Question

The plane passing through the point (4, – 1, 2) and parallel to the lines \[\dfrac{x+2}{3}=\dfrac{y-2}{-1}=\dfrac{z+1}{2}\] and \[\dfrac{x-2}{1}=\dfrac{y-3}{2}=\dfrac{z-4}{3}\] also passes through the point.
(a) (– 1, – 1, – 1)
(b) (– 1, – 1, 1)
(c) (1, 1, – 1)
(d) (1, 1, 1)

Answer 1

Hint: First of all, find the direction ratios of line 1 and 2 and take their cross product to get the vector normal to the plane. Then write the equation of the plane as \[P:\left( \overrightarrow{r}-\overrightarrow{a} \right).\overrightarrow{n}=0\] where \[\overrightarrow{a}\] is the point through which it passes and \[\overrightarrow{n}\] is the normal vector to it. Now substitute each point from options in the equation of the plane and check which satisfies the equation of the plane.

Complete step-by-step answer:
Here, we are given that a plane passes through the point (4, – 1, 2). This plane is also parallel to the lines \[\dfrac{x+2}{3}=\dfrac{y-2}{-1}=\dfrac{z+1}{2}\] and \[\dfrac{x-2}{1}=\dfrac{y-3}{2}=\dfrac{z-4}{3}\]. We have to select the point from the options from which this plane will pass. Let us consider the first line given in the question:
\[{{L }_{1}}:\dfrac{x+2}{3}=\dfrac{y-2}{-1}=\dfrac{z+1}{2}\]
We know that the direction ratios of the line \[\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}\] is \[a\widehat{i}+b\widehat{j}+c\widehat{k}\]. So by using this, we get the direction ratios of the line \[{{L}_{1}}\] as:
\[\overrightarrow{{{D}_{1}}}=3\widehat{i}-\widehat{j}+2\widehat{k}....\left( i \right)\]
Similarly, we get the direction ratios of the line \[{{L}_{2}}\]: \[\dfrac{x-2}{1}=\dfrac{y-3}{2}=\dfrac{z-4}{3}\] as
\[\overrightarrow{{{D}_{2}}}=\widehat{i}+2\widehat{j}+3\widehat{k}....\left( ii \right)\]
We know that the equation of any plane P is given by:
\[P:\left( \overrightarrow{r}-\overrightarrow{a} \right).\overrightarrow{n}=0....\left( iii \right)\]
where \[\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}\] and \[\overrightarrow{a}={{x}_{1}}\widehat{i}+{{y}_{1}}\widehat{j}+{{z}_{1}}\widehat{k}\] where \[\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)\] is the point through which the plane pass. \[\overrightarrow{n}\] is the normal vector to the plane.
We know that the plane is parallel to the lines \[{{L}_{1}}\] and \[{{L}_{2}}\] or plane is parallel to the vectors \[\overrightarrow{{{D}_{1}}}\] and \[\overrightarrow{{{D}_{2}}}\]. Also, we know that the cross product of two vectors gives a vector normal to them, so we get the normal vector to the plane as:
\[\overrightarrow{n}=\overrightarrow{{{D}_{1}}}\times \overrightarrow{{{D}_{2}}}\]
By substituting the value of \[\overrightarrow{{{D}_{1}}}\] and \[\overrightarrow{{{D}_{2}}}\] from equation (i) and (ii) respectively, we get,
\[\overrightarrow{n}=\left( 3\widehat{i}-\widehat{j}+2\widehat{k} \right)\times \left( \widehat{i}+2\widehat{j}+3\widehat{k} \right)\]
We know that if, \[\overrightarrow{n}=\left( a\widehat{i}+b\widehat{j}+c\widehat{k} \right)\times \left( x\widehat{i}+y\widehat{j}+z\widehat{k} \right)\]
Then,
\[\overrightarrow{n}=\left| \begin{matrix}
   \widehat{i} & \widehat{j} & \widehat{k} \\
   a & b & c \\
   x & y & z \\
\end{matrix} \right|\]
\[\overrightarrow{n}=\widehat{i}\left( bz-yc \right)-\widehat{j}\left( az-xc \right)+\widehat{k}\left( ay-bx \right)\]
So, we get, \[\overrightarrow{n}=\left( 3\widehat{i}-\widehat{j}+2\widehat{k} \right)\times \left( \widehat{i}+2\widehat{j}+3\widehat{k} \right)\]
\[\overrightarrow{n}=\left| \begin{matrix}
