Question
Show that the line whose vector equation is $\vec{\text{r}}=2\hat{\text{i}}+5\hat{\text{j}}+7\hat{\text{k}}+\lambda(\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}})$ is parallel to the plane whose vector equation is $\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})=7$ Also, find the distance between thetm.

Answer

The given plane passes through the point with position vector $\vec{\text{a}}=2\hat{\text{i}}+5\hat{\text{j}}+7\hat{\text{k}}$ and is parallel to the vector $\vec{\text{b}}=\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}$
The given plane is $\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})=7$ or
So, the normal vector, $\vec{\text{n}}=\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$ and d = 7
Now, $\vec{\text{b}}\cdot\vec{\text{n}}=(\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}})\cdot(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})$
$=1+3-4$
$=4-4$
$=0$
So, $\vec{\text{b}}$ is perpendicular to $\vec{\text{n}}$
So, the given line is parallel to the given plane.
The distance between the line and the parallel plane. Then,
d = length of the perpendicular from the point $\vec{\text{a}}=2\hat{\text{i}}+5\hat{\text{j}}+7\hat{\text{k}}$ to the plane $\vec{\text{r}}\cdot\vec{\text{n}}=\text{d}$
$\text{d}=\frac{\big|\vec{\text{a}}\cdot\vec{\text{n}}-\text{d}\big|}{|\vec{\text{n}}|}$
$=\frac{\big|(2\hat{\text{i}}+5\hat{\text{j}}+7\hat{\text{k}})\cdot(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})-7\big|}{|\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}|}$
$=\frac{|2+5-7-7|}{\sqrt{1+1+1}}$
$=\frac{7}{\sqrt{3}}\text{ units}$ 

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