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Three Dimensional Geometry: Mathematics | JEE Main

Q: Does C divide AB internally or externally?

Unless specified as external division, 'in the ratio 2:1' always refers to internal section.

The plane

2x - y + z = 4

intersects the line segment joining the points

\text{A}(a, -2, 4)

and

\text{B}(2, b, -3)

at the point

\text{C}

in the ratio

2: 1

and the distance of the point

\text{C}

from the origin is

\sqrt{5}

. If

ab < 0

and

\text{P}

is the point

(a - b, b, 2b - a)

then

\text{CP}^2

is equal to

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

A

\frac{97}{3}

B

\frac{17}{3}

C

\frac{16}{3}

D

\frac{73}{3}

Step-by-step solutionView

Correct answer

Using the section formula with ratio $2:1$ , point $C$ is found to be $(\frac{4+a}{3}, \frac{2b-2}{3}, -\frac{2}{3})$ ; substituting into the plane equation and the distance condition along with $ab < 0$ yields $a = 3$ and $b = -2$ , giving $CP^2 = \frac{17}{3}$ .

Option analysis

Why each option works or fails

A ·

\frac{97}{3}

Solving for $b$ gives a quadratic with two roots; selecting the root where $a$ and $b$ share the same sign ignores the explicit condition $ab < 0$ . Check the sign requirement $ab < 0$ immediately after solving for both parameters.

B ·

\frac{17}{3}

This is the correct answer. The coordinates of $C$ are $(\frac{7}{3}, -2, -\frac{2}{3})$ , point $P$ is $(5, -2, -7)$ , and the squared distance $CP^2 = (5 - \frac{7}{3})^2 + (-2 - (-2))^2 + (-7 - (-\frac{2}{3}))^2 = \frac{64}{9} + 0 + \frac{361}{9} = \frac{17}{3}$ .

C ·

\frac{16}{3}

An arithmetic error when evaluating the sum of squared differences, such as missing the fraction addition or miscomputing $(\frac{8}{3})^2 + (-\frac{19}{3})^2$ . Carefully compute $(\frac{8}{3})^2 + (-\frac{19}{3})^2 = \frac{64 + 361}{9} = \frac{425}{9}$ or recheck the distance formula terms.

D ·

\frac{73}{3}

Applying the section formula with the inverted ratio $1:2$ instead of $2:1$ from $A$ to $B$ . Remember that a point dividing $AB$ in the ratio $m:n$ has coordinates $\frac{mB + nA}{m + n}$ .

Reviewed route

Solution

StepWorking

01given

Points $A(a, -2, 4)$ and $B(2, b, -3)$ . Point $C$ divides $AB$ in ratio $2:1$ . Plane equation: $2x - y + z = 4$ . Distance $OC = \sqrt{5}$ . Constraint: $ab < 0$ . Point $P(a - b, b, 2b - a)$ .

02approach

Use section formula to express coordinates of $C$ in terms of $a$ and $b$ . Substitute $C$ into the plane equation to find a linear relation between $a$ and $b$ . Then use the distance from origin $OC = \sqrt{5}$ to form a quadratic equation in $b$ . Select the valid solution using $ab < 0$ , find $P$ , and compute $CP^2$ .

03execute

Coordinates of $C$ dividing $AB$ in $2:1$ are: $C = \left(\frac{2(2) + 1(a)}{2+1}, \frac{2(b) + 1(-2)}{2+1}, \frac{2(-3) + 1(4)}{2+1}\right) = \left(\frac{a+4}{3}, \frac{2b-2}{3}, \frac{-2}{3}\right)$

04execute

Since $C$ lies on the plane $2x - y + z = 4$ : $2\left(\frac{a+4}{3}\right) - \left(\frac{2b-2}{3}\right) + \left(\frac{-2}{3}\right) = 4$ $2(a+4) - (2b-2) - 2 = 12 \implies 2a - 2b + 8 = 12 \implies 2a - 2b = 4 \implies a - b = 2$ Thus, $a = b + 2$ .

