If the points $\left( 1,1,p \right)$ and \[\left( -3,0,1 \right)\] be equidistant from the plane $\overrightarrow{r}.\left( 3\widehat{i}+4\widehat{j}-12\widehat{k} \right)+13=0$, then find the value of p.
Answer
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Hint: We find the distance of the points $\left( 1,1,p \right)$ and \[\left( -3,0,1 \right)\] from the plane $\overrightarrow{r}.\left( 3\widehat{i}+4\widehat{j}-12\widehat{k} \right)+13=0$. Using the formula of distance \[\left| \dfrac{\overrightarrow{a}.\overrightarrow{n}-d}{\left| \overrightarrow{n} \right|} \right|\], we get the distance based on the variable and plane form according to $\overrightarrow{r}.\overrightarrow{n}=d$. We equal both parts to find a linear equation. We solve that to find possible values of p.
Complete step-by-step solution:
The distance of a point with position vector $\overrightarrow{a}$ from the plane $\overrightarrow{r}.\overrightarrow{n}=d$ is \[\left| \dfrac{\overrightarrow{a}.\overrightarrow{n}-d}{\left| \overrightarrow{n} \right|} \right|\].
The points $\left( 1,1,p \right)$ and \[\left( -3,0,1 \right)\] be equidistant from the plane $\overrightarrow{r}.\left( 3\widehat{i}+4\widehat{j}-12\widehat{k} \right)+13=0$.
We convert the points $\left( 1,1,p \right)$ and \[\left( -3,0,1 \right)\] into their vector form.
$\overrightarrow{{{a}_{1}}}=\widehat{i}+\widehat{j}+p\widehat{k}$, $\overrightarrow{{{a}_{2}}}=-3\widehat{i}+\widehat{k}$.
The dot product value of vectors is either 1 or 0. For same coordinate products like $\widehat{i}.\widehat{i}=\widehat{j}.\widehat{j}=\widehat{k}.\widehat{k}=1$ and for any other cases we have $\widehat{i}.\widehat{j}=\widehat{j}.\widehat{i}=\widehat{j}.\widehat{k}=\widehat{k}.\widehat{j}=\widehat{i}.\widehat{k}=\widehat{k}.\widehat{i}=0$.
Now the plane is
$\begin{align}
& \overrightarrow{r}.\left( 3\widehat{i}+4\widehat{j}-12\widehat{k} \right)+13=0 \\
& \Rightarrow \overrightarrow{r}.\left( 3\widehat{i}+4\widehat{j}-12\widehat{k} \right)=-13 \\
& \Rightarrow \overrightarrow{r}.\left( -3\widehat{i}-4\widehat{j}+12\widehat{k} \right)=13 \\
\end{align}$
Comparing $\overrightarrow{r}.\left( -3\widehat{i}-4\widehat{j}+12\widehat{k} \right)=13$ with $\overrightarrow{r}.\overrightarrow{n}=d$, we get $\overrightarrow{n}=\left( -3\widehat{i}-4\widehat{j}+12\widehat{k} \right),d=13$.
We find modulus of $\overrightarrow{n}=\left( -3\widehat{i}-4\widehat{j}+12\widehat{k} \right)$. \[\left| \overrightarrow{n} \right|=\left| -3\widehat{i}-4\widehat{j}+12\widehat{k} \right|=\sqrt{{{3}^{2}}+{{4}^{2}}+{{12}^{2}}}=13\].
Now we find distance of points $\overrightarrow{{{a}_{1}}}=\widehat{i}+\widehat{j}+p\widehat{k}$, $\overrightarrow{{{a}_{2}}}=-3\widehat{i}+\widehat{k}$ from the plane.
