JEE Main 2022 (29 June Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Three Dimensional Geometry: Let $d$ be the distance between the foot of perpendiculars of the points $P (1, 2, - 1)$ and $Q (2, - 1, 3)$ on the plane $- x + y + z = 1$ . Then $d^{2}$ is equal to ____.

Enter Numerical Value:

Visualized Solution

Visualizing the Problem Setup

Given points: $P (1, 2, - 1)$ and $Q (2, - 1, 3)$
Plane equation: $- x + y + z - 1 = 0$
Objective: Find $d^{2}$ , where $d$ is the distance between the foot of perpendiculars of $P$ and $Q$ on the plane.

Distance of a Point from a Plane

The perpendicular distance $h$ from a point $(x_{1}, y_{1}, z_{1})$ to a plane $a x + b y + cz + d = 0$ is given by:
$h = \frac{∣ a x _{1} + b y _{1} + c z _{1} + d ∣}{a ^{2} + b ^{2} + c ^{2}}$

Setting up Distance for $P$

For point $P (1, 2, - 1)$ and plane $- x + y + z - 1 = 0$ :
$h_{P} = \frac{∣ - ( 1 ) + ( 2 ) + ( - 1 ) - 1∣}{( - 1 ) ^{2} + 1 ^{2} + 1 ^{2}}$

Calculating Distance $h_{P}$

$h_{P} = \frac{∣ - 1 + 2 - 1 - 1∣}{1 + 1 + 1}$
$h_{P} = \frac{∣ - 1∣}{3} = \frac{1}{3}$

Setting up Distance for $Q$

For point $Q (2, - 1, 3)$ and plane $- x + y + z - 1 = 0$ :
$h_{Q} = \frac{∣ - ( 2 ) + ( - 1 ) + ( 3 ) - 1∣}{( - 1 ) ^{2} + 1 ^{2} + 1 ^{2}}$

Calculating Distance $h_{Q}$

$h_{Q} = \frac{∣ - 2 - 1 + 3 - 1∣}{1 + 1 + 1}$
$h_{Q} = \frac{∣ - 1∣}{3} = \frac{1}{3}$

Geometric Deduction: Parallelism

Since $h_{P} = h_{Q} = \frac{1}{3}$ , both points are at the same perpendicular distance from the plane.
This implies the line segment $P Q$ is parallel to the plane.

Equating Distances

Because $P Q$ is parallel to the plane, the quadrilateral formed by $P, Q, Q^{'}, P^{'}$ is a rectangle.
Therefore, the distance between the foot of perpendiculars $d$ is exactly equal to the distance $P Q$ .
$d = P Q ⟹ d^{2} = P Q^{2}$

Setting up $P Q^{2}$

Using the 3D distance formula:
$P Q^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}$
Substitute $P (1, 2, - 1)$ and $Q (2, - 1, 3)$ :
$d^{2} = (2 - 1)^{2} + (- 1 - 2)^{2} + (3 - (- 1))^{2}$

Calculating the Differences

$d^{2} = (1)^{2} + (- 3)^{2} + (4)^{2}$
$d^{2} = 1 + 9 + 16$

Final Conclusion

$d^{2} = 26$
Key Takeaway: Recognizing geometric symmetries (like equal distances implying parallelism) can save you from calculating complex foot of perpendicular coordinates.

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The Sigma Insight: Intersection of a Line and a Plane

Solution Diagram

Analyzing the Setup

The floor is defined by the plane equation

- x + y + z - 1 = 0

. We are given two points in space,

P (1, 2, - 1)

and

Q (2, - 1, 3)

We need to find the square of the distance between the projections

P^{'}

and

Q^{'}

of these points onto the plane.

The Insight

Measuring the Hover

The perpendicular distance

h

from a point

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

to a plane

a x + b y + cz + d = 0

is given by:

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

For point

P (1, 2, - 1)

, the distance

h_{P}

to the plane

- x + y + z - 1 = 0

is:

h_{P} = \frac{∣ - ( 1 ) + ( 2 ) + ( - 1 ) - 1∣}{( - 1 ) ^{2} + 1 ^{2} + 1 ^{2}} = \frac{∣ - 1∣}{3} = \frac{1}{3}

For point

Q (2, - 1, 3)

, the distance

h_{Q}

to the same plane is:

h_{Q} = \frac{∣ - ( 2 ) + ( - 1 ) + ( 3 ) - 1∣}{( - 1 ) ^{2} + 1 ^{2} + 1 ^{2}} = \frac{∣ - 1∣}{3} = \frac{1}{3}

The Moment of Clarity

Since

h_{P} = h_{Q} = \frac{1}{3}

, both points are at the exact same height above the floor. This implies that the line segment

P Q

is parallel to the plane.

Geometrically, the points

P, Q, Q^{'},

and

P^{'}

form a rectangle. In any rectangle, the distance between the projections

P^{'} Q^{'}

is equal to the distance between the original points

P Q

Final Calculation

We now calculate the square of the distance between

P (1, 2, - 1)

and

Q (2, - 1, 3)

using the 3D distance formula:

d^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}

Substituting the coordinates of

P

and

Q

d^{2} = (2 - 1)^{2} + (- 1 - 2)^{2} + (3 - (- 1))^{2}

Simplifying the expression:

d^{2} = (1)^{2} + (- 3)^{2} + (4)^{2}

d^{2} = 1 + 9 + 16

The final result is:

$d^{2} = 26$

Common Traps & FAQs

Why can't I just find the coordinates of the foot of the perpendicular for both points and then use the distance formula?