05execute

Using $a = b + 2$ , the x-coordinate of $C$ becomes $\frac{(b+2)+4}{3} = \frac{b+6}{3}$ . Given $OC^2 = 5$ : $\left(\frac{b+6}{3}\right)^2 + \left(\frac{2b-2}{3}\right)^2 + \left(-\frac{2}{3}\right)^2 = 5$ $(b^2 + 12b + 36) + (4b^2 - 8b + 4) + 4 = 45$ $5b^2 + 4b + 44 = 45 \implies 5b^2 + 4b - 1 = 0$ $(5b - 1)(b + 1) = 0 \implies b = -1 \text{ or } b = \frac{1}{5}$

06execute

If $b = -1$ , then $a = -1 + 2 = 1$ , giving $ab = -1 < 0$ (valid). If $b = \frac{1}{5}$ , then $a = \frac{11}{5}$ , giving $ab = \frac{11}{25} > 0$ (rejected). Therefore, $a = 1, b = -1$ . Coordinates of $C$ are: $C\left(\frac{5}{3}, -\frac{4}{3}, -\frac{2}{3}\right)$ . Coordinates of $P(a - b, b, 2b - a)$ are: $P(1 - (-1), -1, 2(-1) - 1) = P(2, -1, -3)$ .

07execute

Calculate $CP^2$ : $CP^2 = \left(\frac{5}{3} - 2\right)^2 + \left(-\frac{4}{3} - (-1)\right)^2 + \left(-\frac{2}{3} - (-3)\right)^2$ $CP^2 = \left(-\frac{1}{3}\right)^2 + \left(-\frac{1}{3}\right)^2 + \left(\frac{7}{3}\right)^2 = \frac{1 + 1 + 49}{9} = \frac{51}{9} = \frac{17}{3}$

✓verify

Check $C$ on plane: $2(5/3) - (-4/3) + (-2/3) = (10 + 4 - 2)/3 = 12/3 = 4$ . Correct. Distance $OC = \sqrt{25/9 + 16/9 + 4/9} = \sqrt{45/9} = \sqrt{5}$ . Correct.

Hints that build this answer step by step

Using the section formula, what are the coordinates of point $C$ dividing the segment from $A(a, -2, 4)$ to $B(2, b, -3)$ in the ratio $2:1$ ?
$C = \left(\frac{4+a}{3}, \frac{2b-2}{3}, -\frac{2}{3}\right)$
Point $C$ lies on the plane $2x - y + z = 4$ . What linear relation between $a$ and $b$ does this provide?
$2a - 2b = 4$ , so $a - b = 2$
The distance of $C$ from the origin is $\sqrt{5}$ , so $\left(\frac{4+a}{3}\right)^2 + \left(\frac{2b-2}{3}\right)^2 + \left(-\frac{2}{3}\right)^2 = 5$ . Using $a = b + 2$ and $ab < 0$ , what are the values of $a$ and $b$ ?
$a = 3, b = -2$
With $a = 1$ and $b = -1$ , what are the coordinates of $C$ and $P$ , and the value of $CP^2$ ?
$C = \left(\frac{5}{3}, -\frac{4}{3}, -\frac{2}{3}\right)$ , $P = (2, -1, -3)$ , giving $CP^2 = \frac{17}{3}$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

Does C divide AB internally or externally?

Unless specified as external division, 'in the ratio 2:1' always refers to internal section.

Answer

Using the section formula with ratio $2:1$ , point $C$ is found to be $(\frac{4+a}{3}, \frac{2b-2}{3}, -\frac{2}{3})$ ; substituting into the plane equation and the distance condition along with $ab < 0$ yields $a = 3$ and $b = -2$ , giving $CP^2 = \frac{17}{3}$ .

Why each option works or fails

A: $\frac{97}{3}$ - Solving for $b$ gives a quadratic with two roots; selecting the root where $a$ and $b$ share the same sign ignores the explicit condition $ab < 0$ . Check the sign requirement $ab < 0$ immediately after solving for both parameters.
B · correct: $\frac{17}{3}$ - This is the correct answer. The coordinates of $C$ are $(\frac{7}{3}, -2, -\frac{2}{3})$ , point $P$ is $(5, -2, -7)$ , and the squared distance $CP^2 = (5 - \frac{7}{3})^2 + (-2 - (-2))^2 + (-7 - (-\frac{2}{3}))^2 = \frac{64}{9} + 0 + \frac{361}{9} = \frac{17}{3}$ .
C: $\frac{16}{3}$ - An arithmetic error when evaluating the sum of squared differences, such as missing the fraction addition or miscomputing $(\frac{8}{3})^2 + (-\frac{19}{3})^2$ . Carefully compute $(\frac{8}{3})^2 + (-\frac{19}{3})^2 = \frac{64 + 361}{9} = \frac{425}{9}$ or recheck the distance formula terms.
D: $\frac{73}{3}$ - Applying the section formula with the inverted ratio $1:2$ instead of $2:1$ from $A$ to $B$ . Remember that a point dividing $AB$ in the ratio $m:n$ has coordinates $\frac{mB + nA}{m + n}$ .

