Question

The intercept made by plane $\overrightarrow{r}.\overrightarrow{n}=q$ on the x-axis is
(a) $\dfrac{q}{\widehat{i}.\overrightarrow{n}}$
(b) $\dfrac{\widehat{i}.\overrightarrow{n}}{q}$
(c) $\left( \widehat{i}.\overrightarrow{n} \right)q$
(d) $\dfrac{q}{\left| \overrightarrow{n} \right|}$

Answer 1

Hint: Put $\overrightarrow{r}=xi+yj+zk$ to convert the given equation of the plane from vector to Cartesian form. Now relate the equation of the plane with the equation of plane in intercept form. Now, get the intercept in Cartesian form and convert it to vector form to get the correct answer.

Complete step-by-step answer:
Given equation of the plane is
$\overrightarrow{r}.\overrightarrow{n}=q$ ………… (i)
Let us convert the vector form of the plane to Cartesian form by putting $\overrightarrow{r}$ as $\left( x\widehat{i}+y\widehat{j}+z\widehat{k} \right)$ and $\overrightarrow{n}=a\widehat{i}+b\widehat{j}+z\widehat{k}$, so, we get
$\left( x\widehat{i}+y\widehat{j}+z\widehat{k} \right).\left( a\widehat{i}+b\widehat{j}+z\widehat{k} \right)=q..........(ii)$
As we know dot product of two vector $\overrightarrow{A}$ and $\overrightarrow{B}$ is given by relation
$\overrightarrow{A.}\overrightarrow{B}=\left| \overrightarrow{A} \right|\left| \overrightarrow{B} \right|\cos \theta $
Where $\theta $ is the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$.
And we know angle between $\left( \widehat{i},\widehat{j} \right),\left( \widehat{j},\widehat{k} \right),\left( \widehat{k},\widehat{i} \right)$ is ${{90}^{o}}$ and \[\left( \widehat{i},\widehat{i} \right),\left( \widehat{j},\widehat{j} \right),\left( \widehat{k},\widehat{k} \right)\] is ${{90}^{o}}$.
So, we can get dot product of $\overrightarrow{r}$ and $\overrightarrow{n}$from equation (ii); hence, we get

\[\begin{align}
  & ax+by+cz=q \\
 & \Rightarrow \dfrac{qx}{q}+\dfrac{by}{q}+\dfrac{cz}{q}=1 \\
 & \Rightarrow \dfrac{x}{\left( \dfrac{q}{a} \right)}+\dfrac{y}{\left( \dfrac{q}{b} \right)}+\dfrac{z}{\left( \dfrac{q}{c} \right)}=1.........(iii) \\
\end{align}\]
Now, we can compare the equation (iii) with the intercept form of the place which is given as
$\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$ ……………. (iv)
Where (a, b, c) are the intercepts made by the plane on x – axis, y – axis and z – axis respectively.
Hence, intercept made by the given plane on the x – axis from the equation (iii) and (iv) as
\[xintercept\text{ }=~\dfrac{q}{a}\]
Now, we can convert the Cartesian form of x – intercept to vector form as
$\begin{align}
  & x=\dfrac{q}{\widehat{i}.\left( a\widehat{i}+b\widehat{j}+c\widehat{k} \right)} \\
 & x=\dfrac{q}{\widehat{i}.\overrightarrow{n}} \\
\end{align}$
We can observe that $\widehat{i}.\left( a\widehat{i}+b\widehat{j}+c\widehat{k} \right)$ will give ‘a’ on simplification. So, vector form of intercept made by plane $\overrightarrow{r.}\overrightarrow{n}=q$ on the x – axis is $\dfrac{q}{\widehat{i}.\overrightarrow{n}}$ .
Hence, option (a) is correct.

Note: We can directly use the formula of getting intercepts of plane $\overrightarrow{r}. \overrightarrow{n}=d$ on x – axis, y – axis, z – axis as $\dfrac{d}{\widehat{i}.\overrightarrow{n}}, \dfrac{d}{\widehat{j}.\overrightarrow{n}},\dfrac{d}{\widehat{k}.\overrightarrow{n}}$.
Observation for converting Cartesian intercept to vector form is the key point of the solution. One may check all the four options to verify the answer. But it is not required as per the given question; we need to observe for conversion.