You absolutely can, and it will give you the correct answer! However, in a high-stakes exam like JEE Advanced, time is your most valuable resource. Calculating the coordinates of the foot of the perpendicular involves finding the equation of the line passing through the point and perpendicular to the plane, finding the intersection point, and then applying the distance formula. It is a multi-step process prone to arithmetic errors. Recognizing the geometric symmetry here allows you to bypass all that work.

How do I know for sure that the line segment is parallel to the plane?

A line is parallel to a plane if and only if the perpendicular distance from any point on the line to the plane is constant. Since we calculated that both points

P

and

Q

are at the same distance

h = \frac{1}{3}

from the plane, the line segment

P Q

must be parallel to the plane. This is a powerful geometric property.

What if the distances were not equal? Would the distance between the footprints still be the same as the distance between the points?

Great question! No, it would not. If the distances were different, the line

P Q

would be inclined to the plane. In that case, you would have to project the vector

P Q

onto the plane or calculate the coordinates of the footprints explicitly. Always check for symmetry first; it is the hallmark of a well-designed JEE problem.

Animated Solution for Mathematics - Three Dimensional Geometry: Let $d$ be the distance between the foot of perpendiculars of the points $P (1, 2, - 1)$ and $Q (2, - 1, 3)$ on the plane $- x + y + z = 1$ . Then $d^{2}$ is equal to ____.

Enter Numerical Value:

Visualized Solution

Visualizing the Problem Setup

Distance of a Point from a Plane

Setting up Distance for $P$

Calculating Distance $h_{P}$

Setting up Distance for $Q$

Calculating Distance $h_{Q}$

Geometric Deduction: Parallelism

Equating Distances

Setting up $P Q^{2}$

Calculating the Differences

Final Conclusion

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

The Insight

The Moment of Clarity

Final Calculation

Similar Questions

The square of the distance of the point of intersection of the line $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 1}{6}$ and the plane $2 x - y + z = 6$ from the point $(- 1, - 1, 2)$ is .

If for $a > 0$ , the feet of perpendiculars from the points $A (a, - 2 a, 3)$ and $B (0, 4, 5)$ on the plane $l x + m y + n z = 0$ are points $C (0, - a, - 1)$ and $D$ respectively, then the length of line segment $C D$ is equal to :

If the foot of the perpendicular from the point $A (- 1, 4, 3)$ on the plane $P : 2 x + m y + n z = 4$ , is $(- 2, \frac{7}{2}, \frac{3}{2})$ , then the distance of the point $A$ from the plane $P$ , measured parallel to a line with direction ratios $3, - 1, - 4$ , is equal to :

Let the foot of the perpendicular from the point $(1, 2, 4)$ on the line $\frac{x + 2}{4} = \frac{y - 1}{2} = \frac{z + 1}{3}$ be $P$ . Then the distance of $P$ from the plane $3 x + 4 y + 12 z + 23 = 0$

The distance of the point $P (3, 4, 4)$ from the point of intersection of the line joining the points $Q (3, - 4, - 5)$ and $R (2, - 3, 1)$ and the plane $2 x + y + z = 7$ , is equal to ___

If the distance of the point $P (1, - 2, 1)$ from the plane $x + 2 y - 2 z = α$ , where $α > 0$ , is 5, then the foot of the perpendicular from $P$ to the plane is

The distance of the point $(1, 0, 2)$ from the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 16$ , is

The point $P$ is the intersection of the straight line joining the points $Q (2, 3, 5)$ and $R (1, - 1, 4)$ with the plane $5 x - 4 y - z = 1$ . If $S$ is the foot of the perpendicular drawn from the point $T (2, 1, 4)$ to $QR$ , then the length of the line segment $P S$ is

Enter Numerical Value:

Visualized Solution

Visualizing the Problem Setup

Distance of a Point from a Plane

Setting up Distance for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

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

Setting up Distance for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

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

Geometric Deduction: Parallelism

Equating Distances

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

Calculating the Differences

Final Conclusion

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

The Insight

The Moment of Clarity

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The square of the distance of the point of intersection of the line 2x−1​=3y−2​=6z+1​ and the plane 2x−y+z=6 from the point (−1,−1,2) is .

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the foot of the perpendicular from the point (1,2,4) on the line 4x+2​=2y−1​=3z+1​ be P. Then the distance of P from the plane 3x+4y+12z+23=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The distance of the point P(3,4,4) from the point of intersection of the line joining the points Q(3,−4,−5) and R(2,−3,1) and the plane 2x+y+z=7, is equal to ___

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the distance of the point P(1,−2,1) from the plane x+2y−2z=α, where α>0, is 5, then the foot of the perpendicular from P to the plane is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The distance of the point (1,0,2) from the point of intersection of the line 3x−2​=4y+1​=12z−2​ and the plane x−y+z=16, is

Setting up Distance for $P$

Calculating Distance $h_{P}$

Setting up Distance for $Q$

Calculating Distance $h_{Q}$

Setting up $P Q^{2}$

The square of the distance of the point of intersection of the line $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 1}{6}$ and the plane $2 x - y + z = 6$ from the point $(- 1, - 1, 2)$ is .

Let the foot of the perpendicular from the point $(1, 2, 4)$ on the line $\frac{x + 2}{4} = \frac{y - 1}{2} = \frac{z + 1}{3}$ be $P$ . Then the distance of $P$ from the plane $3 x + 4 y + 12 z + 23 = 0$

The distance of the point $P (3, 4, 4)$ from the point of intersection of the line joining the points $Q (3, - 4, - 5)$ and $R (2, - 3, 1)$ and the plane $2 x + y + z = 7$ , is equal to ___

If the distance of the point $P (1, - 2, 1)$ from the plane $x + 2 y - 2 z = α$ , where $α > 0$ , is 5, then the foot of the perpendicular from $P$ to the plane is

The distance of the point $(1, 0, 2)$ from the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 16$ , is