IIT-JEE 2012 · Paper 2 · Mathematics Q57

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MathematicsMultiple correct+4Question 57 of 60

If the straight lines $\dfrac{x-1}2=\dfrac{y+1}k=\dfrac z2$ and $\dfrac{x+1}5=\dfrac{y+1}2=\dfrac zk$ are coplanar, then the plane(s) containing these two lines is(are)

y+2z=-1

y+z=-1

y-z=-1

y-2z=-1

Answer and worked solution

Official answer: B, C

Coplanarity. Points $A(1,-1,0)$ and $B(-1,-1,0)$ , directions $\vec d_1=(2,k,2)$ and $\vec d_2=(5,2,k)$ . With $\overrightarrow{AB}=(-2,0,0)$ :
$\left[\overrightarrow{AB}\ \vec d_1\ \vec d_2\right]=-2(k^2-4)=0\implies k=\pm2.$

Planes.

$k=2$ : $\vec d_1\times\vec d_2=(2,2,2)\times(5,2,2)=(0,6,-6)$ , plane $y-z=c$ ; through A: $y-z=-1$ .
$k=-2$ : $(2,-2,2)\times(5,2,-2)=(0,14,14)$ , plane $y+z=-1$ .

(B), (C)

Topics:Skew lines Equation of a plane

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IIT-JEE 2012 · Paper 2, Mathematics question 57