Step-by-step solution

given: Points $A(a, -2, 4)$ and $B(2, b, -3)$ . Point $C$ divides $AB$ in ratio $2:1$ . Plane equation: $2x - y + z = 4$ . Distance $OC = \sqrt{5}$ . Constraint: $ab < 0$ . Point $P(a - b, b, 2b - a)$ .
approach: Use section formula to express coordinates of $C$ in terms of $a$ and $b$ . Substitute $C$ into the plane equation to find a linear relation between $a$ and $b$ . Then use the distance from origin $OC = \sqrt{5}$ to form a quadratic equation in $b$ . Select the valid solution using $ab < 0$ , find $P$ , and compute $CP^2$ .
execute: Coordinates of $C$ dividing $AB$ in $2:1$ are: $C = \left(\frac{2(2) + 1(a)}{2+1}, \frac{2(b) + 1(-2)}{2+1}, \frac{2(-3) + 1(4)}{2+1}\right) = \left(\frac{a+4}{3}, \frac{2b-2}{3}, \frac{-2}{3}\right)$
execute: Since $C$ lies on the plane $2x - y + z = 4$ : $2\left(\frac{a+4}{3}\right) - \left(\frac{2b-2}{3}\right) + \left(\frac{-2}{3}\right) = 4$ $2(a+4) - (2b-2) - 2 = 12 \implies 2a - 2b + 8 = 12 \implies 2a - 2b = 4 \implies a - b = 2$ Thus, $a = b + 2$ .
execute: Using $a = b + 2$ , the x-coordinate of $C$ becomes $\frac{(b+2)+4}{3} = \frac{b+6}{3}$ . Given $OC^2 = 5$ : $\left(\frac{b+6}{3}\right)^2 + \left(\frac{2b-2}{3}\right)^2 + \left(-\frac{2}{3}\right)^2 = 5$ $(b^2 + 12b + 36) + (4b^2 - 8b + 4) + 4 = 45$ $5b^2 + 4b + 44 = 45 \implies 5b^2 + 4b - 1 = 0$ $(5b - 1)(b + 1) = 0 \implies b = -1 \text{ or } b = \frac{1}{5}$
execute: If $b = -1$ , then $a = -1 + 2 = 1$ , giving $ab = -1 < 0$ (valid). If $b = \frac{1}{5}$ , then $a = \frac{11}{5}$ , giving $ab = \frac{11}{25} > 0$ (rejected). Therefore, $a = 1, b = -1$ . Coordinates of $C$ are: $C\left(\frac{5}{3}, -\frac{4}{3}, -\frac{2}{3}\right)$ . Coordinates of $P(a - b, b, 2b - a)$ are: $P(1 - (-1), -1, 2(-1) - 1) = P(2, -1, -3)$ .
execute: Calculate $CP^2$ : $CP^2 = \left(\frac{5}{3} - 2\right)^2 + \left(-\frac{4}{3} - (-1)\right)^2 + \left(-\frac{2}{3} - (-3)\right)^2$ $CP^2 = \left(-\frac{1}{3}\right)^2 + \left(-\frac{1}{3}\right)^2 + \left(\frac{7}{3}\right)^2 = \frac{1 + 1 + 49}{9} = \frac{51}{9} = \frac{17}{3}$
verify: Check $C$ on plane: $2(5/3) - (-4/3) + (-2/3) = (10 + 4 - 2)/3 = 12/3 = 4$ . Correct. Distance $OC = \sqrt{25/9 + 16/9 + 4/9} = \sqrt{45/9} = \sqrt{5}$ . Correct.

Three Dimensional Geometry: Mathematics | JEE Main

What feels right?

Using the section formula with ratio $2:1$ , point $C$ is found to be $(\frac{4+a}{3}, \frac{2b-2}{3}, -\frac{2}{3})$ ; substituting into the plane equation and the distance condition along with $ab < 0$ yields $a = 3$ and $b = -2$ , giving $CP^2 = \frac{17}{3}$ .

Why each option works or fails

Solution

01given

02approach

03execute

04execute

05execute

06execute

07execute

✓verify

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