Question

If ${{P}_{1}},{{P}_{2}},{{P}_{3}}$ denotes the perpendicular distance of the plane 2x – 3y + 4z + 2 = 0 from the parallel plane 2x – 3y + 4z + 6 = 0, 4x – 6y + 8z + 3 = 0 and 2x – 3y + 4z – 6 = 0 respectively then,
a. ${{P}_{1}}+8{{P}_{2}}-{{P}_{3}}=0$
b. ${{P}_{3}}=16{{P}_{2}}$
c. $8{{P}_{2}}={{P}_{1}}$
d. ${{P}_{1}}+2{{P}_{2}}+3{{P}_{3}}=\sqrt{29}$

Answer 1

Hint: In order to find the solution of this question, we will use the formula of finding the distance between two parallel planes, that is, $P=\dfrac{\left| {{D}_{1}}-{{D}_{2}} \right|}{\sqrt{{{A}^{2}}+{{B}^{2}}+{{C}^{2}}}}$ and we will find the value of ${{P}_{1}},{{P}_{2}},{{P}_{3}}$ and then we will put the values in each option one by one and check the options whether they are correct or not. Hence, we will get the answer.

Complete step-by-step solution -
In this question, we have been asked to find the relation between the perpendicular distances, ${{P}_{1}},{{P}_{2}},{{P}_{3}}$ between the planes. So, to solve this question, we should know that perpendicular distance between two parallel planes $Ax+By+Cz+{{D}_{1}}=0$ and $Ax+By+Cz+{{D}_{2}}=0$ is given by $P=\dfrac{\left| {{D}_{1}}-{{D}_{2}} \right|}{\sqrt{{{A}^{2}}+{{B}^{2}}+{{C}^{2}}}}$. So, we can say that if ${{P}_{1}}$ is the perpendicular distance between the planes 2x – 3y + 4z + 2 = 0 and 2x – 3y + 4z + 6 = 0. Therefore, we can say that for, A = 2, B = -3 and C = 4, ${{D}_{1}}=2$ and ${{D}_{2}}=6$, we get,
$\begin{aligned} & {{P}_{1}}=\dfrac{\left| 2-6 \right|}{\sqrt{{{2}^{2}}+{{\left( -3 \right)}^{2}}+{{4}^{2}}}}=\dfrac{\left| -4 \right|}{\sqrt{4+9+16}} \\\ & {{P}_{1}}=\dfrac{4}{\sqrt{29}}.........\left( i \right) \\\ \end{aligned}$
Similarly, we can say that if ${{P}_{2}}$ is the perpendicular distance between the planes 2x – 3y + 4z + 2 = 0 and 4x – 6y + 8z + 3 = 0 $\Rightarrow 2x-3y+4z+\dfrac{3}{2}=0$. Therefore, we can say that for, A = 2, B = -3 and C = 4, ${{D}_{1}}=2$ and ${{D}_{2}}=\dfrac{3}{2}$, we get,
$\begin{aligned} &{{P}_{2}}=\dfrac{\left| 2-\dfrac{3}{2} \right|}{\sqrt{{{2}^{2}}+{{\left( -3 \right)}^{2}}+{{4}^{2}}}}=\dfrac{\left| \dfrac{1}{2} \right|}{\sqrt{4+9+16}} \\\ & {{P}_{2}}=\dfrac{1}{2\sqrt{29}}.........\left( ii \right) \\\ \end{aligned}$
Similarly, we can say that if ${{P}_{3}}$ is the perpendicular distance between the planes 2x – 3y + 4z + 2 = 0 and 2x – 3y + 4z – 6 = 0 . Therefore, we can say that for, A = 2, B = -3 and C = 4, ${{D}_{1}}=2$ and ${{D}_{2}}=-6$, we get,
$\begin{aligned} &{{P}_{3}}=\dfrac{\left| 2-\left( -6 \right) \right|}{\sqrt{{{2}^{2}}+{{\left( -3 \right)}^{2}}+{{4}^{2}}}}=\dfrac{\left| 8 \right|}{\sqrt{4+9+16}} \\\ & {{P}_{3}}=\dfrac{8}{\sqrt{29}}.........\left( iii \right) \\\ \end{aligned}$
Now, we will put the values of ${{P}_{1}},{{P}_{2}},{{P}_{3}}$ in each option one by one.
So, let us consider option (a) first, that is, ${{P}_{1}}+8{{P}_{2}}-{{P}_{3}}=0$. Therefore, by putting the values, we get,
$\begin{aligned} & \dfrac{4}{\sqrt{29}}+\dfrac{8}{2\sqrt{29}}-\dfrac{8}{\sqrt{29}}=0 \\\ \Rightarrow & \dfrac{4}{\sqrt{29}}+\dfrac{4}{\sqrt{29}}-\dfrac{8}{\sqrt{29}}=0 \\\ \Rightarrow & \dfrac{4+4-8}{\sqrt{29}}=0 \\\ \Rightarrow & \dfrac{8-8}{\sqrt{29}}=0 \\\ \Rightarrow & \dfrac{0}{\sqrt{29}}=0 \\\ \Rightarrow & 0=0 \\\ \end{aligned}$
Hence, we get LHS = RHS. Therefore, we can say that option (a) is correct.
Now, let us consider option (b), that is, ${{P}_{3}}=16{{P}_{2}}$. So, on putting the values, we get,
$\begin{aligned} & \dfrac{8}{\sqrt{29}}=16\times \dfrac{1}{2\sqrt{29}} \\\ & \dfrac{8}{\sqrt{29}}=\dfrac{8}{\sqrt{29}} \\\ \end{aligned}$
Hence, we get LHS = RHS. So, option (b) is correct.
Now, we will consider option (c), that is $8{{P}_{2}}={{P}_{1}}$. So, on substituting the values, we get,
$\begin{aligned} & 8\times \dfrac{1}{2\sqrt{29}}=\dfrac{4}{\sqrt{29}} \\\ \Rightarrow & \dfrac{4}{\sqrt{29}}=\dfrac{4}{\sqrt{29}} \\\ \end{aligned}$
Hence, we get LHS = RHS. So, option (c) is correct.
Now, let us consider option (d), that is, ${{P}_{1}}+2{{P}_{2}}+3{{P}_{3}}=\sqrt{29}$. So, on substituting the values, we get,
$\begin{aligned} & \dfrac{4}{\sqrt{29}}+2\times \dfrac{1}{2\sqrt{29}}+3\times \dfrac{8}{\sqrt{29}}=\sqrt{29} \\\ \Rightarrow & \dfrac{4}{\sqrt{29}}+\dfrac{1}{\sqrt{29}}+\dfrac{24}{\sqrt{29}}=\sqrt{29} \\\ \Rightarrow & \dfrac{29}{\sqrt{29}}=\sqrt{29} \\\ \Rightarrow & \sqrt{29}=\sqrt{29} \\\ \end{aligned}$
Hence, we get LHS = RHS. Therefore, we can say that option (d) is the correct answer.
Therefore, from the above observations, all the options, that is (a), (b), (c) and (d) are correct.

Note: While solving this question, we have to be very careful because if we miss out any of the options, then we might not be able to get all the correct options as the answer and make a mistake and may lose our marks. Also, we have to remember that perpendicular distance between parallel planes, $Ax+By+Cz+{{D}_{1}}=0$ and $Ax+By+Cz+{{D}_{2}}=0$ is given by $P=\dfrac{\left| {{D}_{1}}-{{D}_{2}} \right|}{\sqrt{{{A}^{2}}+{{B}^{2}}+{{C}^{2}}}}$. Also, students must make sure not to make any calculation mistakes.