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Three Dimensional Geometry: Mathematics | JEE Main

The line, that is coplanar to the line

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

, is

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\frac{x + 1}{1} = \frac{y - 2}{2} = \frac{z - 5}{5}

\frac{x + 1}{-1} = \frac{y - 2}{2} = \frac{z - 5}{5}

\frac{x + 1}{-1} = \frac{y - 2}{2} = \frac{z - 5}{4}

\frac{x - 1}{-1} = \frac{y - 2}{2} = \frac{z - 5}{5}

Step-by-step solutionView

Correct answer

The line coplanar to $\frac{x + 3}{-3} = \frac{y - 1}{1} = \frac{z - 5}{5}$ is $\frac{x + 1}{-1} = \frac{y - 2}{2} = \frac{z - 5}{5}$ because the scalar triple product $[\mathbf{a}_2 - \mathbf{a}_1, \mathbf{b}_1, \mathbf{b}_2]$ evaluates to zero.

Option analysis

Why each option works or fails

A ·

\frac{x + 1}{1} = \frac{y - 2}{2} = \frac{z - 5}{5}

Neglecting a negative sign in the direction ratios of the second line when evaluating the coplanarity determinant. Ensure direction ratios carry their proper signs; using direction ratios $\langle 1, 2, 5 \rangle$ yields a non-zero determinant of $-10$ , so the lines are skew.

B ·

\frac{x + 1}{-1} = \frac{y - 2}{2} = \frac{z - 5}{5}

This is the correct option. Evaluating the determinant with row vector $\mathbf{a}_2 - \mathbf{a}_1 = \langle 2, 1, 0 \rangle$ and direction vectors $\langle -3, 1, 5 \rangle$ and $\langle -1, 2, 5 \rangle$ gives $2(5 - 10) - 1(-15 + 5) + 0 = -10 + 10 = 0$ , confirming coplanarity.

C ·

\frac{x + 1}{-1} = \frac{y - 2}{2} = \frac{z - 5}{4}

Assuming that altering the $z$ -component of the direction vector will not disrupt the coplanarity condition. Substitute the altered direction vector $\langle -1, 2, 4 \rangle$ into the determinant to find it equals $-2$ , meaning the lines do not lie in the same plane.

D ·

\frac{x - 1}{-1} = \frac{y - 2}{2} = \frac{z - 5}{5}

Misreading the point on the line as $(1, 2, 5)$ rather than correctly tracking the coordinates in the numerator. For $\frac{x - 1}{-1} = \dots$ , the point is $(1, 2, 5)$ , making $\mathbf{a}_2 - \mathbf{a}_1 = \langle 4, 1, 0 \rangle$ ; this results in a determinant of $-20 \neq 0$ .

Reviewed route

Solution

StepWorking

01given

Given line $L_1$ : $\frac{x + 3}{-3} = rac{y - 1}{1} = rac{z - 5}{5}$ . It passes through $\vec{a}_1 = (-3, 1, 5)$ with direction vector $\vec{d}_1 = \langle -3, 1, 5 \rangle$ .

02goal

Identify which of the given lines in options (0)-(3) is coplanar with line $L_1$ .

03approach

Two lines passing through $\vec{a}_1, \vec{a}_2$ with direction vectors $\vec{d}_1, \vec{d}_2$ are coplanar if and only if $[\vec{a}_2 - \vec{a}_1 \;\; \vec{d}_1 \;\; \vec{d}_2] = 0$ , which is the determinant condition $\begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$ . We test each option sequentially.

04execute

For option (0): $\vec{a}_2 = (-1, 2, 5)$ , $\vec{d}_2 = \langle 1, 2, 5 \rangle$ . Vector $\vec{a}_2 - \vec{a}_1 = \langle -1 - (-3), 2 - 1, 5 - 5 \rangle = \langle 2, 1, 0 \rangle$ . Determinant $\Delta_0 = \begin{vmatrix} 2 & 1 & 0 \\ -3 & 1 & 5 \\ 1 & 2 & 5 \end{vmatrix} = 2(5 - 10) - 1(-15 - 5) + 0 = 2(-5) - (-20) = -10 + 20 = 10 \neq 0$ .

05execute

For option (1): $\vec{a}_2 = (-1, 2, 5)$ , $\vec{d}_2 = \langle -1, 2, 5 \rangle$ . $\vec{a}_2 - \vec{a}_1 = \langle 2, 1, 0 \rangle$ . Determinant $\Delta_1 = \begin{vmatrix} 2 & 1 & 0 \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{vmatrix} = 2(5 - 10) - 1(-15 - (-5)) + 0 = 2(-5) - (-10) = -10 + 10 = 0$ . Thus, line (1) is coplanar with $L_1$ .

✓verify

Check remaining options to be certain: For option (2), $\vec{d}_2 = \langle -1, 2, 4 \rangle$ , determinant is $2(4 - 10) - 1(-12 - (-5)) = -12 + 7 = -5 \neq 0$ . For option (3), $\vec{a}_2 = (1, 2, 5)$ , $\vec{a}_2 - \vec{a}_1 = \langle 4, 1, 0 \rangle$ , determinant is $4(-5) - (-10) = -10 \neq 0$ . Hence, option (1) is uniquely correct.

Hints that build this answer step by step

What algebraic condition must two lines $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$ satisfy to be coplanar?
$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = 0$
For the given line $\frac{x + 3}{-3} = \frac{y - 1}{1} = \frac{z - 5}{5}$ and the candidate line $\frac{x + 1}{-1} = \frac{y - 2}{2} = \frac{z - 5}{5}$ , what are the vector $\vec{a}_2 - \vec{a}_1$ and direction vectors $\vec{b}_1, \vec{b}_2$ ?
$\vec{a}_2 - \vec{a}_1 = \langle 2, 1, 0 \rangle$ , $\vec{b}_1 = \langle -3, 1, 5 \rangle$ , $\vec{b}_2 = \langle -1, 2, 5 \rangle$
What is the value of the determinant $\begin{vmatrix} 2 & 1 & 0 \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{vmatrix}$ ?
$0$

✓ 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 coplanar mean the lines must intersect?

