Question

The plane $x-2y+7z+21=0$?
(a) Contains the line $\dfrac{x+1}{-3}=\dfrac{y-3}{2}=\dfrac{z+2}{1}$
(b) Doesn’t contains the point (0, 7, -1)
(c) Is parallel to the line $\dfrac{x}{1}=\dfrac{y}{-2}=\dfrac{z}{7}$
(d) Is perpendicular to the plane $x-2y+7z=0$

Answer 1

Hint: Assume the given plane as $ax+by+cz+d=0$ and find its direction cosines given as $l=\dfrac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$, $m=\dfrac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$ and $n=\dfrac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$. Now, check each option one by one. For option (a) substitute $\dfrac{x+1}{-3}=\dfrac{y-3}{2}=\dfrac{z+2}{1}=k$ and find the values of x, y and z in terms of k. Substitute these values in the plane and check if it satisfies the equation of the plane. For option (b) do the same process as done in option (a) for point (0, 7, -1). For option (c) consider the direction cosines of the line $\dfrac{x}{1}=\dfrac{y}{-2}=\dfrac{z}{7}$ as $\left( {{l}_{1}},{{m}_{1}},{{n}_{1}} \right)=\left( 1,-2,7 \right)$and check if the relation $\dfrac{l}{{{l}_{1}}}=\dfrac{m}{{{m}_{1}}}=\dfrac{n}{{{n}_{1}}}$ gets satisfied. For option (d) find the direction ratios of the given plane $x-2y+7z=0$ using the formula we used for the plane $x-2y+7z+21=0$ and consider them as \[\left( {{l}_{2}},{{m}_{2}},{{n}_{2}} \right)\], check if the relation $l\times {{l}_{2}}+m\times {{m}_{2}}+n\times {{n}_{2}}=0$ gets satisfied.

Complete step by step solution:
Here we have been provided with the plane $x-2y+7z+21=0$ and we are asked to choose the correct option(s) regarding this plane. Let us check them one by one.
(a) Here we have to check if the line $\dfrac{x+1}{-3}=\dfrac{y-3}{2}=\dfrac{z+2}{1}$ lies on the plane or not.
Now, a line will lie on a plane only when all the points lying on that line also lie on the plane. Now, equating $\dfrac{x+1}{-3}=\dfrac{y-3}{2}=\dfrac{z+2}{1}=k$ and finding the values of x, y and z in terms of k we get $x=-3k-1,y=2k+3,z=k-2$. That means all the points on the line can be written in the coordinate form as $\left( -3k-1,2k+3,k-2 \right)$. Substituting this coordinate in the equation of the plane we get,
$\begin{align}
  & \Rightarrow x-2y+7z+21=-3k-1-2\left( 2k+3 \right)+7\left( k-2 \right)+21 \\
 & \Rightarrow x-2y+7z+21=-3k-1-4k-6+7k-14+21 \\
 & \Rightarrow x-2y+7z+21=0 \\
\end{align}$
Therefore, the plane $x-2y+7z+21=0$ contains the line $\dfrac{x+1}{-3}=\dfrac{y-3}{2}=\dfrac{z+2}{1}$ hence option (a) is correct.
(b) Here we need to check if the point (0, 7, -1) lies on the plane or not.
This point will lie on the plane only when it satisfies the equation of the plane. So substituting this point in the plane we get,
$\begin{align}
  & \Rightarrow x-2y+7z+21=0-2\left( 7 \right)+7\left( -1 \right)+21 \\
 & \Rightarrow x-2y+7z+21=-14-7+21 \\
 & \Rightarrow x-2y+7z+21=0 \\
\end{align}$
Therefore the point (0, 7, -1) lies on the plane hence option (b) is correct.
(c) Here we need to check if the plane is parallel to the line $\dfrac{x}{1}=\dfrac{y}{-2}=\dfrac{z}{7}$.
Now, the line and the plane will only be parallel if the ratios of their direction cosines are equal. The direction cosines of a plane of the form $ax+by+cz+d=0$ is given as $l=\dfrac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$, $m=\dfrac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$ and $n=\dfrac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$. Similarly, for a line of the form $\dfrac{x-{{x}_{1}}}{{{a}_{1}}}=\dfrac{y-{{y}_{1}}}{{{b}_{1}}}=\dfrac{z-{{z}_{1}}}{{{c}_{1}}}$ is given as $\left( {{l}_{1}},{{m}_{1}},{{n}_{1}} \right)=\left( {{a}_{1}},{{b}_{1}},{{c}_{1}} \right)$. So we have,
$\begin{align}
  & l=\dfrac{1}{\sqrt{{{1}^{2}}+{{\left( -2 \right)}^{2}}+{{7}^{2}}}}=\dfrac{1}{\sqrt{54}} \\
 & m=\dfrac{-2}{\sqrt{{{1}^{2}}+{{\left( -2 \right)}^{2}}+{{7}^{2}}}}=\dfrac{-2}{\sqrt{54}} \\
 & n=\dfrac{7}{\sqrt{{{1}^{2}}+{{\left( -2 \right)}^{2}}+{{7}^{2}}}}=\dfrac{7}{\sqrt{54}} \\
\end{align}$
\[\Rightarrow \left( l,m,n \right)=\left( \dfrac{1}{\sqrt{54}},\dfrac{-2}{\sqrt{54}},\dfrac{7}{\sqrt{54}} \right)\] and $\left( {{l}_{1}},{{m}_{1}},{{n}_{1}} \right)=\left( 1,-2,7 \right)$. So considering the ratios of the direction cosines we get,
$\begin{align}
  & \dfrac{l}{{{l}_{1}}}=\dfrac{\dfrac{1}{\sqrt{54}}}{1}=\dfrac{1}{\sqrt{54}} \\
 & \dfrac{m}{{{m}_{1}}}=\dfrac{\dfrac{-2}{\sqrt{54}}}{-2}=\dfrac{1}{\sqrt{54}} \\
 & \dfrac{n}{{{n}_{1}}}=\dfrac{\dfrac{7}{\sqrt{54}}}{7}=\dfrac{1}{\sqrt{54}} \\
\end{align}$
Clearly we can see that we have the condition $\dfrac{l}{{{l}_{1}}}=\dfrac{m}{{{m}_{1}}}=\dfrac{n}{{{n}_{1}}}$ hence option (c) is correct.
(d) Here we have to check if the plane $x-2y+7z=0$ is perpendicular to the given plane or not.
The two planes will only be perpendicular when the sum of their product of direction cosines is equal to 0. Here when we compare the planes $x-2y+7z=0$ and $x-2y+7z+21=0$ we can see that they have the same coefficients of x, y and z. That means they are parallel and not perpendicular.
Hence option (d) is incorrect.

Note: Note that here all the options we related to different conditions and that is why we needed to check all of them one by one. You must remember the conditions of parallel and perpendicular lines in 2 - D as well as 3 – D planes. Note that we need not to find the direction cosines of the planes in the above question; we can also solve the question using the direction ratios. For a plane $ax+by+cz+d=0$ the direction ratios is (a, b, c).