Question

If for some α∈R , the lines$\mathrm{L}_{1}: \dfrac{x+1}{2}=\dfrac{y-2}{-1}=\dfrac{z-1}{1}$ and $\mathrm{L}_{2}: \dfrac{x+2}{\alpha}=\dfrac{y+1}{5-\alpha}=\dfrac{z+1}{1}$ are coplanar, then the line $\mathrm{L}_{2}: \dfrac{x+2}{\alpha}=\dfrac{y+1}{5-\alpha}=\dfrac{z+1}{1}$ passes through the point
A. $(2, -10, -2)$
B. $(10, -2, -2)$
C. $(10, 2, 2)$
D. $(-2, 10, 2)$

Answer 1

Hint: Here we have to find the point where the line passes through. we are using the coplanar condition to find the point. Form a matrix by using two given lines. Then, determine the determinant which will be equivalent to zero as the lines are coplanar. Further solving will give us the required solution.

Formula used:
For, $M_{3\times3}=\begin{bmatrix}a & b & c\\d & e & f\\g & h & i \end{bmatrix}$
$D_{3\times3}= a\lgroup~ei-fh\rgroup-b\lgroup~di-fg\rgroup+c\lgroup~dh-eg\rgroup$

Complete Step-by-step solution:
Given that;
$\mathrm{L}_{1}: \dfrac{x+1}{2}=\dfrac{y-2}{-1}=\dfrac{z-1}{1}$ and
$\mathrm{L}_{2}: \dfrac{x+2}{\alpha}=\dfrac{y+1}{5-\alpha}=\dfrac{z+1}{1}$ are coplanar.
When two lines in a three-dimensional space are located on the same plane, they are said to be coplanar. We have learned how to use vector notations to visualize a line's equation in three dimensions.
Form a matrix with given two lines
For forming a matrix we have to take the denominator values of the lines for the 2nd and 3rd row.
For 1st row let, $a_{1}=1$, $b_{1}=-2$, $c_{1}=-1$ and $a_{2}=2$, $b_{2}=1$, $c_{2}=1$.
Now, $a_{2}-a_{1}=1$, $b_{2}-b_{1}=3$, $c_{2}-c_{1}=2$. This is the 1st row of the matrix.
$\therefore\left|\begin{array}{ccc}
1 & 3 & 2 \\
2 & -1 & 1 \\
\alpha & 5-\alpha & 1
\end{array}\right|=0$
Solving the matrix we get
$\Rightarrow-1(-1+\alpha-5)+3(2-\alpha)-2(10-2 \alpha+\alpha)=0$
$\Rightarrow 6-\alpha+6-3 \alpha+2 \alpha-20=0$
Simplify the expression
$\Rightarrow-8-2 \alpha=0$
$\Rightarrow \alpha=-4$
$\therefore$ Equation of $\mathrm{L}_{2}: \dfrac{x+2}{-4}=\dfrac{y+1}{9}=\dfrac{z+1}{1}$
Check the options whether it satisfies the condition.
The lines are coplanar if the Determinant's value is 0. They are non-coplanar if not.
Check the options by taking the determinant value as 0
So while checking option A, we can see that the condition is satisfied.

So option (A) is the correct answer

Note: Sometimes, collinear is mentioned instead of coplanar in the question. Collinear points are ones that exist along the same line. Coplanar points are all points within the same plane. In the case of collinear points, a person can select one of an infinite number of planes that contain the line across which these points exist. As a result, they can be described as coplanar.