Two lines $L_{1}: x=5, \frac{y}{3-\alpha}=\frac{z}{-2}$ and $L_{2}: x=\alpha, \frac{y}{-1}=\frac{z}{2-\alpha}$ are coplanar. Then $\alpha$ can take value(s)

JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · Three Dimensional Geometry

Two lines

L_{1} : x = 5, \frac{y}{3 - α} = \frac{z}{- 2}

and

L_{2} : x = α, \frac{y}{- 1} = \frac{z}{2 - α}

are coplanar. Then

α

can take value(s)

(A)1✓ Correct
(B)2
(C)3
(D)4✓ Correct

Answer: (A), (D)

Solution

Line

L_{1}

passes through

(5, 0, 0)

with direction ratios

(0, 3 - α, - 2)

.

Line

L_{2}

passes through

(α, 0, 0)

with direction ratios

(0, - 1, 2 - α)

.

For coplanarity, the scalar triple product of the vector joining the two points and the two direction vectors must be zero:

5 - α 00 0 3 - α - 1 0 - 2 2 - α = 0

(5 - α) [(3 - α) (2 - α) - 2] = 0

(5 - α) [6 - 3 α - 2 α + α^{2} - 2] = 0

(5 - α) (α^{2} - 5 α + 4) = 0

(5 - α) (α - 1) (α - 4) = 0

α = 1, 4, 5

. From the given options,

α = 1

4

JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · One or more correct options

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