Question

The perpendicular distance from the origin to the plane containing the two lines, $\dfrac{x+2}{3}=\dfrac{y-2}{5}=\dfrac{z+5}{7}$ and $\dfrac{x-1}{1}=\dfrac{y-4}{4}=\dfrac{z+4}{7}$.

& A)\dfrac{11}{\sqrt{6}} \\\ & B)6\sqrt{11} \\\ & C)\sqrt{11} \\\ & D)11\sqrt{6} \\\ \end{aligned}$$

Answer 1

We know that the plane containing the two lines, $\dfrac{x-{{x}_{1}}}{{{a}_{1}}}=\dfrac{y-{{y}_{1}}}{{{b}_{1}}}=\dfrac{z-{{z}_{1}}}{{{c}_{1}}}$ and $\dfrac{x-{{x}_{2}}}{{{a}_{2}}}=\dfrac{y-{{y}_{2}}}{{{b}_{2}}}=\dfrac{z-{{z}_{2}}}{{{c}_{2}}}$ is $\left| \begin{matrix} x-{{x}_{1}} & y-{{y}_{1}} & z-{{z}_{1}} \\\ {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\\ \end{matrix} \right|=0$. We know that the distance between origin and $ax+by+cz+d=0$ is equal to $\dfrac{\left| d \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$. By using these concepts, we can get the perpendicular distance from the origin to the plane containing the two lines, $\dfrac{x+2}{3}=\dfrac{y-2}{5}=\dfrac{z+5}{7}$ and $\dfrac{x-1}{1}=\dfrac{y-4}{4}=\dfrac{z+4}{7}$.

Complete step-by-step solution:
Now we should find the plane containing the two lines, $\dfrac{x+2}{3}=\dfrac{y-2}{5}=\dfrac{z+5}{7}$ and $\dfrac{x-1}{1}=\dfrac{y-4}{4}=\dfrac{z+4}{7}$.
We know that the plane containing the two lines, $\dfrac{x-{{x}_{1}}}{{{a}_{1}}}=\dfrac{y-{{y}_{1}}}{{{b}_{1}}}=\dfrac{z-{{z}_{1}}}{{{c}_{1}}}$ and $\dfrac{x-{{x}_{2}}}{{{a}_{2}}}=\dfrac{y-{{y}_{2}}}{{{b}_{2}}}=\dfrac{z-{{z}_{2}}}{{{c}_{2}}}$ is $\left| \begin{matrix} x-{{x}_{1}} & y-{{y}_{1}} & z-{{z}_{1}} \\\ {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\\ \end{matrix} \right|=0$.
By using the above concept, we can find the plane containing the two lines, $\dfrac{x+2}{3}=\dfrac{y-2}{5}=\dfrac{z+5}{7}$ and $\dfrac{x-1}{1}=\dfrac{y-4}{4}=\dfrac{z+4}{7}$.
Let us assume this plane containing the two lines, $\dfrac{x+2}{3}=\dfrac{y-2}{5}=\dfrac{z+5}{7}$ and $\dfrac{x-1}{1}=\dfrac{y-4}{4}=\dfrac{z+4}{7}$ as plane P.

x+2 & y-2 & z+5 \\\ 3 & 5 & 7 \\\ 1 & 4 & 7 \\\ \end{matrix} \right|=0$$ We know that the value of $$\left| \begin{matrix} a & b & c \\\ d & e & f \\\ g & h & i \\\ \end{matrix} \right|=a\left( ei-hf \right)-b\left( di-fg \right)+c\left( dh-ge \right)$$. Now by using this concept, we should find the equation of plane P. $$\begin{aligned} & \Rightarrow P\equiv \left| \begin{matrix} x+2 & y-2 & z+5 \\\ 3 & 5 & 7 \\\ 1 & 4 & 7 \\\ \end{matrix} \right|=0 \\\ & \Rightarrow P\equiv \left( x+2 \right)\left( (5)(7)-(4)(7) \right)-\left( y-2 \right)\left( (3)(7)-(7)(1) \right)+\left( z+5 \right)\left( (3)(7)-(1)(7) \right)=0 \\\ & \Rightarrow P\equiv \left( x+2 \right)\left( 35-28 \right)-\left( y-2 \right)\left( 21-7 \right)+\left( z+5 \right)\left( 21-14 \right)=0 \\\ & \Rightarrow P\equiv \left( x+2 \right)\left( 7 \right)-\left( y-2 \right)\left( 14 \right)+\left( z+5 \right)\left( 7 \right)=0 \\\ & \Rightarrow P\equiv 7x+14-14y+28+7z+35=0 \\\ & \Rightarrow P\equiv 7x-14y+7z+77=0 \\\ & \Rightarrow P\equiv 7\left( x-2y+z+11 \right)=0 \\\ & \Rightarrow P\equiv x-2y+z+11=0.....(1) \\\ \end{aligned}$$ From equation (1), it is clear that $$x-2y+z+11=0$$ is the equation of plane containing the two lines, $$\dfrac{x+2}{3}=\dfrac{y-2}{5}=\dfrac{z+5}{7}$$ and $$\dfrac{x-1}{1}=\dfrac{y-4}{4}=\dfrac{z+4}{7}$$. Now we should find the distance between origin and $$x-2y+z+11=0$$. We know that the distance between origin and $$ax+by+cz+d=0$$ is equal to $$\dfrac{\left| d \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$$. Now by using this concept, we can find the distance between origin and $$x-2y+z+11=0$$. Let us assume this distance is equal to D. $$\begin{aligned} & \Rightarrow D=\dfrac{\left| 11 \right|}{\sqrt{{{1}^{2}}+{{2}^{2}}+{{1}^{2}}}} \\\ & \Rightarrow D=\dfrac{11}{\sqrt{6}}....(2) \\\ \end{aligned}$$ From equation (2), it is clear that the distance between origin and $$x-2y+z+11=0$$ is equal to $$\dfrac{11}{\sqrt{6}}$$. **Hence, option A is correct.** **Note:** Students may have a misconception that the distance between origin and $$ax+by+cz+d=0$$ is equal to $$\dfrac{\left| d \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}}}$$. If this misconception is followed, then we cannot get the exact value of D. So, this misconception should be avoided. Otherwise, we cannot get correct answer at any cost.