JEE Main 2019 (08 April Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Three Dimensional Geometry: If a point R(4,y,z) lies on the line segment joining the points P(2,-3,4) and Q(8,0,10), then the distance of R from the origin is :

Q: Is there a way to do this using vectors instead of these fractions?

Yes! You could define the position vector $\vec{R} = \vec{P} + t(\vec{Q} - \vec{P})$. By equating the $x$-component to $4$, you would find the parameter $t$. This is actually the 'pro' way to solve it. If you are comfortable with vectors, I highly encourage you to try this method—it builds a deeper intuition for how points move along a line.

Select Answer:

Progressive Hints

Hint 1

Visualize the line segment

P Q

as a rigid rod in 3D space. Since point

R

lies on this segment, it is constrained by the same direction ratios that define the line connecting

P

and

Q

. Do not think of

R

as a random point; think of it as a point locked into a specific path.

Hint 2

Recall the symmetric form of the equation of a line passing through two points

(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}}

. This is your most powerful tool to link the unknown coordinates of

R

to the known coordinates of

P

and

Q

Hint 3

Since you know the

x

-coordinate of

R

4

, substitute

x = 4

into the first part of your line equation:

\frac{4 - 2}{8 - 2}

. This will yield a constant ratio. Set the other parts of the equation equal to this constant to solve for

y

and

z

individually.

Visualized Solution

Visualizing the Points in 3D Space

Given points: $P (2, - 3, 4)$ and $Q (8, 0, 10)$
These points define a line segment in 3D space.

Introducing Point $R$

Point $R (4, y, z)$ lies exactly on the line segment $P Q$ .
Goal: Find the distance of $R$ from the origin $(0, 0, 0)$ .

Equation of a Line in 3D

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 $P$ and $Q$

Substitute $P (2, - 3, 4)$ and $Q (8, 0, 10)$ into the formula:
$\frac{x - 2}{8 - 2} = \frac{y - ( - 3 )}{0 - ( - 3 )} = \frac{z - 4}{10 - 4}$

Simplifying the Line Equation

Simplifying the denominators (direction ratios):
$\frac{x - 2}{6} = \frac{y + 3}{3} = \frac{z - 4}{6}$

Applying the Condition for Point $R$

Since $R (4, y, z)$ lies on the line, it must satisfy the equation:
$\frac{4 - 2}{6} = \frac{y + 3}{3} = \frac{z - 4}{6}$

Finding the Common Ratio

Calculate the first ratio using the known $x$ -coordinate:
$\frac{4 - 2}{6} = \frac{2}{6} = \frac{1}{3}$
Therefore, all parts of the equation equal $\frac{1}{3}$ .

Solving for $y$

Equate the $y$ -part to the common ratio:
$\frac{y + 3}{3} = \frac{1}{3}$
$y + 3 = 1 ⟹ y = - 2$

Solving for $z$

Equate the $z$ -part to the common ratio:
$\frac{z - 4}{6} = \frac{1}{3}$
$z - 4 = \frac{6}{3} = 2 ⟹ z = 6$

The Complete Coordinates of $R$

The exact coordinates of point $R$ are $(4, - 2, 6)$ .
We need the distance from the origin $O (0, 0, 0)$ to $R (4, - 2, 6)$ .

Distance Formula from Origin

Distance $d$ of a point $(x, y, z)$ from the origin $(0, 0, 0)$ :
$d = x^{2} + y^{2} + z^{2}$

Substituting and Squaring

Substitute $x = 4$ , $y = - 2$ , $z = 6$ into the distance formula:
$d = 4^{2} + (- 2)^{2} + 6^{2}$
Evaluate the squares:
$d = 16 + 4 + 36$

Summing and Simplifying

Sum the values: $d = 16 + 4 + 36 = 56$
Factorize to simplify: $d = 4 \times 14$
$d = 214$

Final Conclusion

The distance of point $R$ from the origin is $214$ .
Key Takeaway: Collinear points share the same direction ratios, allowing us to find unknown coordinates.

