If the straight lines \,x - 1 \over 2 = y + 1 \over k = z \over 2 and x + 1 \over 5 = y + 1 \over 2 = z \over k are coplanar, then the plane (s) containing these two lines is (are)

3D Geometry PYQ — JEE Advanced 2012 · Shift 2 · Q22 with Solution

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Correct answer: B, C

Write the two lines in parametric form

The given lines are

$\frac{x-1}{2}=\frac{y+1}{k}=\frac{z}{2}$ and $\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}.$

So their parametric forms are:

For line $L_1$ : $x=1+2\lambda,\quad y=-1+k\lambda,\quad z=2\lambda.$ Hence a point on $L_1$ is $P_1=(1,-1,0)$ and direction vector is $\vec d_1=(2,k,2).$
For line $L_2$ : $x=-1+5\mu,\quad y=-1+2\mu,\quad z=k\mu.$ Hence a point on $L_2$ is $P_2=(-1,-1,0)$ and direction vector is $\vec d_2=(5,2,k).$

Condition for two lines to be coplanar

Two lines are coplanar iff $[\overrightarrow{P_1P_2},\vec d_1,\vec d_2]=0,$ that is, the scalar triple product is zero.

Now, $\overrightarrow{P_1P_2}=P_2-P_1=(-2,0,0).$

So we compute

-2 & 0 & 0\\ 2 & k & 2\\ 5 & 2 & k \end{pmatrix}=0.$$ Expanding along the first row: $$-2\det\begin{pmatrix}k & 2\\ 2 & k\end{pmatrix}=0.$$ Thus, $$-2(k^2-4)=0$$ $$k^2-4=0$$ $$k=\pm 2.$$ So there are **two possible planes**, corresponding to $k=2$ and $k=-2$. --- 3. **Case 1: $k=2$** Then the lines become: $$L_1: \frac{x-1}{2}=\frac{y+1}{2}=\frac{z}{2}$$ So direction vector is $$\vec d_1=(2,2,2)\propto (1,1,1).$$ $$L_2: \frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{2}$$ So direction vector is $$\vec d_2=(5,2,2).$$ A vector joining points on the two lines is $$\vec{P_1P_2}=(-2,0,0).$$ A normal to the plane is $$\vec n=\vec d_1\times \vec d_2 =\begin{vmatrix} \hat i & \hat j & \hat k\\ 2 & 2 & 2\\ 5 & 2 & 2 \end{vmatrix}.$$ Compute: $$\vec n=\hat i(4-4)-\hat j(4-10)+\hat k(4-10) =(0,6,-6)\propto (0,1,-1).$$ So plane equation is of form $$y-z+c=0.$$ Since point $(1,-1,0)$ lies on the plane, $$-1-0+c=0\Rightarrow c=1.$$ Hence, $$y-z+1=0$$ $$\boxed{y-z=-1.}$$ This is **Option C**. --- 4. **Case 2: $k=-2$** Then the lines become: $$L_1: \frac{x-1}{2}=\frac{y+1}{-2}=\frac{z}{2}$$ So direction vector is $$\vec d_1=(2,-2,2).$$ $$L_2: \frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{-2}$$ So direction vector is $$\vec d_2=(5,2,-2).$$ Now, $$\vec n=\vec d_1\times \vec d_2 =\begin{vmatrix} \hat i & \hat j & \hat k\\ 2 & -2 & 2\\ 5 & 2 & -2 \end{vmatrix}.$$ Compute: $$\vec n= \hat i(4-4)-\hat j(-4-10)+\hat k(4+10) =(0,14,14)\propto (0,1,1).$$ So plane equation is of form $$y+z+c=0.$$ Using point $(1,-1,0)$: $$-1+0+c=0\Rightarrow c=1.$$ Hence, $$y+z+1=0$$ $$\boxed{y+z=-1.}$$ This is **Option B**. --- 5. **Check all options** - **A:** $y+2z=-1$ → not obtained. - **B:** $y+z=-1$ → correct for $k=-2$. - **C:** $y-z=-1$ → correct for $k=2$. - **D:** $y-2z=-1$ → not obtained. --- 6. **Final answer** The possible planes are $$\boxed{y+z=-1 \text{ and } y-z=-1}.$$ So the correct options are: $$\boxed{B,\ C}.$$

