Question

The equation of the plane through the point (-1, 2, 0) and parallel to the lines $\dfrac{x}{3}=\dfrac{y+1}{0}=\dfrac{z-2}{-1}$ and $\dfrac{x-1}{1}=\dfrac{2y+1}{2}=\dfrac{z+1}{-1}$ is:
(a) $2x+3y+6z-4=0$
(b) $x-2y+3z+5=0$
(c) $x+2y+3z-3=0$
(d) $x+y+3z-1=0$

Answer 1

Hint: First of all find the vector perpendicular to the given plane which is denoted by $\overset{\hat{\ }}{\mathop{n}}\,$ which we will find by taking the vector cross product of the two direction vectors of the two given lines. Now, assume a point (x, y, z) on the given plane and find the vector between the point (x, y, z) and the given point on the plane (-1, 2, 0) which is $\left( x+1 \right)\overset{\hat{\ }}{\mathop{i}}\,+\left( y-2 \right)\overset{\hat{\ }}{\mathop{j}}\,+z\overset{\hat{\ }}{\mathop{k}}\,$ and then take the scalar product of this vector that we have just calculated with the $\overset{\hat{\ }}{\mathop{n}}\,$ and we know that when the two vectors are perpendicular to each other then their scalar product is 0 so equating the scalar product between $\left( x+1 \right)\overset{\hat{\ }}{\mathop{i}}\,+\left( y-2 \right)\overset{\hat{\ }}{\mathop{j}}\,+z\overset{\hat{\ }}{\mathop{k}}\,$ and $\overset{\hat{\ }}{\mathop{n}}\,$ to 0.

Complete step-by-step answer:
In the below figure, we have shown a plane on which a vector is perpendicular to the plane and we have also shown two points L (-1, 2, 0) and M (x, y, z) which lies on the plane.

