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Mathematics Question on Three Dimensional Geometry

If the lines x+12=y−11=z+13\frac{x+1}{2} = \frac{y-1}{1} = \frac{z+1}{3} and x+22=y−k3=z4\frac{x+2}{2} = \frac{y-k}{3} =\frac{z}{4} are coplanar, then the value of k is :

A

112\frac{11}{2}

B

−112 - \frac{11}{2}

C

92\frac{9}{2}

D

−92 - \frac{9}{2}

Answer

112\frac{11}{2}

Explanation

Solution

Two given planes are coplanar, if ∣−2−(−1)k−10−(−1) 213 234∣=0\begin{vmatrix}-2-\left(-1\right)&k-1&0-\left(-1\right)\\\ 2&1&3\\\ 2&3&4\end{vmatrix}= 0 ⇒∣−1k−11 213 234∣=0 \Rightarrow \begin{vmatrix}-1&k-1&1\\\ 2&1&3\\\ 2&3&4\end{vmatrix} = 0 ⇒(−1)(4−9)−(k−1)(8−6)+6−2=0\Rightarrow \left(- 1\right) \left(4 - 9\right) - \left(k - 1\right) \left(8 - 6\right) + 6 - 2 = 0 ⇒k=112\Rightarrow k = \frac{11}{2}