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JEE Main 202313 AprilEvening Shift
Exam
JEE Main
Year
2023
Date
13 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2023 · 13 April · Evening Shift · Mathematics · Three Dimensional Geometry

The line, that is coplanar to the line (x + 3)/(- 3) = (y - 1)/(1)...

The line, that is coplanar to the line x+3−3=y−11=z−55\dfrac{x + 3}{- 3} = \dfrac{y - 1}{1} = \dfrac{z - 5}{5}, is

Options

  1. Ax+1−1=y−22=z−54\dfrac{x + 1}{- 1} = \dfrac{y - 2}{2} = \dfrac{z - 5}{4}
  2. Bx+1−1=y−22=z−55\dfrac{x + 1}{- 1} = \dfrac{y - 2}{2} = \dfrac{z - 5}{5}
  3. Cx−1−1=y−22=z−55\dfrac{x - 1}{- 1} = \dfrac{y - 2}{2} = \dfrac{z - 5}{5}
  4. Dx+11=y−22=z−55\dfrac{x + 1}{1} = \dfrac{y - 2}{2} = \dfrac{z - 5}{5}

Correct answer

B. x+1−1=y−22=z−55\dfrac{x + 1}{- 1} = \dfrac{y - 2}{2} = \dfrac{z - 5}{5}

Step-by-step solution

We know that two lines x−x1a1=y−y1b1=z−z1c1\dfrac{x - {x}_{1}}{{a}_{1}} = \dfrac{y - {y}_{1}}{{b}_{1}} = \dfrac{z - {z}_{1}}{{c}_{1}} and x−x2a2=y−y2b2=z−z2c2\dfrac{x - {x}_{2}}{{a}_{2}} = \dfrac{y - {y}_{2}}{{b}_{2}} = \dfrac{z - {z}_{2}}{{c}_{2}} are coplanar if \left|\begin{cases}{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{cases}\right| = 0

Let us verify this option.

The given lines are x+3−3=y−11=z−55\dfrac{x + 3}{- 3} = \dfrac{y - 1}{1} = \dfrac{z - 5}{5} and x+1−1=y−22=z−55\dfrac{x + 1}{- 1} = \dfrac{y - 2}{2} = \dfrac{z - 5}{5}.

= \left|\begin{cases}- 1 - \left(- 3\right) & 2 - 1 & 5 - 5 \\ - 3 & 1 & 5 \\ - 1 & 2 & 5\end{cases}\right|

= \left|\begin{cases}2 & 1 & 0 \\ - 3 & 1 & 5 \\ - 1 & 2 & 5\end{cases}\right|

=2(5−10)−(1)(−15−(−5))+0= 2\left(5 - 10\right) - \left(1\right)\left(- 15 - \left(- 5\right)\right) + 0

= 2\left(- 5\right) + \left(10\right) = 0

Therefore, this is the correct option.

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