NVAMediumJEE 2024Skew Lines & Shortest Distance

JEE Mathematics 2024 Question with Solution

Let OO be the origin, and MM and NN be the points on the lines: x54=y41=z53\frac{x-5}{4} = \frac{y-4}{1} = \frac{z-5}{3} and x+812=y+25=z+119\frac{x+8}{12} = \frac{y+2}{5} = \frac{z+11}{9}, respectively, such that MNMN is the shortest distance between the given lines. Then OMONOM \cdot ON is equal to:

Answer

Correct answer:9

Step-by-step solution

Standard Method

Given: The points MM and NN lie on the lines

x54=y41=z53\frac{x-5}{4} = \frac{y-4}{1} = \frac{z-5}{3}

and

x+812=y+25=z+119\frac{x+8}{12} = \frac{y+2}{5} = \frac{z+11}{9}

with MNMN as the shortest distance between the lines.

Find: OMONOM \cdot ON

Parametrize the lines:

M(4λ+5,λ+4,3λ+5)M(4\lambda + 5, \lambda + 4, 3\lambda + 5) N(12μ8,5μ2,9μ11)N(12\mu - 8, 5\mu - 2, 9\mu - 11)

Then

MN=(4λ12μ+13,λ5μ+6,3λ9μ+16)\overrightarrow{MN} = \left(4\lambda - 12\mu + 13, \lambda - 5\mu + 6, 3\lambda - 9\mu + 16\right)

The direction vectors are

b1=(4,1,3),b2=(12,5,9)\overrightarrow{b_1} = (4,1,3), \qquad \overrightarrow{b_2} = (12,5,9)

Their cross product is

b1×b2=i^j^k^4131259=6i^+8k^=(6,0,8)\overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 1 & 3 \\ 12 & 5 & 9 \end{vmatrix} = -6\hat{i} + 8\hat{k} = (-6,0,8)

Since MNMN is the shortest distance between the skew lines, MN\overrightarrow{MN} is parallel to b1×b2\overrightarrow{b_1} \times \overrightarrow{b_2}. Hence,

4λ12μ+136=λ5μ+60=3λ9μ+168\frac{4\lambda - 12\mu + 13}{-6} = \frac{\lambda - 5\mu + 6}{0} = \frac{3\lambda - 9\mu + 16}{8}

So we get

λ5μ+6=0\lambda - 5\mu + 6 = 0

and

λ3μ+4=0\lambda - 3\mu + 4 = 0

Solving these equations,

λ=1,μ=1\lambda = -1, \qquad \mu = 1

Therefore,

M=(1,3,2),N=(4,3,2)M = (1,3,2), \qquad N = (4,3,-2)

So,

OM=(1,3,2),ON=(4,3,2)\overrightarrow{OM} = (1,3,2), \qquad \overrightarrow{ON} = (4,3,-2)

Now compute the dot product:

OMON=1×4+3×3+2×(2)=4+94=9OM \cdot ON = 1 \times 4 + 3 \times 3 + 2 \times (-2) = 4 + 9 - 4 = 9

Therefore, the value of OMONOM \cdot ON is 99.

Shortest Distance Condition in Vector Form

Given: Points MM and NN lie on the two given lines, and MNMN is the shortest distance between them.

Find: OMONOM \cdot ON

For the first line, let the common ratio be λ\lambda. Then

x54=y41=z53=λ\frac{x-5}{4} = \frac{y-4}{1} = \frac{z-5}{3} = \lambda

which gives

x=4λ+5,y=λ+4,z=3λ+5x = 4\lambda + 5, \qquad y = \lambda + 4, \qquad z = 3\lambda + 5

Hence,

M(4λ+5,λ+4,3λ+5)M(4\lambda + 5, \lambda + 4, 3\lambda + 5)

For the second line, let the common ratio be μ\mu. Then

x+812=y+25=z+119=μ\frac{x+8}{12} = \frac{y+2}{5} = \frac{z+11}{9} = \mu

which gives

x=12μ8,y=5μ2,z=9μ11x = 12\mu - 8, \qquad y = 5\mu - 2, \qquad z = 9\mu - 11

Hence,

N(12μ8,5μ2,9μ11)N(12\mu - 8, 5\mu - 2, 9\mu - 11)

Now,

MN=ONOM\overrightarrow{MN} = \overrightarrow{ON} - \overrightarrow{OM}

Using the coordinates listed in the extracted solution, this is written as

MN=(4λ12μ+13,λ5μ+6,3λ9μ+16)\overrightarrow{MN} = \left(4\lambda - 12\mu + 13, \lambda - 5\mu + 6, 3\lambda - 9\mu + 16\right)

The line directions are

b1=(4,1,3),b2=(12,5,9)\overrightarrow{b_1} = (4,1,3), \qquad \overrightarrow{b_2} = (12,5,9)

Then

b1×b2=(6,0,8)\overrightarrow{b_1} \times \overrightarrow{b_2} = (-6,0,8)

For shortest distance between two skew lines, the joining vector is parallel to this cross product. Therefore the corresponding component relations give

λ5μ+6=0\lambda - 5\mu + 6 = 0

and

λ3μ+4=0\lambda - 3\mu + 4 = 0

Subtracting the second equation from the first,

2μ+2=0-2\mu + 2 = 0

so

μ=1\mu = 1

Substitute into

λ3μ+4=0\lambda - 3\mu + 4 = 0

to get

λ3+4=0λ=1\lambda - 3 + 4 = 0 \Rightarrow \lambda = -1

Now substitute back:

M=(1,3,2),N=(4,3,2)M = (1,3,2), \qquad N = (4,3,-2)

Therefore,

OMON=(1)(4)+(3)(3)+(2)(2)=9OM \cdot ON = (1)(4) + (3)(3) + (2)(-2) = 9

So the required numerical value is 99.

Common mistakes

  • Taking the direction ratios incorrectly from the symmetric form. In x54=y41=z53\frac{x-5}{4} = \frac{y-4}{1} = \frac{z-5}{3}, the direction vector is (4,1,3)(4,1,3), not (5,4,5)(5,4,5). Use the denominators as direction ratios and the constants in numerators to locate a point on the line.

  • Forgetting the shortest-distance condition. The segment joining the two required points is not arbitrary; it must be parallel to b1×b2\overrightarrow{b_1} \times \overrightarrow{b_2}. Do not equate points directly without using this perpendicularity condition.

  • Making a sign error while parametrizing the second line. From x+812=y+25=z+119=μ\frac{x+8}{12} = \frac{y+2}{5} = \frac{z+11}{9} = \mu, we get x=12μ8x = 12\mu - 8, y=5μ2y = 5\mu - 2, z=9μ11z = 9\mu - 11. The constants are negative after rearrangement.

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