JEE Main 2021 (27 July Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Three Dimensional Geometry: 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 ___

Q: Can I use the vector form of the line instead of the symmetric form?

Absolutely! The vector form $\vec{r} = \vec{a} + \lambda \vec{b}$ is mathematically identical to the symmetric form. If you prefer working with vectors, you can define $\vec{a}$ as the position vector of point $Q$ and $\vec{b}$ as the vector $\vec{QR}$. The logic remains exactly the same: substitute the components of $\vec{r}$ into the plane equation $2x + y + z = 7$ and solve for $\lambda$. Use whichever method makes you feel more confident!

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1: To find the intersection of a line and a plane, you must first represent any arbitrary point on the line in terms of a single parameter, say

λ

. This allows you to treat the line as a collection of points

(x, y, z)

that depend on this one variable.

Hint 2

Hint 2: Once you have the parametric coordinates

(x, y, z)

in terms of

λ

, remember that the intersection point

T

must satisfy the equation of the plane

2 x + y + z = 7

. Substitute your parametric expressions for

x

y

, and

z

into this plane equation to solve for the specific value of

λ

that corresponds to the intersection.

Hint 3

Hint 3: After finding the value of

λ

, plug it back into your parametric expressions to find the exact coordinates of the intersection point

T

. Finally, use the 3D distance formula

P T = (x_{T} - x_{P})^{2} + (y_{T} - y_{P})^{2} + (z_{T} - z_{P})^{2}

to calculate the distance between point

P

and point

T

Visualized Solution

Visualizing the $3$ D Geometry

Given Plane: $2 x + y + z = 7$
Given Points: $P (3, 4, 4)$ , $Q (3, - 4, - 5)$ , $R (2, - 3, 1)$
Goal: Find the distance $P T$ , where $T$ is the intersection of line $QR$ and the plane.

Equation of Line $QR$

Equation of a line passing through $(x_{1}, y_{1}, z_{1})$ and $(x_{2}, y_{2}, z_{2})$ :
$\frac{x - x _{1}}{x _{2} - x _{1}} = \frac{y - y _{1}}{y _{2} - y _{1}} = \frac{z - z _{1}}{z _{2} - z _{1}}$

Substituting Coordinates of $Q$ and $R$

Substituting $Q (3, - 4, - 5)$ and $R (2, - 3, 1)$ :
$\frac{x - 3}{2 - 3} = \frac{y - ( - 4 )}{- 3 - ( - 4 )} = \frac{z - ( - 5 )}{1 - ( - 5 )}$

Simplifying the Line Equation

Simplifying the denominators:
$\frac{x - 3}{- 1} = \frac{y + 4}{1} = \frac{z + 5}{6} = λ$

General Point on the Line

Expressing $x, y, z$ in terms of $λ$ :
$x = - λ + 3$
$y = λ - 4$
$z = 6 λ - 5$

Intersection with the Plane

The intersection point $T$ must lie on the plane $2 x + y + z = 7$ .

Substituting into the Plane Equation

Substitute $(- λ + 3, λ - 4, 6 λ - 5)$ into $2 x + y + z = 7$ :
$2 (- λ + 3) + (λ - 4) + (6 λ - 5) = 7$

Solving for $λ$

Expanding the equation:
$- 2 λ + 6 + λ - 4 + 6 λ - 5 = 7$
$5 λ - 3 = 7 ⟹ 5 λ = 10 ⟹ λ = 2$

Finding Intersection Point $T$

Substitute $λ = 2$ back into the coordinates:
$x = - 2 + 3 = 1$
$y = 2 - 4 = - 2$
$z = 6 (2) - 5 = 7$
Intersection Point $T = (1, - 2, 7)$

The Distance Formula

Distance formula between $P (x_{1}, y_{1}, z_{1})$ and $T (x_{2}, y_{2}, z_{2})$ :
$P T = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}$

Substituting Points $P$ and $T$

Substituting $P (3, 4, 4)$ and $T (1, - 2, 7)$ :
$P T = (3 - 1)^{2} + (4 - (- 2))^{2} + (4 - 7)^{2}$

Final Calculation

$P T = 2^{2} + 6^{2} + (- 3)^{2}$
$P T = 4 + 36 + 9 = 49 = 7$

Conclusion & Key Takeaway

Final Answer: The distance is $7$ units.
Key Takeaway: To find the intersection of a line and a plane, use the parametric form of the line $(λ)$ and substitute it into the plane equation.