From the point $\overrightarrow{{{a}_{1}}}=\widehat{i}+\widehat{j}+p\widehat{k}$, the distance is \[\left| \dfrac{\overrightarrow{{{a}_{1}}}.\overrightarrow{n}-d}{\left| \overrightarrow{n} \right|} \right|\]. We now find the value as
\[\begin{align}
& \left| \dfrac{\overrightarrow{{{a}_{1}}}.\overrightarrow{n}-d}{\left| \overrightarrow{n} \right|} \right| \\
& =\left| \dfrac{\left( \widehat{i}+\widehat{j}+p\widehat{k} \right).\left( -3\widehat{i}-4\widehat{j}+12\widehat{k} \right)-13}{13} \right| \\
\end{align}\]
Here, we apply the dot product formulas to get
\[\begin{align}
& \left| \dfrac{\left( \widehat{i}+\widehat{j}+p\widehat{k} \right).\left( -3\widehat{i}-4\widehat{j}+12\widehat{k} \right)-13}{13} \right| \\
& =\left| \dfrac{-3-4+12p-13}{13} \right| \\
& =\left| \dfrac{12p-20}{13} \right| \\
\end{align}\]
From the point $\overrightarrow{{{a}_{2}}}=-3\widehat{i}+\widehat{k}$, the distance is \[\left| \dfrac{\overrightarrow{{{a}_{2}}}.\overrightarrow{n}-d}{\left| \overrightarrow{n} \right|} \right|\]. We now find the value as
\[\begin{align}
& \left| \dfrac{\overrightarrow{{{a}_{2}}}.\overrightarrow{n}-d}{\left| \overrightarrow{n} \right|} \right| \\
& =\left| \dfrac{\left( -3\widehat{i}+\widehat{k} \right).\left( -3\widehat{i}-4\widehat{j}+12\widehat{k} \right)-13}{13} \right| \\
\end{align}\].
Here, we apply the dot product formulas to get
\[\begin{align}
& \left| \dfrac{\left( -3\widehat{i}+\widehat{k} \right).\left( -3\widehat{i}-4\widehat{j}+12\widehat{k} \right)-13}{13} \right| \\
& =\left| \dfrac{9+12-13}{13} \right| \\
& =\dfrac{8}{13} \\
\end{align}\]
It’s given the points $\left( 1,1,p \right)$ and \[\left( -3,0,1 \right)\] be equidistant from the plane $\overrightarrow{r}.\left( 3\widehat{i}+4\widehat{j}-12\widehat{k} \right)+13=0$
So, \[\left| \dfrac{12p-20}{13} \right|=\dfrac{8}{13}\].
We solve the equation for p.
\[\begin{align}
& \left| \dfrac{12p-20}{13} \right|=\dfrac{8}{13} \\
& \Rightarrow \dfrac{12p-20}{13}=\pm \dfrac{8}{13} \\
& \Rightarrow 12p-20=\pm 8 \\
& \Rightarrow 12p=20\pm 8=28,12 \\
& \Rightarrow p=\dfrac{7}{3},1 \\
\end{align}\]
Therefore, the value so p is \[p=\dfrac{7}{3},1\].
Note: The plane needs to be changed into its normal form to equate with $\overrightarrow{r}.\overrightarrow{n}=d$. The modulus form needs to be kept as it is. That’s why we will get two possible values. We equated the plane formula of $\overrightarrow{r}.\left( 3\widehat{i}+4\widehat{j}-12\widehat{k} \right)+13=0$ with accordance to the form $\overrightarrow{r}.\overrightarrow{n}=d$ where d defines the distance of the plane from the origin. We need to keep it positive always.
Complete step-by-step solution:
The distance of a point with position vector $\overrightarrow{a}$ from the plane $\overrightarrow{r}.\overrightarrow{n}=d$ is \[\left| \dfrac{\overrightarrow{a}.\overrightarrow{n}-d}{\left| \overrightarrow{n} \right|} \right|\].
The points $\left( 1,1,p \right)$ and \[\left( -3,0,1 \right)\] be equidistant from the plane $\overrightarrow{r}.\left( 3\widehat{i}+4\widehat{j}-12\widehat{k} \right)+13=0$.
We convert the points $\left( 1,1,p \right)$ and \[\left( -3,0,1 \right)\] into their vector form.
$\overrightarrow{{{a}_{1}}}=\widehat{i}+\widehat{j}+p\widehat{k}$, $\overrightarrow{{{a}_{2}}}=-3\widehat{i}+\widehat{k}$.
The dot product value of vectors is either 1 or 0. For same coordinate products like $\widehat{i}.\widehat{i}=\widehat{j}.\widehat{j}=\widehat{k}.\widehat{k}=1$ and for any other cases we have $\widehat{i}.\widehat{j}=\widehat{j}.\widehat{i}=\widehat{j}.\widehat{k}=\widehat{k}.\widehat{j}=\widehat{i}.\widehat{k}=\widehat{k}.\widehat{i}=0$.
Now the plane is
$\begin{align}
& \overrightarrow{r}.\left( 3\widehat{i}+4\widehat{j}-12\widehat{k} \right)+13=0 \\
& \Rightarrow \overrightarrow{r}.\left( 3\widehat{i}+4\widehat{j}-12\widehat{k} \right)=-13 \\
& \Rightarrow \overrightarrow{r}.\left( -3\widehat{i}-4\widehat{j}+12\widehat{k} \right)=13 \\
\end{align}$
Comparing $\overrightarrow{r}.\left( -3\widehat{i}-4\widehat{j}+12\widehat{k} \right)=13$ with $\overrightarrow{r}.\overrightarrow{n}=d$, we get $\overrightarrow{n}=\left( -3\widehat{i}-4\widehat{j}+12\widehat{k} \right),d=13$.