Coplanar lines in 3D can be either parallel (same direction) or intersecting. The determinant condition $[\vec{a}_2 - \vec{a}_1 \;\; \vec{d}_1 \;\; \vec{d}_2] = 0$ covers both cases.

Answer

The line coplanar to $\frac{x + 3}{-3} = \frac{y - 1}{1} = \frac{z - 5}{5}$ is $\frac{x + 1}{-1} = \frac{y - 2}{2} = \frac{z - 5}{5}$ because the scalar triple product $[\mathbf{a}_2 - \mathbf{a}_1, \mathbf{b}_1, \mathbf{b}_2]$ evaluates to zero.

Why each option works or fails

A: $\frac{x + 1}{1} = \frac{y - 2}{2} = \frac{z - 5}{5}$ - Neglecting a negative sign in the direction ratios of the second line when evaluating the coplanarity determinant. Ensure direction ratios carry their proper signs; using direction ratios $\langle 1, 2, 5 \rangle$ yields a non-zero determinant of $-10$ , so the lines are skew.
B · correct: $\frac{x + 1}{-1} = \frac{y - 2}{2} = \frac{z - 5}{5}$ - This is the correct option. Evaluating the determinant with row vector $\mathbf{a}_2 - \mathbf{a}_1 = \langle 2, 1, 0 \rangle$ and direction vectors $\langle -3, 1, 5 \rangle$ and $\langle -1, 2, 5 \rangle$ gives $2(5 - 10) - 1(-15 + 5) + 0 = -10 + 10 = 0$ , confirming coplanarity.
C: $\frac{x + 1}{-1} = \frac{y - 2}{2} = \frac{z - 5}{4}$ - Assuming that altering the $z$ -component of the direction vector will not disrupt the coplanarity condition. Substitute the altered direction vector $\langle -1, 2, 4 \rangle$ into the determinant to find it equals $-2$ , meaning the lines do not lie in the same plane.
D: $\frac{x - 1}{-1} = \frac{y - 2}{2} = \frac{z - 5}{5}$ - Misreading the point on the line as $(1, 2, 5)$ rather than correctly tracking the coordinates in the numerator. For $\frac{x - 1}{-1} = \dots$ , the point is $(1, 2, 5)$ , making $\mathbf{a}_2 - \mathbf{a}_1 = \langle 4, 1, 0 \rangle$ ; this results in a determinant of $-20 \neq 0$ .

Step-by-step solution

given: Given line $L_1$ : $\frac{x + 3}{-3} = rac{y - 1}{1} = rac{z - 5}{5}$ . It passes through $\vec{a}_1 = (-3, 1, 5)$ with direction vector $\vec{d}_1 = \langle -3, 1, 5 \rangle$ .
goal: Identify which of the given lines in options (0)-(3) is coplanar with line $L_1$ .
approach: Two lines passing through $\vec{a}_1, \vec{a}_2$ with direction vectors $\vec{d}_1, \vec{d}_2$ are coplanar if and only if $[\vec{a}_2 - \vec{a}_1 \;\; \vec{d}_1 \;\; \vec{d}_2] = 0$ , which is the determinant condition $\begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$ . We test each option sequentially.
execute: For option (0): $\vec{a}_2 = (-1, 2, 5)$ , $\vec{d}_2 = \langle 1, 2, 5 \rangle$ . Vector $\vec{a}_2 - \vec{a}_1 = \langle -1 - (-3), 2 - 1, 5 - 5 \rangle = \langle 2, 1, 0 \rangle$ . Determinant $\Delta_0 = \begin{vmatrix} 2 & 1 & 0 \\ -3 & 1 & 5 \\ 1 & 2 & 5 \end{vmatrix} = 2(5 - 10) - 1(-15 - 5) + 0 = 2(-5) - (-20) = -10 + 20 = 10 \neq 0$ .
execute: For option (1): $\vec{a}_2 = (-1, 2, 5)$ , $\vec{d}_2 = \langle -1, 2, 5 \rangle$ . $\vec{a}_2 - \vec{a}_1 = \langle 2, 1, 0 \rangle$ . Determinant $\Delta_1 = \begin{vmatrix} 2 & 1 & 0 \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{vmatrix} = 2(5 - 10) - 1(-15 - (-5)) + 0 = 2(-5) - (-10) = -10 + 10 = 0$ . Thus, line (1) is coplanar with $L_1$ .
verify: Check remaining options to be certain: For option (2), $\vec{d}_2 = \langle -1, 2, 4 \rangle$ , determinant is $2(4 - 10) - 1(-12 - (-5)) = -12 + 7 = -5 \neq 0$ . For option (3), $\vec{a}_2 = (1, 2, 5)$ , $\vec{a}_2 - \vec{a}_1 = \langle 4, 1, 0 \rangle$ , determinant is $4(-5) - (-10) = -10 \neq 0$ . Hence, option (1) is uniquely correct.

Three Dimensional Geometry: Mathematics | JEE Main

What feels right?

The line coplanar to $\frac{x + 3}{-3} = \frac{y - 1}{1} = \frac{z - 5}{5}$ is $\frac{x + 1}{-1} = \frac{y - 2}{2} = \frac{z - 5}{5}$ because the scalar triple product $[\mathbf{a}_2 - \mathbf{a}_1, \mathbf{b}_1, \mathbf{b}_2]$ evaluates to zero.

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

✓verify

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