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The Sigma Insight: Intersection of a Line and a Plane

Solution Diagram

Analyzing the Setup

We are given a plane defined by the equation

2 x + y + z = 7

and a point

P (3, 4, 4)

. Additionally, a line passes through the points

Q (3, - 4, - 5)

and

R (2, - 3, 1)

Our objective is to determine the point of intersection

T

where the line

QR

pierces the plane, and subsequently calculate the distance between point

P

and point

T

Defining the Path

To define the line

QR

, we use the symmetric form of the line equation:

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

Substituting the coordinates of

Q (3, - 4, - 5)

and

R (2, - 3, 1)

, we obtain:

\frac{x - 3}{2 - 3} = \frac{y - ( - 4 )}{- 3 - ( - 4 )} = \frac{z - ( - 5 )}{1 - ( - 5 )}

Simplifying the denominators, the equation of the line becomes:

\frac{x - 3}{- 1} = \frac{y + 4}{1} = \frac{z + 5}{6}

By setting these ratios equal to a parameter

λ

, we express any point on the line as a function of

λ

x = - λ + 3, y = λ - 4, z = 6 λ - 5

The Moment of Impact

The point

T

lies on the plane

2 x + y + z = 7

. Therefore, the parametric coordinates must satisfy this equation:

2 (- λ + 3) + (λ - 4) + (6 λ - 5) = 7

Expanding the terms, we get:

- 2 λ + 6 + λ - 4 + 6 λ - 5 = 7

Combining the

λ

terms and the constants:

5 λ - 3 = 7

5 λ = 10 ⟹ λ = 2

Substituting

λ = 2

back into our parametric equations, we find the coordinates of

T

x = - 2 + 3 = 1

y = 2 - 4 = - 2

z = 6 (2) - 5 = 7

Thus, the intersection point is $T (1, - 2, 7)$ .

Final Calculation

We now calculate the distance between

P (3, 4, 4)

and

T (1, - 2, 7)

using the 3D distance formula:

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

Substituting the coordinates:

P T = (3 - 1)^{2} + (4 - (- 2))^{2} + (4 - 7)^{2}

P T = 2^{2} + 6^{2} + (- 3)^{2}

Calculating the squares:

P T = 4 + 36 + 9 = 49

The final distance is $7$ .

Common Traps & FAQs

Wait, why do we use

λ

instead of just solving for

x, y, z

directly?

Great question! A line in 3D space contains infinitely many points. We cannot solve for a single

(x, y, z)

because the line doesn't have one coordinate; it has a continuum of them. By introducing the parameter

λ

, we create a 'slider' that allows us to navigate to any point on that line. When we set that expression equal to the plane's equation, we are essentially asking, 'At what value of the slider does the line touch the plane?' It is the most efficient way to bridge the gap between a line and a plane.

What if the line is parallel to the plane? Will this method still work?

If the line is parallel to the plane, the direction vector of the line will be perpendicular to the normal vector of the plane. Mathematically, when you substitute the parametric form into the plane equation, the

λ

terms will cancel out, leaving you with a contradiction (like

0 = 5

) or an identity (like

7 = 7

). In this specific problem, the line is not parallel, so you will get a unique, solvable value for

λ

Can I use the vector form of the line instead of the symmetric form?

Absolutely! The vector form

r = a + λ b

is mathematically identical to the symmetric form. If you prefer working with vectors, you can define

a

as the position vector of point

Q

and

b

as the vector

QR

. The logic remains exactly the same: substitute the components of

r

into the plane equation

2 x + y + z = 7

and solve for

λ

. Use whichever method makes you feel more confident!