We find modulus of $\overrightarrow{n}=\left( -3\widehat{i}-4\widehat{j}+12\widehat{k} \right)$. \[\left| \overrightarrow{n} \right|=\left| -3\widehat{i}-4\widehat{j}+12\widehat{k} \right|=\sqrt{{{3}^{2}}+{{4}^{2}}+{{12}^{2}}}=13\].
Now we find distance of points $\overrightarrow{{{a}_{1}}}=\widehat{i}+\widehat{j}+p\widehat{k}$, $\overrightarrow{{{a}_{2}}}=-3\widehat{i}+\widehat{k}$ from the plane.
From the point $\overrightarrow{{{a}_{1}}}=\widehat{i}+\widehat{j}+p\widehat{k}$, the distance is \[\left| \dfrac{\overrightarrow{{{a}_{1}}}.\overrightarrow{n}-d}{\left| \overrightarrow{n} \right|} \right|\]. We now find the value as
\[\begin{align}
& \left| \dfrac{\overrightarrow{{{a}_{1}}}.\overrightarrow{n}-d}{\left| \overrightarrow{n} \right|} \right| \\
& =\left| \dfrac{\left( \widehat{i}+\widehat{j}+p\widehat{k} \right).\left( -3\widehat{i}-4\widehat{j}+12\widehat{k} \right)-13}{13} \right| \\
\end{align}\]
Here, we apply the dot product formulas to get
\[\begin{align}
& \left| \dfrac{\left( \widehat{i}+\widehat{j}+p\widehat{k} \right).\left( -3\widehat{i}-4\widehat{j}+12\widehat{k} \right)-13}{13} \right| \\
& =\left| \dfrac{-3-4+12p-13}{13} \right| \\
& =\left| \dfrac{12p-20}{13} \right| \\
\end{align}\]
From the point $\overrightarrow{{{a}_{2}}}=-3\widehat{i}+\widehat{k}$, the distance is \[\left| \dfrac{\overrightarrow{{{a}_{2}}}.\overrightarrow{n}-d}{\left| \overrightarrow{n} \right|} \right|\]. We now find the value as
\[\begin{align}
& \left| \dfrac{\overrightarrow{{{a}_{2}}}.\overrightarrow{n}-d}{\left| \overrightarrow{n} \right|} \right| \\
& =\left| \dfrac{\left( -3\widehat{i}+\widehat{k} \right).\left( -3\widehat{i}-4\widehat{j}+12\widehat{k} \right)-13}{13} \right| \\
\end{align}\].
Here, we apply the dot product formulas to get
\[\begin{align}
& \left| \dfrac{\left( -3\widehat{i}+\widehat{k} \right).\left( -3\widehat{i}-4\widehat{j}+12\widehat{k} \right)-13}{13} \right| \\
& =\left| \dfrac{9+12-13}{13} \right| \\
& =\dfrac{8}{13} \\
\end{align}\]
It’s given the points $\left( 1,1,p \right)$ and \[\left( -3,0,1 \right)\] be equidistant from the plane $\overrightarrow{r}.\left( 3\widehat{i}+4\widehat{j}-12\widehat{k} \right)+13=0$
So, \[\left| \dfrac{12p-20}{13} \right|=\dfrac{8}{13}\].
We solve the equation for p.
\[\begin{align}
& \left| \dfrac{12p-20}{13} \right|=\dfrac{8}{13} \\
& \Rightarrow \dfrac{12p-20}{13}=\pm \dfrac{8}{13} \\
& \Rightarrow 12p-20=\pm 8 \\
& \Rightarrow 12p=20\pm 8=28,12 \\
& \Rightarrow p=\dfrac{7}{3},1 \\
\end{align}\]
Therefore, the value so p is \[p=\dfrac{7}{3},1\].
Note: The plane needs to be changed into its normal form to equate with $\overrightarrow{r}.\overrightarrow{n}=d$. The modulus form needs to be kept as it is. That’s why we will get two possible values. We equated the plane formula of $\overrightarrow{r}.\left( 3\widehat{i}+4\widehat{j}-12\widehat{k} \right)+13=0$ with accordance to the form $\overrightarrow{r}.\overrightarrow{n}=d$ where d defines the distance of the plane from the origin. We need to keep it positive always.
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