JEE Main 2024 (04 Apr Shift 1)

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Animated Solution for Mathematics - Three Dimensional Geometry: Let the point, on the line passing through the points $P (1, - 2, 3)$ and $Q (5, - 4, 7)$ , farther from the origin and at distance of 9 units from the point $P$ , be $(α, β, γ)$ . Then $α^{2} + β^{2} + γ^{2}$ is equal to :

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

Absolutely! In fact, using the vector form $\vec{r} = \vec{a} + t\vec{v}$ is often more elegant. Here, $\vec{a} = \vec{OP} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{v} = \vec{PQ} = 4\hat{i} - 2\hat{j} + 4\hat{k}$. The distance condition $|\vec{r} - \vec{a}| = 9$ simplifies to $|t\vec{v}| = 9$, which leads to $|t| |\vec{v}| = 9$. Since $|\vec{v}| = \sqrt{4^2 + (-2)^2 + 4^2} = 6$, you get $|t| = \frac{9}{6} = 1.5$. It's the same result, just a different, perhaps faster, path to the summit!

Select Answer:

Progressive Hints

Hint 1

Visualize the line as a path in 3D space. Instead of thinking about the line as a static object, think of it as a trajectory defined by a parameter

t

. If you start at point

P

and move along the direction vector

P Q

, any point on this line can be represented as

P + t P Q

. What does the parameter

t

represent in terms of distance?

Hint 2

The distance constraint is your key. You are looking for points at a distance of

9

units from

P

. If the vector

P Q

has a magnitude of

6

, how does the parameter

t

relate to the distance

9

? Remember, the distance formula in 3D is

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

Hint 3

Set up the parametric coordinates of the point as

(4 t + 1, - 2 t - 2, 4 t + 3)

. Now, substitute these into the distance formula from

P (1, - 2, 3)

. You will get an equation in

t^{2}

. Solve for

t

, find the two candidate points, and then calculate their squared distances from the origin

(0, 0, 0)

to identify the one that is farther away.

Visualized Solution

Identify Points $P$ and $Q$

Given points $P (1, - 2, 3)$ and $Q (5, - 4, 7)$ .
We need to find a point $(α, β, γ)$ on the line $P Q$ at a distance of $9$ units from $P$ .

Direction Ratios of Line $P Q$

Direction ratios of line $P Q$ :
$(5 - 1, - 4 - (- 2), 7 - 3) = (4, - 2, 4)$ .

Formulate Line Equation

Equation of line $P Q$ :
$\frac{x - 1}{4} = \frac{y + 2}{- 2} = \frac{z - 3}{4} = t$ .

General Point on the Line

Any point on the line is:
$(x, y, z) = (4 t + 1, - 2 t - 2, 4 t + 3)$ .

Apply Distance Constraint

Distance from $P (1, - 2, 3)$ is $9$ units.
Using distance formula:
$(4 t + 1 - 1)^{2} + (- 2 t - 2 - (- 2))^{2} + (4 t + 3 - 3)^{2} = 9^{2}$ .

Simplify Distance Equation

Simplify the terms:
$(4 t)^{2} + (- 2 t)^{2} + (4 t)^{2} = 81$ .
$16 t^{2} + 4 t^{2} + 16 t^{2} = 81$ .

Solve for Parameter $t$

$36 t^{2} = 81 ⟹ t^{2} = \frac{81}{36} = \frac{9}{4}$ .
$t = \pm \frac{3}{2}$ .

Calculate First Candidate Point

For $t = \frac{3}{2}$ :
$x = 4 (\frac{3}{2}) + 1 = 7$ , $y = - 2 (\frac{3}{2}) - 2 = - 5$ , $z = 4 (\frac{3}{2}) + 3 = 9$ .
Point $A = (7, - 5, 9)$ .

Calculate Second Candidate Point

For $t = - \frac{3}{2}$ :
$x = 4 (- \frac{3}{2}) + 1 = - 5$ , $y = - 2 (- \frac{3}{2}) - 2 = 1$ , $z = 4 (- \frac{3}{2}) + 3 = - 3$ .
Point $B = (- 5, 1, - 3)$ .

Check Distance from Origin

The required point must be farther from the origin $(0, 0, 0)$ .
We calculate the squared distance for both points.