In the above figure, let us assume that the vector which is pointing upwards and perpendicular to the plane be $\overset{\hat{\ }}{\mathop{n}}\,$.
It is given that the two lines are parallel to the plane. The two lines which are parallel to the plane are:
$\dfrac{x}{3}=\dfrac{y+1}{0}=\dfrac{z-2}{-1}$
$\dfrac{x-1}{1}=\dfrac{2y+1}{2}=\dfrac{z+1}{-1}$
The direction vector for the first line i.e. $\dfrac{x}{3}=\dfrac{y+1}{0}=\dfrac{z-2}{-1}$ is (3, 0, -1).
To find the direction vector for this line $\dfrac{x-1}{1}=\dfrac{2y+1}{2}=\dfrac{z+1}{-1}$ we need to make the coefficient of x, y and z as 1 which we are showing below.
$\dfrac{x-1}{1}=\dfrac{2y+1}{2}=\dfrac{z+1}{-1}$
In the above equation of a line, divide the numerator and the denominator of the expression $\dfrac{2y+1}{2}$ by 2.
$\begin{aligned} & \dfrac{x-1}{1}=\dfrac{\dfrac{2y+1}{2}}{\dfrac{2}{2}}=\dfrac{z+1}{-1} \\\ & \Rightarrow \dfrac{x-1}{1}=\dfrac{y+\dfrac{1}{2}}{1}=\dfrac{z+1}{-1} \\\ \end{aligned}$
From the above equation, the direction vector of the above line $\dfrac{x-1}{1}=\dfrac{2y+1}{2}=\dfrac{z+1}{-1}$ is (1, 1, -1).
Now, it is given that the plane is parallel to the given two lines so taking the vector cross – product of direction vectors of the two given lines we can find the vector which is perpendicular to the two lines and hence perpendicular to the plane.
We are representing the cross-product by $\overset{\hat{\ }}{\mathop{n}}\,$. Let us name the direction vectors for the two lines as:
The direction vector for the line $\dfrac{x}{3}=\dfrac{y+1}{0}=\dfrac{z-2}{-1}$ is (3, 0, -1) which we are denoted by $\overrightarrow{a}=3\overset{\hat{\ }}{\mathop{i}}\,-\overset{\hat{\ }}{\mathop{k}}\,$.
The direction vector for the line $\dfrac{x-1}{1}=\dfrac{2y+1}{2}=\dfrac{z+1}{-1}$ is (1, 1, -1) and which we are denoted by $\overrightarrow{b}=\overset{\hat{\ }}{\mathop{i}}\,+\overset{\hat{\ }}{\mathop{j}}\,-\overset{\hat{\ }}{\mathop{k}}\,$.
Taking cross-product of the vector a and b we get,
$\overrightarrow{a}\times \overrightarrow{b}=\left| \begin{aligned} & \overset{\hat{\ }}{\mathop{i}}\,\text{ }\overset{\hat{\ }}{\mathop{j}}\,\text{ }\overset{\hat{\ }}{\mathop{k}}\, \\\ & 3\text{ 0 -1} \\\ & \text{1 1 -1} \\\ \end{aligned} \right|$
Solving the above determinant we get,
$\begin{aligned} & \overset{\hat{\ }}{\mathop{i}}\,\left( 0-\left( -1 \right) \right)-\overset{\hat{\ }}{\mathop{j}}\,\left( -3+1 \right)+\overset{\hat{\ }}{\mathop{k}}\,\left( 3 \right) \\\ & =\overset{\hat{\ }}{\mathop{i}}\,+2\overset{\hat{\ }}{\mathop{j}}\,+\overset{\hat{\ }}{\mathop{3k}}\, \\\ \end{aligned}$
This cross-product is equal to $\overset{\hat{\ }}{\mathop{n}}\,$ so $\overset{\hat{\ }}{\mathop{n}}\,=\overset{\hat{\ }}{\mathop{i}}\,+2\overset{\hat{\ }}{\mathop{j}}\,+\overset{\hat{\ }}{\mathop{3k}}\,$.
Now, we are drawing a vector from the point L (-1, 2, 0) to M (x, y, z).
Let us name the vector as:
$\begin{aligned} & \overrightarrow{ML}=\left( x-\left( -1 \right) \right)\overset{\hat{\ }}{\mathop{i}}\,+\left( y-2 \right)\overset{\hat{\ }}{\mathop{j}}\,+\left( z-0 \right)\overset{\hat{\ }}{\mathop{k}}\, \\\ & \Rightarrow \overrightarrow{ML}=\left( x+1 \right)\overset{\hat{\ }}{\mathop{i}}\,+\left( y-2 \right)\overset{\hat{\ }}{\mathop{j}}\,+\left( z \right)\overset{\hat{\ }}{\mathop{k}}\, \\\ \end{aligned}$
Now, this vector lies on the plane and $\overset{\hat{\ }}{\mathop{n}}\,$is perpendicular to the vector so we can do the dot product of $\overrightarrow{ML}\And \overset{\hat{\ }}{\mathop{n}}\,$ and put it to 0 because we know that scalar product of two perpendicular vectors is equal to 0.
$\begin{aligned} & \overrightarrow{ML}\cdot \overset{\hat{\ }}{\mathop{n}}\,=0 \\\ & \Rightarrow \left( \left( x+1 \right)\overset{\hat{\ }}{\mathop{i}}\,+\left( y-2 \right)\overset{\hat{\ }}{\mathop{j}}\,+z\overset{\hat{\ }}{\mathop{k}}\, \right)\cdot \left( \overset{\hat{\ }}{\mathop{i}}\,+2\overset{\hat{\ }}{\mathop{j}}\,+3\overset{\hat{\ }}{\mathop{k}}\, \right)=0 \\\ & \Rightarrow x+1+2\left( y-2 \right)+3z=0 \\\ & \Rightarrow x+2y+3z+1-4=0 \\\ & \Rightarrow x+2y+3z-3=0 \\\ \end{aligned}$
From the above solution, we have derived the equation of the plane as $x+2y+3z-3=0$.
Hence, the correct option is (c).

Note: While writing the direction vector of the line $\dfrac{x-1}{1}=\dfrac{2y+1}{2}=\dfrac{z+1}{-1}$ people tend to forget to make the coefficient of y equal to 1 so be careful while finding the direction vector of the line. As in this problem only the coefficient of y is not equal to 1 but you might see questions in which all the coefficients of x, y and z are not equal to 1 so before finding the direction vector, make sure that all the coefficients of x, y and z are 1 and if not 1 then make it one.