Question

Find the vector equation of the plane passing through the point $A\left( a,0,0 \right)$, $B\left( 0,b,0 \right)$ and $C\left( 0,0,c \right)$. Reduce it to normal form. If plane ABC is at distance p from the origin, prove that $\dfrac{1}{{{p}^{2}}}=\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}+\dfrac{1}{{{c}^{2}}}$.

Answer 1

We need to first find the vector points of A, B, C. then we need to form the equation of the plane. Then using the modulus value, we find the normal form of the plane. This gives us the distance of the plane from the origin.

Complete step-by-step solution:
We need to find the equation of the plane passing through the point $A\left( a,0,0 \right)$, $B\left( 0,b,0 \right)$ and $C\left( 0,0,c \right)$.
The required plane passes through the point $A\left( a,0,0 \right)$ whose position vector is $\overrightarrow{a}=a\widehat{i}$ and normal to the vector given by $\overrightarrow{n}=\overrightarrow{AB}\times \overrightarrow{AC}$. We also have $\overrightarrow{b}=b\widehat{j}$, $\overrightarrow{c}=c\widehat{k}$.
Now, $\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=-a\widehat{i}+b\widehat{j}$ and $\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=-a\widehat{i}+c\widehat{k}$.
We find the value of $\overrightarrow{n}=\overrightarrow{AB}\times \overrightarrow{AC}$.
We can express the vector product $\overrightarrow{n}=\overrightarrow{AB}\times \overrightarrow{AC}$ as a determinant form.
$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=-a\widehat{i}+b\widehat{j}$ and $\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=-a\widehat{i}+c\widehat{k}$.
$\overrightarrow{n}=\overrightarrow{AB}\times \overrightarrow{AC}=\left| \begin{matrix} \widehat{i} & \widehat{j} & \widehat{k} \\\ -a & b & 0 \\\ -a & 0 & c \\\ \end{matrix} \right|=\left( bc-0 \right)\widehat{i}-\left( -ac-0 \right)\widehat{j}+\left( 0+ab \right)\widehat{k}$
So, $\overrightarrow{n}=\overrightarrow{AB}\times \overrightarrow{AC}=\left( bc \right)\widehat{i}+\left( ac \right)\widehat{j}+\left( ab \right)\widehat{k}$.
We use the scalar multiplication and use $\widehat{i}.\widehat{i}=\widehat{j}.\widehat{j}=\widehat{k}.\widehat{k}=1,\widehat{i}.\widehat{j}=\widehat{j}.\widehat{i}=\widehat{i}.\widehat{k}=\widehat{j}.\widehat{k}=\widehat{k}.\widehat{j}=\widehat{k}.\widehat{i}=0$
The vector equation of the required plane is $\begin{aligned} & \overrightarrow{r}.\overrightarrow{n}=\overrightarrow{a}.\overrightarrow{n} \\\ & \Rightarrow \overrightarrow{r}.\left( \left( bc \right)\widehat{i}+\left( ac \right)\widehat{j}+\left( ab \right)\widehat{k} \right)=\left( a\widehat{i} \right)\left( \left( bc \right)\widehat{i}+\left( ac \right)\widehat{j}+\left( ab \right)\widehat{k} \right) \\\ & \Rightarrow \overrightarrow{r}.\left( bc\widehat{i}+ac\widehat{j}+ab\widehat{k} \right)=abc.........(i) \\\ \end{aligned}$
Now we find the vector value of $\overrightarrow{n}=\overrightarrow{AB}\times \overrightarrow{AC}$.
So, $\left| \overrightarrow{n} \right|=\sqrt{{{\left( bc \right)}^{2}}+{{\left( ac \right)}^{2}}+{{\left( ab \right)}^{2}}}=\sqrt{{{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}+{{c}^{2}}{{a}^{2}}}$.
To reduce the equation (i) in normal form we apply $\overrightarrow{r}.\dfrac{\overrightarrow{n}}{\left| \overrightarrow{n} \right|}=\dfrac{\overrightarrow{a}.\overrightarrow{n}}{\left| \overrightarrow{n} \right|}$.
So, the normal form becomes

& \overrightarrow{r}.\dfrac{\overrightarrow{n}}{\left| \overrightarrow{n} \right|}=\dfrac{\overrightarrow{a}.\overrightarrow{n}}{\left| \overrightarrow{n} \right|} \\\ & \Rightarrow \overrightarrow{r}.\left( \dfrac{bc\widehat{i}+ac\widehat{j}+ab\widehat{k}}{\sqrt{{{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}+{{c}^{2}}{{a}^{2}}}} \right)=\dfrac{abc}{\sqrt{{{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}+{{c}^{2}}{{a}^{2}}}} \\\ \end{aligned}$$ The right part of the equality of $$\overrightarrow{r}.\dfrac{\overrightarrow{n}}{\left| \overrightarrow{n} \right|}=\dfrac{\overrightarrow{a}.\overrightarrow{n}}{\left| \overrightarrow{n} \right|}$$ gives us the distance of the plane from the origin. So, if p be the distance then $$p=\dfrac{abc}{\sqrt{{{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}+{{c}^{2}}{{a}^{2}}}}$$. We solve it to get $$\begin{aligned} & p=\dfrac{abc}{\sqrt{{{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}+{{c}^{2}}{{a}^{2}}}} \\\ & \Rightarrow \dfrac{1}{p}=\dfrac{\sqrt{{{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}+{{c}^{2}}{{a}^{2}}}}{abc} \\\ & \Rightarrow \dfrac{1}{{{p}^{2}}}=\dfrac{{{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}+{{c}^{2}}{{a}^{2}}}{{{a}^{2}}{{b}^{2}}{{c}^{2}}} \\\ & \Rightarrow \dfrac{1}{{{p}^{2}}}=\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}+\dfrac{1}{{{c}^{2}}} \\\ \end{aligned}$$ Thus proved. **Note:** We don’t need to find the distance exclusively. The normal form has been asked to find the distance easily. We can use matrix form to find the scalar multiplication. We know that Two or more vectors are coplanar if they are linearly dependent, therefore their components are proportional. Two or more points are coplanar if the vectors determined by them are also coplanar. Also, we can find out the coplanarity if the scalar triple product of any three vectors is zero then they are coplanar.