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Three Dimensional Geometry — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsThree Dimensional Geometry4 Marks
Question
State when the line $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$ is parallel to the plane $\vec{\text{r}}\cdot\vec{\text{n}}=\text{d}.$ Show that the line $\vec{\text{r}}=\hat{\text{i}}+\hat{\text{j}}+\lambda(3\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})$ is parallel to the plane $\vec{\text{r}}\cdot(2\hat{\text{i}}+\hat{\text{k}})=3.$ Also, find the distance between the line and the plane.
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Answer

We know that line $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$ is paralle to plane $\vec{\text{r}}\cdot\vec{\text{n}}=\text{d}$ if
$\vec{\text{b}}\cdot\vec{\text{n}}=0$
Given, line is $\vec{\text{r}}=(\hat{\text{i}}+\hat{\text{j}})+\lambda(3\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})$ and plane is $\vec{\text{r}}\cdot(2\hat{\text{i}}+\hat{\text{k}})=3,$
$\vec{\text{b}}=3\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}},\vec{\text{a}}=(\hat{\text{i}}+\hat{\text{j}})$ and $\vec{\text{n}}=(2\hat{\text{j}}+\hat{\text{k}})$
Now, $\vec{\text{b}}\cdot\vec{\text{n}}=(3\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})(2\hat{\text{j}}+\hat{\text{k}})$
$=(3)(0)+(-1)(2)+(2)(1)$
$=0-2+2$
$=0$
Since, $\vec{\text{b}}\cdot\vec{\text{n}}=0$ So line is parallel to plane.
Distance between point $\vec{\text{a}}$ and plane $​​\vec{\text{r}}\cdot\vec{\text{n}}-\text{d}=0$ is given by
$\text{D}=\Bigg|\frac{\vec{\text{a}}\vec{\text{n}}-\text{d}}{|\vec{\text{n}}|}\Bigg|\ ...(\text{i})$
$\vec{\text{a}}$ is a point on the line. So diatance between line and plane is equal to the distance between $​​\vec{\text{a}}=(\hat{\text{i}}+\hat{\text{j}})$ and plane $\vec{\text{r}}\cdot(2\hat{\text{i}}+\hat{\text{k}})=3,$ so using (i)
$\text{D}=\Bigg|\frac{(\hat{\text{i}}+\hat{\text{j}})(2\hat{\text{j}}+\hat{\text{k}})-3}{\sqrt{(2)^2+(1)^2}}\Bigg|$
$=\bigg|\frac{(1)(0)+(1)(2)+(0)(1)-3}{\sqrt{4+1}}\bigg|$
$=\Big|\frac{0+2+0-3}{\sqrt{5}}\Big|$
$=\Big|\frac{-1}{\sqrt{5}}\Big|$
$=\frac{1}{\sqrt{5}}\text{ unit}$
So, required distance $=\frac{1}{\sqrt{5}}\text{ unit}$

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