, - 5 t)$ and $P_{1} : x + 2 y - z + 1 = 0$ :
$\frac{x - t}{1} = \frac{y - ( - 3 t )}{2} = \frac{z - ( - 5 t )}{- 1} = - \frac{( t ) + 2 ( - 3 t ) - ( - 5 t ) + 1}{1 ^{2} + 2 ^{2} + ( - 1 ) ^{2}}$

Evaluating the Constant Ratio

Numerator of RHS: $t - 6 t + 5 t + 1 = 1$
Denominator of RHS: $1^{2} + 2^{2} + (- 1)^{2} = 6$
The ratio simplifies to $- \frac{1}{6}$ .
So, $\frac{x - t}{1} = \frac{y + 3 t}{2} = \frac{z + 5 t}{- 1} = - \frac{1}{6}$

Coordinates of Locus $M$

$x - t = - \frac{1}{6} ⟹ x = t - \frac{1}{6}$
$\frac{y + 3 t}{2} = - \frac{1}{6} ⟹ y = - 3 t - \frac{1}{3}$
$\frac{z + 5 t}{- 1} = - \frac{1}{6} ⟹ z = - 5 t + \frac{1}{6}$
General point on $M$ : $(t - \frac{1}{6}, - 3 t - \frac{1}{3}, - 5 t + \frac{1}{6})$

Testing Option A

Option A: $(0, - \frac{5}{6}, - \frac{2}{3})$
Set $x = 0 ⟹ t - \frac{1}{6} = 0 ⟹ t = \frac{1}{6}$
Check $y$ : $- 3 (\frac{1}{6}) - \frac{1}{3} = - \frac{1}{2} - \frac{1}{3} = - \frac{5}{6}$
Check $z$ : $- 5 (\frac{1}{6}) + \frac{1}{6} = - \frac{4}{6} = - \frac{2}{3}$
Option A lies on $M$ .

Testing Option B

Let's check the point on $M$ when $t = 0$ .
Substitute $t = 0$ into the general point $(t - \frac{1}{6}, - 3 t - \frac{1}{3}, - 5 t + \frac{1}{6})$ .
We get the point $(- \frac{1}{6}, - \frac{1}{3}, \frac{1}{6})$ .
This matches the intended Option B.

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The Sigma Insight: Equation of a Line in Space

Solution Diagram

Analyzing the Setup

We are exploring the geometry of a line

L

that maintains a constant distance from two planes:

P_{1} : x + 2 y - z + 1 = 0

P_{2} : 2 x - y + z - 1 = 0

Since the distance from the line to each plane is constant, the line

L

must be parallel to both planes. This implies that the direction vector

v

of the line

L

is perpendicular to the normal vectors of both planes,

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

and

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

Determining the Direction of $L$

To find the direction vector

v

, we compute the cross product of the normals:

v = n_{1} \times n_{2} = \hat{i} 12 \hat{j} 2 - 1 \hat{k} - 1 1

Expanding the determinant, we obtain:

v = \hat{i} (2 - 1) - \hat{j} (1 - (- 2)) + \hat{k} (- 1 - 4) = \hat{i} - 3 \hat{j} - 5 \hat{k}

Given that the line passes through the origin, any point

Q

on the line

L

can be represented as:

Q = (t, - 3 t, - 5 t)

The Locus Construction

We define

M

as the locus of the feet of the perpendiculars from points on

L

to the plane

P_{1}

. For a point

Q (x_{1}, y_{1}, z_{1})

L

, the foot of the perpendicular

(x, y, z)

onto

P_{1}

is given by:

\frac{x - x _{1}}{a} = \frac{y - y _{1}}{b} = \frac{z - z _{1}}{c} = - \frac{a x _{1} + b y _{1} + c z _{1} + d}{a ^{2} + b ^{2} + c ^{2}}

Substituting

Q (t, - 3 t, - 5 t)

and the coefficients of

P_{1}

(

a = 1, b = 2, c = - 1, d = 1

), we get:

