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Three Dimensional Geometry — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsThree Dimensional Geometry4 Marks
Question
Show that the plane vector equation is $\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})=1$ and the line whose vector equation is $\vec{\text{r}}=(-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})+\lambda(2\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}})$ are parallel. Also, find the distance between them.
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Answer

The given plane passes through the point with position vector $\vec{\text{a}}=-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ and is parallel to the vector $\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}}$
The given plane is $\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})=1$ or $\vec{\text{r}}\cdot\vec{\text{n}}=\text{d}.$
So, normal vector $\vec{\text{n}}=\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$ and d = 1
Now, $\vec{\text{b}}\cdot\vec{\text{n}}=(2\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}})\cdot(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})$
$=2+2-4=0$
So, $\vec{\text{b}}$ is perpendicular to $\vec{\text{n}}.$
So, the given line is parallel to the given plane.
$\text{d}=\frac{\big|\vec{\text{a}}\cdot\vec{\text{n}}-\text{d}\big|}{|\vec{\text{n}}|}$
$=\frac{(-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})\cdot(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})-1}{|\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}|}$
$=\frac{|-1+2-1-1|}{\sqrt{1+4+1}}$
$=\frac{1}{\sqrt{6}}\text{ units}$

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