Evaluate Distances

For $A (7, - 5, 9)$ : Distance $^{2} = 7^{2} + (- 5)^{2} + 9^{2} = 49 + 25 + 81 = 155$ .
For $B (- 5, 1, - 3)$ : Distance $^{2} = (- 5)^{2} + 1^{2} + (- 3)^{2} = 25 + 1 + 9 = 35$ .

Final Calculation

The required point is $(α, β, γ) = (7, - 5, 9)$ .
We need to find $α^{2} + β^{2} + γ^{2}$ .
$α^{2} + β^{2} + γ^{2} = 7^{2} + (- 5)^{2} + 9^{2} = 155$ .

00:00 / 00:00

The Sigma Insight: Equation of a Line in Space

Solution Diagram

Analyzing the Setup

You are positioned at point

P (1, - 2, 3)

in a three-dimensional space. The infinite highway is defined by the line passing through

P

and

Q (5, - 4, 7)

To define the direction of this line, we calculate the direction vector

P Q

P Q = (5 - 1, - 4 - (- 2), 7 - 3) = (4, - 2, 4)

We can represent any point on this line using a parameter

t

. The parametric equations for the line are:

x = 4 t + 1

y = - 2 t - 2

z = 4 t + 3

The Master Equation

We seek a point

(α, β, γ)

on this line that is exactly

9

units away from

P (1, - 2, 3)

. Using the distance formula, we set the distance between the general point and

P

9

(4 t + 1 - 1)^{2} + (- 2 t - 2 + 2)^{2} + (4 t + 3 - 3)^{2} = 9

Simplifying the expression inside the square root, we obtain:

(4 t)^{2} + (- 2 t)^{2} + (4 t)^{2} = 9

Squaring both sides leads to:

16 t^{2} + 4 t^{2} + 16 t^{2} = 81

36 t^{2} = 81

Solving for

t

, we find:

t^{2} = \frac{81}{36} = \frac{9}{4} \Rightarrow t = \pm \frac{3}{2}

Final Calculation

We now determine the two candidate points by substituting

t = \frac{3}{2}

and

t = - \frac{3}{2}

into our parametric equations.

For

t = \frac{3}{2}

x = 4 (\frac{3}{2}) + 1 = 7, y = - 2 (\frac{3}{2}) - 2 = - 5, z = 4 (\frac{3}{2}) + 3 = 9

This gives point

A (7, - 5, 9)

For

t = - \frac{3}{2}

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

This gives point

B (- 5, 1, - 3)

To identify the point farther from the origin

(0, 0, 0)

, we calculate the squared distance

D^{2} = α^{2} + β^{2} + γ^{2}

for both:

For point

A

D_{A}^{2} = 7^{2} + (- 5)^{2} + 9^{2} = 49 + 25 + 81 = 155

For point

B

D_{B}^{2} = (- 5)^{2} + 1^{2} + (- 3)^{2} = 25 + 1 + 9 = 35

Since

155 > 35

, the point farther from the origin is

A (7, - 5, 9)

. The final value is 155.

Common Traps & FAQs

Wait, why do we get two points? Is the question flawed or did I miss something?

That is a fantastic observation! You haven't missed anything; you've actually uncovered a fundamental geometric truth. When you define a distance constraint from a point

P

on a line, you are essentially intersecting a sphere of radius

9

centered at

P

with that line. A line passing through the center of a sphere will always intersect it at two points—one in the direction of the vector

P Q

and one in the opposite direction. Both are valid mathematical solutions, which is why the problem provides the extra condition of being 'farther from the origin' to help you select the unique point required.

Do I really need to find the coordinates of the points to compare their distances from the origin?

While you could technically use vector properties to compare distances, calculating the coordinates is actually the most robust and error-proof method here. Since the question asks for

α^{2} + β^{2} + γ^{2}

, which is exactly the square of the distance from the origin, you are essentially calculating the final answer while performing the check. It's a beautiful synergy where the verification step and the final calculation merge into one.

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

Absolutely! In fact, using the vector form

r = a + t v

is often more elegant. Here,

a = O P = \hat{i} - 2 \hat{j} + 3 \hat{k}

and

v = P Q = 4 \hat{i} - 2 \hat{j} + 4 \hat{k}

. The distance condition

∣ r - a ∣ = 9

simplifies to

∣ t v ∣ = 9

, which leads to

∣ t ∣∣ v ∣ = 9

. Since

∣ v ∣ = 4^{2} + (- 2)^{2} + 4^{2} = 6

, you get

∣ t ∣ = \frac{9}{6} = 1.5

. It's the same result, just a different, perhaps faster, path to the summit!