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View written solutionFree

Correct answer: B, C

Write the two lines in parametric form

The given lines are

$\frac{x-1}{2}=\frac{y+1}{k}=\frac{z}{2}$ and $\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}.$

So their parametric forms are:

For line $L_1$ : $x=1+2\lambda,\quad y=-1+k\lambda,\quad z=2\lambda.$ Hence a point on $L_1$ is $P_1=(1,-1,0)$ and direction vector is $\vec d_1=(2,k,2).$
For line $L_2$ : $x=-1+5\mu,\quad y=-1+2\mu,\quad z=k\mu.$ Hence a point on $L_2$ is $P_2=(-1,-1,0)$ and direction vector is $\vec d_2=(5,2,k).$

Condition for two lines to be coplanar

Two lines are coplanar iff $[\overrightarrow{P_1P_2},\vec d_1,\vec d_2]=0,$ that is, the scalar triple product is zero.

Now, $\overrightarrow{P_1P_2}=P_2-P_1=(-2,0,0).$

So we compute

-2 & 0 & 0\\ 2 & k & 2\\ 5 & 2 & k \end{pmatrix}=0.$$ Expanding along the first row: $$-2\det\begin{pmatrix}k & 2\\ 2 & k\end{pmatrix}=0.$$ Thus, $$-2(k^2-4)=0$$ $$k^2-4=0$$ $$k=\pm 2.$$ So there are **two possible planes**, corresponding to $k=2$ and $k=-2$. --- 3. **Case 1: $k=2$** Then the lines become: $$L_1: \frac{x-1}{2}=\frac{y+1}{2}=\frac{z}{2}$$ So direction vector is $$\vec d_1=(2,2,2)\propto (1,1,1).$$ $$L_2: \frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{2}$$ So direction vector is $$\vec d_2=(5,2,2).$$ A vector joining points on the two lines is $$\vec{P_1P_2}=(-2,0,0).$$ A normal to the plane is $$\vec n=\vec d_1\times \vec d_2 =\begin{vmatrix} \hat i & \hat j & \hat k\\ 2 & 2 & 2\\ 5 & 2 & 2 \end{vmatrix}.$$ Compute: $$\vec n=\hat i(4-4)-\hat j(4-10)+\hat k(4-10) =(0,6,-6)\propto (0,1,-1).$$ So plane equation is of form $$y-z+c=0.$$ Since point $(1,-1,0)$ lies on the plane, $$-1-0+c=0\Rightarrow c=1.$$ Hence, $$y-z+1=0$$ $$\boxed{y-z=-1.}$$ This is **Option C**. --- 4. **Case 2: $k=-2$** Then the lines become: $$L_1: \frac{x-1}{2}=\frac{y+1}{-2}=\frac{z}{2}$$ So direction vector is $$\vec d_1=(2,-2,2).$$ $$L_2: \frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{-2}$$ So direction vector is $$\vec d_2=(5,2,-2).$$ Now, $$\vec n=\vec d_1\times \vec d_2 =\begin{vmatrix} \hat i & \hat j & \hat k\\ 2 & -2 & 2\\ 5 & 2 & -2 \end{vmatrix}.$$ Compute: $$\vec n= \hat i(4-4)-\hat j(-4-10)+\hat k(4+10) =(0,14,14)\propto (0,1,1).$$ So plane equation is of form $$y+z+c=0.$$ Using point $(1,-1,0)$: $$-1+0+c=0\Rightarrow c=1.$$ Hence, $$y+z+1=0$$ $$\boxed{y+z=-1.}$$ This is **Option B**. --- 5. **Check all options** - **A:** $y+2z=-1$ → not obtained. - **B:** $y+z=-1$ → correct for $k=-2$. - **C:** $y-z=-1$ → correct for $k=2$. - **D:** $y-2z=-1$ → not obtained. --- 6. **Final answer** The possible planes are $$\boxed{y+z=-1 \text{ and } y-z=-1}.$$ So the correct options are: $$\boxed{B,\ C}.$$

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