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Question: If the points \(\left( 1,1,p \right)\) and \[\left( -3,0,1 \right)\] be equidistant from the plane \...

If the points (1,1,p)\left( 1,1,p \right) and (−3,0,1)\left( -3,0,1 \right) be equidistant from the plane r→.(3i^+4j^−12k^)+13=0\overrightarrow{r}.\left( 3\widehat{i}+4\widehat{j}-12\widehat{k} \right)+13=0, then find the value of p.

Explanation

Solution

We find the distance of the points (1,1,p)\left( 1,1,p \right) and (−3,0,1)\left( -3,0,1 \right) from the plane r→.(3i^+4j^−12k^)+13=0\overrightarrow{r}.\left( 3\widehat{i}+4\widehat{j}-12\widehat{k} \right)+13=0. Using the formula of distance ∣a→.n→−d∣n→∣∣\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 r→.n→=d\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 a→\overrightarrow{a} from the plane r→.n→=d\overrightarrow{r}.\overrightarrow{n}=d is ∣a→.n→−d∣n→∣∣\left| \dfrac{\overrightarrow{a}.\overrightarrow{n}-d}{\left| \overrightarrow{n} \right|} \right|.
The points (1,1,p)\left( 1,1,p \right) and (−3,0,1)\left( -3,0,1 \right) be equidistant from the plane r→.(3i^+4j^−12k^)+13=0\overrightarrow{r}.\left( 3\widehat{i}+4\widehat{j}-12\widehat{k} \right)+13=0.
We convert the points (1,1,p)\left( 1,1,p \right) and (−3,0,1)\left( -3,0,1 \right) into their vector form.
a1→=i^+j^+pk^\overrightarrow{{{a}_{1}}}=\widehat{i}+\widehat{j}+p\widehat{k}, a2→=−3i^+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 i^.i^=j^.j^=k^.k^=1\widehat{i}.\widehat{i}=\widehat{j}.\widehat{j}=\widehat{k}.\widehat{k}=1 and for any other cases we have i^.j^=j^.i^=j^.k^=k^.j^=i^.k^=k^.i^=0\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
r→.(3i^+4j^−12k^)+13=0 ⇒r→.(3i^+4j^−12k^)=−13 ⇒r→.(−3i^−4j^+12k^)=13 \begin{aligned} & \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{aligned}
Comparing r→.(−3i^−4j^+12k^)=13\overrightarrow{r}.\left( -3\widehat{i}-4\widehat{j}+12\widehat{k} \right)=13 with r→.n→=d\overrightarrow{r}.\overrightarrow{n}=d, we get n→=(−3i^−4j^+12k^),d=13\overrightarrow{n}=\left( -3\widehat{i}-4\widehat{j}+12\widehat{k} \right),d=13.
We find modulus of n→=(−3i^−4j^+12k^)\overrightarrow{n}=\left( -3\widehat{i}-4\widehat{j}+12\widehat{k} \right). ∣n→∣=∣−3i^−4j^+12k^∣=32+42+122=13\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 a1→=i^+j^+pk^\overrightarrow{{{a}_{1}}}=\widehat{i}+\widehat{j}+p\widehat{k}, a2→=−3i^+k^\overrightarrow{{{a}_{2}}}=-3\widehat{i}+\widehat{k} from the plane.
From the point a1→=i^+j^+pk^\overrightarrow{{{a}_{1}}}=\widehat{i}+\widehat{j}+p\widehat{k}, the distance is ∣a1→.n→−d∣n→∣∣\left| \dfrac{\overrightarrow{{{a}_{1}}}.\overrightarrow{n}-d}{\left| \overrightarrow{n} \right|} \right|. We now find the value as

& \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{aligned}$$ Here, we apply the dot product formulas to get $$\begin{aligned} & \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{aligned}$$ From the point $\overrightarrow{{{a}_{2}}}=-3\widehat{i}+\widehat{k}$, the distance is $$\left| \dfrac{\overrightarrow{{{a}_{2}}}.\overrightarrow{n}-d}{\left| \overrightarrow{n} \right|} \right|$$. We now find the value as $$\begin{aligned} & \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{aligned}$$. Here, we apply the dot product formulas to get $$\begin{aligned} & \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{aligned}$$ It’s given the points $\left( 1,1,p \right)$ and $$\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{aligned} & \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{aligned}$$ Therefore, the value so p is $$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.