   \widehat{i} & \widehat{j} & \widehat{k} \\
   3 & -1 & 2 \\
   1 & 2 & 3 \\
\end{matrix} \right|\]
\[\overrightarrow{n}=\widehat{i}\left[ \left( -1 \right)\left( 3 \right)-\left( 2\times 2 \right) \right]-\widehat{j}\left[ \left( 3\times 3 \right)-\left( 2\times 1 \right) \right]+\widehat{k}\left[ \left( 3\times 2 \right)-\left( -1\times 1 \right) \right]\]
\[\overrightarrow{n}=\widehat{i}\left( -3-4 \right)-\widehat{j}\left( 9-2 \right)+\widehat{k}\left( 6+1 \right)\]
\[\overrightarrow{n}=-7\widehat{i}-7\widehat{j}+7\widehat{k}\]
Also, we are given that the plane passes through (4, – 1, 2). So, we get,
\[\overrightarrow{a}=4\widehat{i}-\widehat{j}+2\widehat{k}\]
By substituting the value of \[\overrightarrow{r},\overrightarrow{a}\] and \[\overrightarrow{n}\] in equation (iii), we get,
\[P:\left[ \left( x\widehat{i}+y\widehat{j}+z\widehat{k} \right)-\left( 4\widehat{i}-\widehat{j}+2\widehat{k} \right) \right].\left( -7\widehat{i}-7\widehat{j}+7\widehat{k} \right)=0\]
\[\Rightarrow P:\left[ \left( x-4 \right)\widehat{i}+\left( y+1 \right)\widehat{j}+\left( z-2 \right)\widehat{k} \right].\left( -7\widehat{i}-7\widehat{j}+7\widehat{k} \right)=0\]
We know that \[\left( a\widehat{i}+b\widehat{j}+c\widehat{k} \right).\left( x\widehat{i}+y\widehat{j}+z\widehat{k} \right)\]\[=ax+bj+cz\]
So, by using this is in the above plane, we get,
\[P:-7\left( x-4 \right)-7\left( y+1 \right)+7\left( z-2 \right)=0\]
By dividing – 7 on both the sides of the above equation, we get,
\[P:\left( x-4 \right)+\left( y+1 \right)-\left( z-2 \right)=0\]
\[\Rightarrow P:x+y-z-1=0\]
By substituting option (a) (– 1, – 1, – 1) in the plane P, we get,
\[\begin{align}
  & \left( -1 \right)+\left( -1 \right)-\left( -1 \right)-1=0 \\
 & -1-1+1-1=0 \\
 & -2\ne 0 \\
\end{align}\]
Hence LHS \[\ne \] RHS. So this is not correct.
By substituting option (b) (– 1, – 1, 1) in the plane P, we get,
\[\begin{align}
  & \left( -1 \right)+\left( -1 \right)-\left( 1 \right)-1=0 \\
 & -1-1-1-1=0 \\
 & -4\ne 0 \\
\end{align}\]
Hence LHS \[\ne \] RHS. So this is not correct.
By substituting option (c) ( 1, 1, – 1) in the plane P, we get,
\[\begin{align}
  & \left( 1 \right)+\left( 1 \right)-\left( -1 \right)-1=0 \\
 & 1+1+1-1=0 \\
 & 2\ne 0 \\
\end{align}\]
Hence LHS \[\ne \] RHS. So this is not correct.
By substituting option (d) (1, 1, 1) in the plane P, we get,
\[\begin{align}
  & \left( 1 \right)+\left( 1 \right)-\left( 1 \right)-1=0 \\
 & 1+1-1-1=0 \\
 & 2-2=0 \\
 & 0=0 \\
\end{align}\]
Hence LHS = RHS. So this is correct.
Hence, Plane also passes through the point (1, 1, 1).
Therefore, option (d) is the right answer.

Note: In this question, some students get confused with dot product and cross product. So, they must remember that whenever we need to find a vector perpendicular to two vectors, then the cross product of these two vectors will give the vector perpendicular to both of them. Also, the dot product of two vectors that are perpendicular to each other is 0. If we have two vectors \[\overrightarrow{A}\] and \[\overrightarrow{B}\], then,
The dot product of \[\overrightarrow{A}\] and \[\overrightarrow{B}\], \[\overrightarrow{A}.\overrightarrow{B}=\left( A \right)\left( B \right)\cos \theta \]
Cross product of \[\overrightarrow{A}\] and \[\overrightarrow{B}\]\[=\left| \overrightarrow{A} \right|\left| \overrightarrow{B} \right|\sin \theta \widehat{n}\] where \[\theta \] is the angle between \[\overrightarrow{A}\] and \[\overrightarrow{B}\] and \[\widehat{n}\] is the vector perpendicular to both of them.