Q116 / 180 · KCET 2016 · Mathematics
The vector equation of the plane which is at a distance of \( \frac{3}{\sqrt{14}} \) from the origin and the normal
from the origin is \( 2 \hat{i}-3 \hat{j}+\hat{k} \) is
from the origin is \( 2 \hat{i}-3 \hat{j}+\hat{k} \) is
- Correct answerA\( \rightarrow \vec{r} \cdot(2 \hat{i}-3 \hat{j}+\hat{k})=3 \)
- B\( \vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=9 \)
- C\( \stackrel{\vec{r}} \cdot(\hat{i}+2 \hat{j})=3 \)
- D\( \rightarrow \vec{r} \cdot(2 \hat{i}+\hat{k})=3 \)
Solution
Given that \( \vec{N}=2 \hat{i}-3 \hat{j}+\hat{k} \)
So, \( |\vec{N}|=\sqrt{4+9+1}=\sqrt{14} \)
Unit vector is given by
\( \hat{n}=\frac{\vec{N}}{\sqrt{14}} \)
and from origin is \( d=\frac{3}{\sqrt{14}} \)
So, required equation, \( \vec{r} \cdot \hat{n}=d \)
\( \Rightarrow \vec{r} \cdot 2 \hat{i}-3 \hat{j}+\hat{k}=\frac{3}{\sqrt{14}} \) \( \Rightarrow \vec{r} \cdot(2 \hat{i}-3 \hat{j}+\hat{k})=3 \)
So, \( |\vec{N}|=\sqrt{4+9+1}=\sqrt{14} \)
Unit vector is given by
\( \hat{n}=\frac{\vec{N}}{\sqrt{14}} \)
and from origin is \( d=\frac{3}{\sqrt{14}} \)
So, required equation, \( \vec{r} \cdot \hat{n}=d \)
\( \Rightarrow \vec{r} \cdot 2 \hat{i}-3 \hat{j}+\hat{k}=\frac{3}{\sqrt{14}} \) \( \Rightarrow \vec{r} \cdot(2 \hat{i}-3 \hat{j}+\hat{k})=3 \)
Hence, the correct answer is option: 'A'.