Animated Solution for Mathematics - Three Dimensional Geometry: 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 ___

Enter Numerical Value:

Visualized Solution

Visualizing the $3$ D Geometry

Equation of Line $QR$

Substituting Coordinates of $Q$ and $R$

Simplifying the Line Equation

General Point on the Line

Intersection with the Plane

Substituting into the Plane Equation

Solving for $λ$

Finding Intersection Point $T$

The Distance Formula

Substituting Points $P$ and $T$

Final Calculation

Conclusion & Key Takeaway

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

Defining the Path

The Moment of Impact

Final Calculation

Similar Questions

The distance of the point $(1, 1, 9)$ from the point of intersection of the line $\frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z - 5}{2}$ and the plane $x + y + z = 17$ 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 distance of the point $(1, - 2, 3)$ from the plane $x - y + z = 5$ measured parallel to the line $\frac{x}{2} = \frac{y}{3} = \frac{z}{- 6}$ is:

If the line, $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 2}{4}$ meets the plane, $x + 2 y + 3 z = 15$ at a point $P$ , then the distance of $P$ from the origin is

The distance of the point $(- 1, 9, - 16)$ from the plane $2 x + 3 y - z = 5$ measured parallel to the line $\frac{x + 4}{3} = \frac{2 - y}{4} = \frac{z - 3}{12}$ is

If the lines $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z + 3}{1}$ and $\frac{x - a}{2} = \frac{y + 2}{3} = \frac{z - 3}{1}$ intersects at the point $P$ , then the distance of the point $P$ from the plane $z = a$ is :

The distance of the point $(1, - 2, 3)$ from the plane $x - y + z = 5$ measured parallel to a line, whose direction ratios are $2, 3, - 6$ is :

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 .

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 .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3D Geometry

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

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

Simplifying the Line Equation

General Point on the Line

Intersection with the Plane

Substituting into the Plane Equation

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

Finding Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T

The Distance Formula

Substituting Points .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T

Final Calculation

Conclusion & Key Takeaway

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

Defining the Path

The Moment of Impact

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The distance of the point (1,1,9) from the point of intersection of the line 1x−3​=2y−4​=2z−5​ and the plane x+y+z=17 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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The distance of the point (1,−2,3) from the plane x−y+z=5 measured parallel to the line 2x​=3y​=−6z​ is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the line, 2x−1​=3y+1​=4z−2​ meets the plane, x+2y+3z=15 at a point P, then the distance of P from the origin is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The distance of the point (−1,9,−16) from the plane 2x+3y−z=5 measured parallel to the line 3x+4​=42−y​=12z−3​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the lines 1x−1​=2y−2​=1z+3​ and 2x−a​=3y+2​=1z−3​ intersects at the point P, then the distance of the point P from the plane z=a 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,−2,3) from the plane x−y+z=5 measured parallel to a line, whose direction ratios are 2,3,−6 is :

.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 .

Visualizing the $3$ D Geometry

Equation of Line $QR$

Substituting Coordinates of $Q$ and $R$

Solving for $λ$

Finding Intersection Point $T$

Substituting Points $P$ and $T$

The distance of the point $(1, 1, 9)$ from the point of intersection of the line $\frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z - 5}{2}$ and the plane $x + y + z = 17$ 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 distance of the point $(1, - 2, 3)$ from the plane $x - y + z = 5$ measured parallel to the line $\frac{x}{2} = \frac{y}{3} = \frac{z}{- 6}$ is:

If the line, $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 2}{4}$ meets the plane, $x + 2 y + 3 z = 15$ at a point $P$ , then the distance of $P$ from the origin is

The distance of the point $(- 1, 9, - 16)$ from the plane $2 x + 3 y - z = 5$ measured parallel to the line $\frac{x + 4}{3} = \frac{2 - y}{4} = \frac{z - 3}{12}$ is

If the lines $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z + 3}{1}$ and $\frac{x - a}{2} = \frac{y + 2}{3} = \frac{z - 3}{1}$ intersects at the point $P$ , then the distance of the point $P$ from the plane $z = a$ is :

The distance of the point $(1, - 2, 3)$ from the plane $x - y + z = 5$ measured parallel to a line, whose direction ratios are $2, 3, - 6$ is :

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 .