\frac{x - t}{1} = \frac{y + 3 t}{2} = \frac{z + 5 t}{- 1} = - \frac{t + 2 ( - 3 t ) - ( - 5 t ) + 1}{1 ^{2} + 2 ^{2} + ( - 1 ) ^{2}}

The Moment of Elegance

Simplifying the numerator of the right-hand side, we observe:

t - 6 t + 5 t + 1 = 1

The denominator is

1 + 4 + 1 = 6

. Thus, the ratio simplifies to:

\frac{x - t}{1} = \frac{y + 3 t}{2} = \frac{z + 5 t}{- 1} = - \frac{1}{6}

Solving for the coordinates

(x, y, z)

in terms of the parameter

t

x = t - \frac{1}{6}

y = - 3 t - \frac{1}{3}

z = - 5 t + \frac{1}{6}

This parametric representation defines the locus $M$ , which is a straight line in 3D space. By varying

t

, we can identify any point on this locus, confirming the path of the foot of the perpendicular as the line

L

moves through space.

Common Traps & FAQs

Wait, why do we assume the line is parallel to the plane? Couldn't it be skew or just not intersecting?

That is a fantastic question! In 3D geometry, the distance between a line and a plane is defined as the perpendicular distance from any point on the line to the plane. If this distance is constant, the line cannot be getting closer to or further from the plane. If it weren't parallel, the line would eventually intersect the plane (where the distance would be zero) or diverge to infinity. Thus, parallelism is the only geometric configuration that satisfies the 'constant distance' condition.

Why do we take the cross product of the normals to find the line's direction?

Think about the geometry: the line

L

is parallel to plane

P_{1}

, so its direction vector

v

is perpendicular to the normal

n_{1}

. Similarly,

L

is parallel to

P_{2}

, so

v

is perpendicular to

n_{2}

. If a vector is perpendicular to two other vectors, it must be parallel to their cross product,

n_{1} \times n_{2}

. It is the most elegant way to lock down the direction of

L

in space.

Is the locus

M

always a line? How do we know?

Yes, absolutely! Think of it this way: you are projecting a line

L

onto a plane

P_{1}

. The projection of a straight line onto a plane is always another straight line (unless the line is perpendicular to the plane, which is not the case here). By using the foot of the perpendicular formula, we are essentially mapping every point

Q (t)

L

to a point on

M

. Since the mapping is linear in terms of the parameter

t

, the resulting locus

M

is indeed a line.

Select Answer:

Visualized Solution

Understanding the Geometry

Implication of Constant Distance

Normal Vectors of the Planes

Direction Vector of Line $L$

Parametric Equation of Line $L$

Defining the Locus $M$

Foot of Perpendicular Formula

Applying the Formula for $M$

Evaluating the Constant Ratio

Coordinates of Locus $M$

Testing Option A

Testing Option B

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

Determining the Direction of $L$

The Locus Construction

The Moment of Elegance

Similar Questions

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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 the line $L$ pass through $(1, 1, 1)$ and intersect the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z}{1}$ . Then, which of the following points lies on the line $L$ ?

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

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:

Three lines $L_{1} : r = λ \hat{i}, λ \in R$ , $L_{2} : r = \hat{k} + μ \hat{j}, μ \in R$ and $L_{3} : r = \hat{i} + \hat{j} + ν \hat{k}, ν \in R$ are given. For which point(s) $Q$ on $L_{2}$ can we find a point $P$ on $L_{1}$ and a point $R$ on $L_{3}$ so that $P, Q$ and $R$ are collinear ?

Select Answer:

Visualized Solution

Understanding the Geometry

Implication of Constant Distance

Normal Vectors of the Planes

Direction Vector of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L

Parametric Equation of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L

Defining the Locus .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M

Foot of Perpendicular Formula

Applying the Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M

Evaluating the Constant Ratio

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

Testing Option A

Testing Option B

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

Determining the Direction of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L

The Locus Construction

The Moment of Elegance

Similar Questions

Direction Vector of Line $L$

Parametric Equation of Line $L$

Defining the Locus $M$

Applying the Formula for $M$

Coordinates of Locus $M$

Determining the Direction of $L$