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The Sigma Insight: Equation of a Line in Space

Solution Diagram

Analyzing the Setup

We are given two fixed points in 3D space:

P (2, - 3, 4)

and

Q (8, 0, 10)

. These points define a unique line segment. We seek a point

R (4, y, z)

that lies on this segment.

For points to be collinear, they must share the same direction ratios. We utilize 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}}

The Master Equation

By substituting the coordinates of

P

and

Q

into the symmetric form, we obtain the equation of the line:

\frac{x - 2}{8 - 2} = \frac{y - ( - 3 )}{0 - ( - 3 )} = \frac{z - 4}{10 - 4}

Simplifying the denominators, we get:

\frac{x - 2}{6} = \frac{y + 3}{3} = \frac{z - 4}{6}

Determining the Coordinates of R

We know the

x

-coordinate of

R

4

. Substituting

x = 4

into the first part of the equation yields our ratio:

\frac{4 - 2}{6} = \frac{2}{6} = \frac{1}{3}

This ratio,

\frac{1}{3}

, dictates the position of

R

along the segment. We now equate the remaining components to this value to solve for

y

and

z

For

y

\frac{y + 3}{3} = \frac{1}{3} \Rightarrow y + 3 = 1 \Rightarrow y = - 2

For

z

\frac{z - 4}{6} = \frac{1}{3} \Rightarrow z - 4 = 2 \Rightarrow z = 6

Thus, the coordinates of the point are

R (4, - 2, 6)

Final Calculation

To find the distance of

R

from the origin

(0, 0, 0)

, we apply the 3D distance formula:

d = x^{2} + y^{2} + z^{2}

Substituting our values:

d = 4^{2} + (- 2)^{2} + 6^{2} = 16 + 4 + 36 = 56

Simplifying the radical, we arrive at the final result:

$d = 214$

Common Traps & FAQs

Wait, why do we use the symmetric form of the line equation? Can't we just use the section formula?

You absolutely can! The section formula is a fantastic alternative. However, the symmetric form is often more intuitive when you are given the coordinates of two points and need to find a third point on the line. It treats

x

y

, and

z

symmetrically, which reduces the chance of algebraic errors. Both paths lead to the same destination—choose the one that feels most natural to you!

I got

y = - 2

. When I calculate the distance, should I worry about the negative sign?

That is a brilliant question! In the distance formula, we square the coordinates:

x^{2} + y^{2} + z^{2}

. Because

(- 2)^{2}

becomes positive

4

, the negative sign effectively vanishes. Always remember: in geometry, the magnitude of the coordinate matters more than its sign when calculating distances from the origin.

Is there a way to do this using vectors instead of these fractions?

Yes! You could define the position vector

R = P + t (Q - P)

. By equating the

x

-component to

4

, you would find the parameter

t

. This is actually the 'pro' way to solve it. If you are comfortable with vectors, I highly encourage you to try this method—it builds a deeper intuition for how points move along a line.

Animated Solution for Mathematics - Three Dimensional Geometry: If a point R(4,y,z) lies on the line segment joining the points P(2,-3,4) and Q(8,0,10), then the distance of R from the origin is :

Select Answer:

Visualized Solution

Visualizing the Points in 3D Space

Introducing Point $R$

Equation of a Line in 3D

Substituting Coordinates of $P$ and $Q$

Simplifying the Line Equation

Applying the Condition for Point $R$

Finding the Common Ratio

Solving for $y$

Solving for $z$

The Complete Coordinates of $R$

Distance Formula from Origin

Substituting and Squaring

Summing and Simplifying

Final Conclusion

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Master Equation

Determining the Coordinates of R

Final Calculation

Similar Questions

The distance of the line $\frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4}$ from the point (1,4,0) along the line $\frac{x}{1} = \frac{y - 2}{2} = \frac{z + 3}{3}$ is:

The perpendicular distance, of the line $\frac{x - 1}{2} = \frac{y + 2}{- 1} = \frac{z + 3}{2}$ from the point $P (2, - 10, 1)$ , is :

