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Question: If the straight lines \(\dfrac{x-1}{2}=\dfrac{y+1}{K}=\dfrac{z}{2}\) and \(\dfrac{x+1}{5}=\dfrac{y+1...

If the straight lines x−12=y+1K=z2\dfrac{x-1}{2}=\dfrac{y+1}{K}=\dfrac{z}{2} and x+15=y+12=zK\dfrac{x+1}{5}=\dfrac{y+1}{2}=\dfrac{z}{K} are coplanar, then the planes containing these two lines are
A. y+2z=−1y+2z=-1
B. y+z=−1y+z=-1
C. y−z=−1y-z=-1
D. y−2z=−1y-2z=-1

Explanation

Solution

We first find the points and the direction vectors of the lines. From the cross product we find the normal vectors. We also assume the plane and put the points on the plane to find the final solutions for the plane.

Complete step by step solution:
It’s given that the straight lines x−12=y+1K=z2\dfrac{x-1}{2}=\dfrac{y+1}{K}=\dfrac{z}{2} and x+15=y+12=zK\dfrac{x+1}{5}=\dfrac{y+1}{2}=\dfrac{z}{K} are coplanar.
We can consider the first line as l1:x−12=y+1K=z2{{l}_{1}}:\dfrac{x-1}{2}=\dfrac{y+1}{K}=\dfrac{z}{2}. It goes through the point (1,−1,0)\left( 1,-1,0 \right) and the direction vector is a→=2i^+Kj^+2k^\overrightarrow{a}=2\widehat{i}+K\widehat{j}+2\widehat{k}.
Similarly, we can consider the second line as l2:x+15=y+12=zK{{l}_{2}}:\dfrac{x+1}{5}=\dfrac{y+1}{2}=\dfrac{z}{K}. It goes through the point (−1,−1,0)\left( -1,-1,0 \right) and the direction vector is b→=5i^+2j^+Kk^\overrightarrow{b}=5\widehat{i}+2\widehat{j}+K\widehat{k}.
The direction vectors and the vectors itself lie on the same plane.
Now we find the normal vector n→\overrightarrow{n} along the vectors a→=2i^+Kj^+2k^\overrightarrow{a}=2\widehat{i}+K\widehat{j}+2\widehat{k} and b→=5i^+2j^+Kk^\overrightarrow{b}=5\widehat{i}+2\widehat{j}+K\widehat{k}.
Therefore, n→=a→×b→\overrightarrow{n}=\overrightarrow{a}\times \overrightarrow{b}. Completing the vector multiplication, we get
n→=(2i^+Kj^+2k^)×(5i^+2j^+Kk^)=(K2−4)i^+(10−2K)j^+(4−5K)k^\overrightarrow{n}=\left( 2\widehat{i}+K\widehat{j}+2\widehat{k} \right)\times \left( 5\widehat{i}+2\widehat{j}+K\widehat{k} \right)=\left( {{K}^{2}}-4 \right)\widehat{i}+\left( 10-2K \right)\widehat{j}+\left( 4-5K \right)\widehat{k}.
Now we find the equation of the plane from the normal vector of these two lines.
The plane equation containing these two lines is (K2−4)x+(10−2K)y+(4−5K)z=d\left( {{K}^{2}}-4 \right)x+\left( 10-2K \right)y+\left( 4-5K \right)z=d.
We have two variables K,dK,d. The points (1,−1,0)\left( 1,-1,0 \right) and (−1,−1,0)\left( -1,-1,0 \right) lie on the same plane.
They will satisfy the plane equation (K2−4)x+(10−2K)y+(4−5K)z=d\left( {{K}^{2}}-4 \right)x+\left( 10-2K \right)y+\left( 4-5K \right)z=d.
We place the value (1,−1,0)\left( 1,-1,0 \right) to get
(K2−4)−(10−2K)=d ⇒K2+2k−14=d \begin{aligned} & \left( {{K}^{2}}-4 \right)-\left( 10-2K \right)=d \\\ & \Rightarrow {{K}^{2}}+2k-14=d \\\ \end{aligned}
And the value of (−1,−1,0)\left( -1,-1,0 \right) to get
−(K2−4)−(10−2K)=d ⇒−K2+2k−6=d \begin{aligned} & -\left( {{K}^{2}}-4 \right)-\left( 10-2K \right)=d \\\ & \Rightarrow -{{K}^{2}}+2k-6=d \\\ \end{aligned}
Equating value of dd we get K2+2k−14=−K2+2k−6{{K}^{2}}+2k-14=-{{K}^{2}}+2k-6.
Simplifying we get
K2+2k−14=−K2+2k−6 ⇒2K2=8 ⇒K=±2 \begin{aligned} & {{K}^{2}}+2k-14=-{{K}^{2}}+2k-6 \\\ & \Rightarrow 2{{K}^{2}}=8 \\\ & \Rightarrow K=\pm 2 \\\ \end{aligned}
The value of dd is d=22+2×2−14=−6d={{2}^{2}}+2\times 2-14=-6 or d=(−2)2+2×(−2)−14=−14d={{\left( -2 \right)}^{2}}+2\times \left( -2 \right)-14=-14.
The planes’ equations will be 6y−6z=−6⇒y−z=−16y-6z=-6\Rightarrow y-z=-1 or 14y+14z=−14⇒y+z=−114y+14z=-14\Rightarrow y+z=-1.
Therefore, the correct options are B and C.

Note: We need to always take the constants in the equation of planes as the normal vectors create the plane’s equation and the points would satisfy the expression. The straight lines being coplanar, the normal of the direction vectors gives the equation of the plane.