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Question: The equation of the plane containing the two lines \[\dfrac{x-1}{2}=\dfrac{y+1}{-1}=\dfrac{z}{3}\] a...

The equation of the plane containing the two lines x12=y+11=z3\dfrac{x-1}{2}=\dfrac{y+1}{-1}=\dfrac{z}{3} and x2=y21=z+13\dfrac{x}{2}=\dfrac{y-2}{-1}=\dfrac{z+1}{3} is
A. 8x + y - 5z – 7 = 0
B. 8x + y + 5z – 7 = 0
C. 8x – y – 5z – 7 = 0
D. None of these

Explanation

Solution

Equation in this form xal=ybm=zcn\dfrac{x-a}{l}=\dfrac{y-b}{m}=\dfrac{z-c}{n} is a line passing through point (a, b, c) with direction cosines (l, m, n). Then the equation of the plane passing through a line will be of the form a’(x-a) + b’(y-b) + c’(z-c) = 0 where a’, b’ and c’ are the direction cosines of the normal line of the plane.

Complete step-by-step solution:
From the question, the given two lines are x12=y+11=z3\dfrac{x-1}{2}=\dfrac{y+1}{-1}=\dfrac{z}{3} and x2=y21=z+13\dfrac{x}{2}=\dfrac{y-2}{-1}=\dfrac{z+1}{3}.
Here, we can see that the given two lines have same direction cosines which implies that they are parallel to each other.
Now, let’s find the equation of the plane.
Equation of any plane containing the first line is given by
p (x – 1) + q (y + 1) + r(z) = 0 -------------(1)
where 2p – q + 3r = 0 -----------------(2)
We know that the plane passes through the second line if (0, 2, -1), which is the point on the second line, lies on it.
So, on substituting the point (0, 2, -1) in the equation (1), we get
-p + 3q – r = 0 -------------(3)
Then on solving equations (2) and (3), we get
p8=q1=r5\dfrac{p}{-8}=\dfrac{q}{-1}=\dfrac{r}{5} or p8=q1=r5\dfrac{p}{8}=\dfrac{q}{1}=\dfrac{r}{-5}
Hence, the required equation of plane is
8 (x – 1) + (y + 1) – 5z = 0
\Rightarrow 8x + y – 5z – 7 = 0.

Note: While solving such types of questions, it is necessary to check whether they are parallel to each other or perpendicular to each other. If the two lines are perpendicular then we cannot form a plane from these two lines. If the two lines are non-parallel coplanar like xx1b1=yy1b2=zz1b3\dfrac{x-x_1}{b_1}=\dfrac{y-y_1}{b_2}=\dfrac{z-z_1}{b_3} and xx2d1=yy2d2=zz2d3\dfrac{x-x_2}{d_1}=\dfrac{y-y_2}{d_2}=\dfrac{z-z_2}{d_3}, then the equation of plane is given by

x-x_1 & y-y_1 & z-z_1 \\\ b_1 & b_2 & b_3 \\\ d_1 & d_2 & d_3 \\\ \end{matrix} \right|=0$$ or $$\left| \begin{matrix} x-x_2 & y-y_2 & z-z_2 \\\ b_1 & b_2 & b_3 \\\ d_1 & d_2 & d_3 \\\ \end{matrix} \right|=0$$ .