Two lines L_1 : x = 5 , y/(3 - α) = z/(- 2) and L_2 : x = α , y/(- 1) = z/(2 - α) are coplanar. Then α can take value(s)

Correct answer: (A) 1; (D) 4. (x - 5)/0 = (- y)/(α - 3) = z/(- 2) (x - α)/0 = y/(- 1) = z/(2 - α) 5 - α 0 0 0 3 - α - 2 0 - 1 2 - α = 0 5 - α 3 - α 2 - α - 2 = 0 α - 5 α^2 - 5 α + 6 - 2 = 0 α - 5 α^2 - 5 α + 4 = 0 α = 1 , 4 , 5

Q27 / 60 · JEE Advanced 2013 (Paper 2) · Mathematics

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)

Correct answer
A
1
B
2
C
3
Correct answer
D
4

Solution

\frac{x - 5}{0} = \frac{- y}{α - 3} = \frac{z}{- 2}

\frac{x - α}{0} = \frac{y}{- 1} = \frac{z}{2 - α}

|\begin{matrix} 5 - α & 0 & 0 \\ 0 & 3 - α & - 2 \\ 0 & - 1 & 2 - α \end{matrix}| = 0

(5 - α) ((3 - α) (2 - α) - 2) = 0

(α - 5)

(α^{2} - 5 α + 6 - 2) = 0

(α - 5) (α^{2} - 5 α + 4) = 0

α = 1, 4, 5

Hence, the correct answer is options: 'A', 'D'.

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Two lines L 1 : x = 5 , y 3 - α = z - 2 and L 2 : x = α , y - 1 = z 2 - α are coplanar. Then α can take value(s)

Solution

Relevant Questions

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)