Animated Solution for Mathematics - Three Dimensional Geometry: Let the point, on the line passing through the points $P (1, - 2, 3)$ and $Q (5, - 4, 7)$ , farther from the origin and at distance of 9 units from the point $P$ , be $(α, β, γ)$ . Then $α^{2} + β^{2} + γ^{2}$ is equal to :

Select Answer:

Visualized Solution

Identify Points $P$ and $Q$

Direction Ratios of Line $P Q$

Formulate Line Equation

General Point on the Line

Apply Distance Constraint

Simplify Distance Equation

Solve for Parameter $t$

Calculate First Candidate Point

Calculate Second Candidate Point

Check Distance from Origin

Evaluate Distances

Final Calculation

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

The square of the distance of the point $(\frac{15}{7}, \frac{32}{7}, 7)$ from the line $\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$ in the direction of the vector $\hat{i} + 4 \hat{j} + 7 \hat{k}$ is:

Let a line $L$ pass through the point $P (2, 3, 1)$ and be parallel to the line $x + 3 y - 2 z - 2 = 0 = x - y + 2 z$ . If the distance of $L$ from the point $(5, 3, 8)$ is $α$ , then $3 α^{2}$ is equal to ______.

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If the distance of the point $P (43, α, β), β < 0$ , from the line $r = 4 \hat{i} - \hat{k} + μ (2 \hat{i} + 3 \hat{k}), μ \in R$ along a line with direction ratios $3, - 1, 0$ is $1310$ , then $α^{2} + β^{2}$ is equal to ____

Let $Q (a, b, c)$ be the image of the point $P (3, 2, 1)$ in the line $\frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{1}$ . Then the distance of $Q$ from the line $\frac{x - 9}{3} = \frac{y - 9}{2} = \frac{z - 5}{- 2}$ is

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:

Consider the line $L$ passing through the points $(1, 2, 3)$ and $(2, 3, 5)$ . The distance of the point $(\frac{11}{3}, \frac{11}{3}, \frac{19}{3})$ from the line $L$ along the line $\frac{3 x - 11}{2} = \frac{3 y - 11}{1} = \frac{3 z - 19}{2}$ is equal to

Let $L$ be the line $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z + 3}{6}$ and let $S$ be the set of all points $(a, b, c)$ on $L$ , whose distance from the line $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z - 9}{0}$ along the line $L$ is 7. Then $\sum_{(a, b, c) \in S} (a + b + c)$ is equal to :

Let the line passing through the points $(- 1, 2, 1)$ and parallel to the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ intersect the line $\frac{x+2}{3}= rac{y-3}{2}= rac{z-4}{1}$ at the point P. Then the distance of P from the point $Q (4, - 5, 1)$ is

Select Answer:

Visualized Solution

Identify 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; } Q

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

Formulate Line Equation

General Point on the Line

Apply Distance Constraint

Simplify Distance Equation

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

Calculate First Candidate Point

Calculate Second Candidate Point

Check Distance from Origin

Evaluate Distances

Final Calculation

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Master Equation

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 (715​,732​,7) from the line 3x+1​=5y+3​=7z+5​ in the direction of the vector i^+4j^​+7k^ is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let Q(a,b,c) be the image of the point P(3,2,1) in the line 1x−1​=2y​=1z−1​. Then the distance of Q from the line 3x−9​=2y−9​=−2z−5​ is

Identify Points $P$ and $Q$

Direction Ratios of Line $P Q$

Solve for Parameter $t$

The square of the distance of the point $(\frac{15}{7}, \frac{32}{7}, 7)$ from the line $\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$ in the direction of the vector $\hat{i} + 4 \hat{j} + 7 \hat{k}$ is:

Let $Q (a, b, c)$ be the image of the point $P (3, 2, 1)$ in the line $\frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{1}$ . Then the distance of $Q$ from the line $\frac{x - 9}{3} = \frac{y - 9}{2} = \frac{z - 5}{- 2}$ is