If for some α ∈ R , the lines L 1: ( x +1/2)=( y -2/-1)=( z -1/1) and L 2: ( x +2/α)=( y +1/5-α)=( z +1/1) are coplanar, then the line L2 passes through the point:

Question

If for some α ∈ R , the lines L 1: ( x +1/2)=( y -2/-1)=( z -1/1) and L 2: ( x +2/α)=( y +1/5-α)=( z +1/1) are coplanar, then the line L2 passes through the point: (A) (-2,10,2) (B) (10,2,2) (C) (10,-2,-2) (D) (2,-10,-2)

Tardigrade Admin · Accepted Answer

L 1 ≡ ( x +1/2)=( y -2/-1)=( z -1/1) L 2 ≡ ( x +2/α)=( y +1/5-α)=( z +1/1) Point A (-1,2,1) B (-2,-1,-1) ∵ L 1 and L 2 are coplanar ⇒ |2&-1&1 α&5-α&1 1&3&2| = 0 α=-4 L 2 ≡ ( x +2/-4)=( y +1/9)=( z +1/1) Check options (2,-10,-2) lies on L2

Q. If for some $α \in R,$ the lines $L_{1} : \frac{x + 1}{2} = \frac{y - 2}{- 1} = \frac{z - 1}{1}$ and $L_{2} : \frac{x + 2}{α} = \frac{y + 1}{5 - α} = \frac{z + 1}{1}$ are coplanar, then the line $L_{2}$ passes through the point:

$(- 2, 10, 2)$

$(10, 2, 2)$

$(10, - 2, - 2)$

$(2, - 10, - 2)$

Q. If for some α∈R, the lines L1​:2x+1​=−1y−2​=1z−1​ and L2​:αx+2​=5−αy+1​=1z+1​ are coplanar, then the line L2​ passes through the point:

(−2,10,2)

(10,2,2)

(10,−2,−2)

(2,−10,−2)

L1​≡2x+1​=−1y−2​=1z−1​ L2​≡αx+2​=5−αy+1​=1z+1​ Point A(−1,2,1)B(−2,−1,−1) ∵L1​ and L2​ are coplanar ⇒∣∣​2α1​−15−α3​112​∣∣​=0 α=−4 L2​≡−4x+2​=9y+1​=1z+1​ Check options (2,-10,-2) lies on L2​

Q. If for some $α \in R,$ the lines $L_{1} : \frac{x + 1}{2} = \frac{y - 2}{- 1} = \frac{z - 1}{1}$ and $L_{2} : \frac{x + 2}{α} = \frac{y + 1}{5 - α} = \frac{z + 1}{1}$ are coplanar, then the line $L_{2}$ passes through the point:

$(- 2, 10, 2)$

$(10, 2, 2)$

$(10, - 2, - 2)$

$(2, - 10, - 2)$