Three Dimensional Geometry — Maths STD 12 Science — Question
Gujarat BoardEnglish MediumSTD 12 ScienceMathsThree Dimensional Geometry5 Marks
Question
Find the vector equations of the coordinate planes.
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Answer
We have to find vector equation of coordinate planes.
For xy-plane.
It passes through origin and is perpendicular to z-axis, so
Put $\vec{\text{a}}=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$ and $\vec{\text{n}}=\hat{\text{k}}$ in the vector equation of plane passing through point $\vec{\text{a}}$ and perpendicular to vector $\vec{\text{n}}$
$(\vec{\text{r}}-\vec{\text{n}})\vec{\text{n}}=0$
$(\vec{\text{r}}-0\hat{\text{i}}-0\hat{\text{j}}-0\hat{\text{k}})\hat{\text{k}}=0$
$\vec{\text{r}}\cdot\vec{\text{k}}=0\ ...(\text{i})$
For xz-plane,
It passes throught origin and perpendicular to y-axis, so
$\vec{\text{a}}=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$ and $\vec{\text{n}}=\hat{\text{j}}$
Equation of xz-plane is given by
$(\vec{\text{r}}-\vec{\text{a}})\vec{\text{n}}=0$
$(\vec{\text{r}}-0\hat{\text{i}}-0\hat{\text{j}}-0\hat{\text{k}})\hat{\text{j}}=0$
$\vec{\text{r}}\cdot\hat{\text{j}}=0$
For yz-plane,
It passes throught origin and is perpendicular to x-axis, so
$\vec{\text{a}}=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}},\vec{ \text{n}}=\hat{\text{i}}$
$(\vec{\text{r}}-\vec{\text{a}})\vec{\text{n}}=0$
$(\vec{\text{r}}-0\hat{\text{i}}-0\hat{\text{j}}-0\hat{\text{k}})\hat{\text{i}}=0$
$\vec{\text{r}}\cdot\hat{\text{i}}=0$
Hence, equation of xy, yz, zx-plane are given by
$\vec{\text{r}}\cdot\vec{\text{k}}=0$
$\vec{\text{r}}\cdot\hat{\text{i}}=0$
$\vec{\text{r}}\cdot\hat{\text{j}}=0$
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