If the lines x - 2 \over 1 = y - 3 \over 1 = z - 4 \over - k and x - 1 \over k = y - 4 \over 2 = z - 5 \over 1 are coplanar, then k can have :

3D Geometry PYQ — JEE Main 2013 · Shift 0 · Q47 with Solution

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Correct answer: C

Write the lines in vector form

The given lines are $\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}$ and $\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}.$

So we can identify:

A point on line $L_1$ : $A(2,3,4)$
Direction vector of $L_1$ : $\vec d_1=(1,1,-k)$
A point on line $L_2$ : $B(1,4,5)$
Direction vector of $L_2$ : $\vec d_2=(k,2,1)$

Condition for two lines to be coplanar

Two lines with points $A,B$ and direction vectors $\vec d_1,\vec d_2$ are coplanar iff $[\overrightarrow{AB},\vec d_1,\vec d_2]=0,$ where $[\cdot,\cdot,\cdot]$ denotes the scalar triple product.

Here, $\overrightarrow{AB}=B-A=(1-2,4-3,5-4)=(-1,1,1).$

So we need

-1 & 1 & 1\\ 1 & 1 & -k\\ k & 2 & 1 \end{pmatrix}=0.$$ 3. **Compute the determinant** Expanding along the first row:

\begin{aligned} \Delta&=-1\begin{vmatrix}1 & -k\ 2 & 1\end{vmatrix} -1\begin{vmatrix}1 & -k\ k & 1\end{vmatrix} +1\begin{vmatrix}1 & 1\ k & 2\end{vmatrix}\ &=-1(1+2k)-1(1+k^2)+(2-k). \end{aligned}

Thus

\Delta=-(1+2k)-(1+k^2)+(2-k).

Simplify:

\Delta=-1-2k-1-k^2+2-k=-k^2-3k.

Hence coplanarity requires $$-k^2-3k=0$$ $$k^2+3k=0$$ $$k(k+3)=0.$$ Therefore, $$k=0 \quad \text{or} \quad k=-3.$$ 4. **Number of possible values** There are **exactly two values** of $k$. So the correct option is: $$\boxed{\text{C}}$$ 5. **Comparison with stored answer** Stored correct answer: **C** Our derived answer: **C** They agree.

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View written solutionFree

Correct answer: C

Write the lines in vector form

The given lines are $\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}$ and $\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}.$

So we can identify:

A point on line $L_1$ : $A(2,3,4)$
Direction vector of $L_1$ : $\vec d_1=(1,1,-k)$
A point on line $L_2$ : $B(1,4,5)$
Direction vector of $L_2$ : $\vec d_2=(k,2,1)$

Condition for two lines to be coplanar

Two lines with points $A,B$ and direction vectors $\vec d_1,\vec d_2$ are coplanar iff $[\overrightarrow{AB},\vec d_1,\vec d_2]=0,$ where $[\cdot,\cdot,\cdot]$ denotes the scalar triple product.

Here, $\overrightarrow{AB}=B-A=(1-2,4-3,5-4)=(-1,1,1).$

So we need

-1 & 1 & 1\\ 1 & 1 & -k\\ k & 2 & 1 \end{pmatrix}=0.$$ 3. **Compute the determinant** Expanding along the first row:

\begin{aligned} \Delta&=-1\begin{vmatrix}1 & -k\ 2 & 1\end{vmatrix} -1\begin{vmatrix}1 & -k\ k & 1\end{vmatrix} +1\begin{vmatrix}1 & 1\ k & 2\end{vmatrix}\ &=-1(1+2k)-1(1+k^2)+(2-k). \end{aligned}

Thus

\Delta=-(1+2k)-(1+k^2)+(2-k).

Simplify:

\Delta=-1-2k-1-k^2+2-k=-k^2-3k.

Hence coplanarity requires $$-k^2-3k=0$$ $$k^2+3k=0$$ $$k(k+3)=0.$$ Therefore, $$k=0 \quad \text{or} \quad k=-3.$$ 4. **Number of possible values** There are **exactly two values** of $k$. So the correct option is: $$\boxed{\text{C}}$$ 5. **Comparison with stored answer** Stored correct answer: **C** Our derived answer: **C** They agree.

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