Question
A plane passes through the point (1, -2, 5) and is perpendicular to the line joining the origin to the point $2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}.$ Find the vector and cartesian forms of the equation of the plane.
✓
Answer
The normal is passing throught the point A(0, 0, 0) and B(3, 1, -1). So,
$\vec{\text{n}}=\overrightarrow{\text{OP}}$
$=(3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})-(0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}})$
$=3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$
Since the plane passes throught (1, -2, 5)
$\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}}+5\hat{\text{k}}$
We know that the vector equation of the plane passing through a point $\vec{\text{a}}$ and normal to $\vec{\text{n}}$ is,
$\vec{\text{r}}\cdot\vec{\text{n}}=\vec{\text{a}}\cdot\vec{\text{n}}$
Substituting $\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{n}}=4\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}},$ we get
$\vec{\text{r}}\cdot(3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})=(\hat{\text{i}}-2\hat{\text{j}}+5\hat{\text{k}})\cdot(3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}\cdot(3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})=3-2-5$
$\Rightarrow\vec{\text{r}}\cdot(3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})=-4$
Substituting $\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$ in the vector equation we get
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})=-4$
$\Rightarrow3\text{x}+\text{y}-\text{z}=-4$
$\vec{\text{n}}=\overrightarrow{\text{OP}}$
$=(3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})-(0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}})$
$=3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$
Since the plane passes throught (1, -2, 5)
$\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}}+5\hat{\text{k}}$
We know that the vector equation of the plane passing through a point $\vec{\text{a}}$ and normal to $\vec{\text{n}}$ is,
$\vec{\text{r}}\cdot\vec{\text{n}}=\vec{\text{a}}\cdot\vec{\text{n}}$
Substituting $\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{n}}=4\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}},$ we get
$\vec{\text{r}}\cdot(3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})=(\hat{\text{i}}-2\hat{\text{j}}+5\hat{\text{k}})\cdot(3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}\cdot(3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})=3-2-5$
$\Rightarrow\vec{\text{r}}\cdot(3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})=-4$
Substituting $\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$ in the vector equation we get
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})=-4$
$\Rightarrow3\text{x}+\text{y}-\text{z}=-4$
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