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MCQMediumJEE 2023Equation of Plane

JEE Mathematics 2023 Question with Solution

Let the line x1=6−y2=z+85\frac{x}{1} = \frac{6 - y}{2} = \frac{z + 8}{5} intersect the lines x−54=y−73=z+21\frac{x - 5}{4} = \frac{y - 7}{3} = \frac{z + 2}{1} and x+36=3−y3=z−61\frac{x + 3}{6} = \frac{3 - y}{3} = \frac{z - 6}{1} at the points A and B respectively. Then the distance of the mid-point of the line segment AB from the plane 2x−2y+z=142x - 2y + z = 14 is:

  • A

    33

  • B

    103\frac{10}{3}

  • C

    44

  • D

    113\frac{11}{3}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The lines are

x1=y−6−2=z+85=λ\frac{x}{1} = \frac{y - 6}{-2} = \frac{z + 8}{5} = \lambda x−54=y−73=z+21=μ\frac{x - 5}{4} = \frac{y - 7}{3} = \frac{z + 2}{1} = \mu x+36=3−y3=z−61=γ\frac{x + 3}{6} = \frac{3 - y}{3} = \frac{z - 6}{1} = \gamma

And the plane is 2x−2y+z=142x - 2y + z = 14.

Find: The perpendicular distance of the midpoint of ABAB from the plane.

For intersection of the first and second lines, solving the system gives

λ=−1,μ=−1\lambda = -1, \mu = -1

so

A(1,4,−3)A(1,4,-3)

For intersection of the first and third lines, solving the system gives

λ=3,γ=1\lambda = 3, \gamma = 1

so

B(0,7,7)B(0,7,7)

The midpoint of A(1,4,−3)A(1,4,-3) and B(0,7,7)B(0,7,7) is

(1+02,4+72,−3+72)=(0.5,5.5,2)\left( \frac{1+0}{2}, \frac{4+7}{2}, \frac{-3+7}{2} \right) = \left(0.5, 5.5, 2\right)

Now use the perpendicular distance formula from the plane:

∣2(0.5)−2(5.5)+2−14∣22+(−2)2+12\frac{|2(0.5) - 2(5.5) + 2 - 14|}{\sqrt{2^2 + (-2)^2 + 1^2}} =∣1−11+2−14∣4+4+1=∣−22∣3=4= \frac{|1 - 11 + 2 - 14|}{\sqrt{4 + 4 + 1}} = \frac{|-22|}{3} = 4

Therefore, the distance of the midpoint from the plane is 44. Hence, the correct option is C.

Common mistakes

  • Using the line 6−y2\frac{6-y}{2} incorrectly as y−62\frac{y-6}{2}. This changes the direction ratio sign. Rewrite it correctly as y−6−2\frac{y-6}{-2} before solving.

  • Finding the midpoint incorrectly by not averaging each coordinate separately. The midpoint of A(x1,y1,z1)A(x_1,y_1,z_1) and B(x2,y2,z2)B(x_2,y_2,z_2) must be (x1+x22,y1+y22,z1+z22)\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\right).

  • Applying the distance formula from a plane without converting it to the form ax+by+cz+d=0ax+by+cz+d=0. For 2x−2y+z=142x-2y+z=14, use 2x−2y+z−14=02x-2y+z-14=0 in the numerator.

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