The distance of the point having position vector $- \hat{i} + 2 \hat{j} + 6 \hat{k}$ from the straight line passing through the point $(2, 3, - 4)$ and parallel to the vector, $6 \hat{i} + 3 \hat{j} - 4 \hat{k}$ is :

Let $P$ be the point $(10, - 2, - 1)$ and $Q$ be the foot of the perpendicular drawn from the point $R (1, 7, 6)$ on the line passing through the points $(2, - 5, 11)$ and $(- 6, 7, - 5)$ . Then the length of the line segment $P Q$ 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

The distance of the point $P (4, 6, - 2)$ from the line passing through the point $(- 3, 2, 3)$ and parallel to a line with direction ratios $3, 3, - 1$ is equal to :

The distance of line $3 y - 2 z - 1 = 0 = 3 x - z + 4$ from the point $(2, - 1, 6)$ is :

The length of the perpendicular from the point $(2, - 1, 4)$ on the straight line, $\frac{x + 3}{10} = \frac{y - 2}{- 7} = \frac{z}{1}$ is :

The distance, of the point $(7, - 2, 11)$ from the line $\frac{x - 6}{1} = \frac{y - 4}{0} = \frac{z - 8}{3}$ along the line $\frac{x - 5}{2} = \frac{y - 1}{- 3} = \frac{z - 5}{6}$ , is :

Animated Solution for Mathematics - Three Dimensional Geometry: If a point R(4,y,z) lies on the line segment joining the points P(2,-3,4) and Q(8,0,10), then the distance of R from the origin is :

Select Answer:

Visualized Solution

Visualizing the Points in 3D Space

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

Equation of a Line in 3D

Substituting Coordinates of .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

Simplifying the Line Equation

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

Finding the Common Ratio

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

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

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

Distance Formula from Origin

Substituting and Squaring

Summing and Simplifying

Final Conclusion

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Master Equation

Determining the Coordinates of R

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 line 2x−2​=3y−6​=4z−3​ from the point (1,4,0) along the line 1x​=2y−2​=3z+3​ is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The perpendicular distance, of the line 2x−1​=−1y+2​=2z+3​ from the point P(2,−10,1), is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The distance of the point P(4,6,−2) from the line passing through the point (−3,2,3) and parallel to a line with direction ratios 3,3,−1 is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The distance of line 3y−2z−1=0=3x−z+4 from the point (2,−1,6) is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The length of the perpendicular from the point (2,−1,4) on the straight line, 10x+3​=−7y−2​=1z​ is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The distance, of the point (7,−2,11) from the line 1x−6​=0y−4​=3z−8​ along the line 2x−5​=−3y−1​=6z−5​, is :

Introducing Point $R$

Substituting Coordinates of $P$ and $Q$

Applying the Condition for Point $R$

Solving for $y$

Solving for $z$

The Complete Coordinates of $R$

The distance of the line $\frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4}$ from the point (1,4,0) along the line $\frac{x}{1} = \frac{y - 2}{2} = \frac{z + 3}{3}$ is:

The perpendicular distance, of the line $\frac{x - 1}{2} = \frac{y + 2}{- 1} = \frac{z + 3}{2}$ from the point $P (2, - 10, 1)$ , is :

The distance of the point $P (4, 6, - 2)$ from the line passing through the point $(- 3, 2, 3)$ and parallel to a line with direction ratios $3, 3, - 1$ is equal to :

The distance of line $3 y - 2 z - 1 = 0 = 3 x - z + 4$ from the point $(2, - 1, 6)$ is :

The length of the perpendicular from the point $(2, - 1, 4)$ on the straight line, $\frac{x + 3}{10} = \frac{y - 2}{- 7} = \frac{z}{1}$ is :

The distance, of the point $(7, - 2, 11)$ from the line $\frac{x - 6}{1} = \frac{y - 4}{0} = \frac{z - 8}{3}$ along the line $\frac{x - 5}{2} = \frac{y - 1}{- 3} = \frac{z - 5}{6}$ , is :