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The plane — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsThe plane5 Marks
Question
Find the vector equation of the plane passing through three point with position vectors $\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}},2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}.$ Also, find coordinates of the point of intersection of this plane and the line $\vec{\text{r}}=3\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}+\lambda(2\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}).$
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Answer

Let A(1, 1, -2), B(2, -1, 1) and C(1, 2, 1) be the point represented by the given position vectors.
The required planes through the point A(1, 1, -1) whose position vector is
$\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$ and is normal to the vector $\vec{\text{n}}$ given by
$\vec{\text{n}}=\overrightarrow{\text{AB}}\times\overrightarrow{\text{AC}}$
Clearly, $\overrightarrow{\text{AB}}=\overrightarrow{\text{OB}}-\overrightarrow{\text{OA}}$
$=(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})-(\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}})$
$=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$
$\overrightarrow{\text{AC}}=\overrightarrow{\text{OC}}-\overrightarrow{\text{OA}}$
$=(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}})-(\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}})$
$=0\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}$
$\vec{\text{n}}=\overrightarrow{\text{AB}}\times\overrightarrow{\text{AC}}$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&-2&3\\0&1&3\end{vmatrix}$
$=-9\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}}$
The vector equation of the required plane is
$\vec{\text{r}}\cdot\vec{\text{n}}=\vec{\text{a}}\cdot\vec{\text{n}}$
$\Rightarrow\vec{\text{r}}\cdot(-9\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}})=(\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}})\cdot(-9\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}\cdot(9\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}})=14$
To find the point of intersection of the plane
The given equation of the line is
$\vec{\text{r}}=(3\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}})+\lambda(2\hat{\text{i}}-2\text{j}+\hat{\text{k}})$
$\vec{\text{r}}=(3+2\lambda)\hat{\text{i}}+(-1-2\lambda)\hat{\text{j}}+(-1+\lambda)\hat{\text{k}}$
The coordinates of any point on this line are in the form of
$(3+2\lambda)\hat{\text{i}}+(-1-2\lambda)\hat{\text{j}}+(-1+\lambda)\hat{\text{k}}$
Since this point lies on the plane $\vec{\text{r}}\cdot(9\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}})=14,$
$\Big[(3+2\lambda)\hat{\text{i}}+(-1-2\lambda)\hat{\text{j}}+(-1+\lambda)\hat{\text{k}}\Big]\cdot\big(9\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}\big)=14$
$\Rightarrow27+18\lambda-3-6\lambda+1-\lambda=14$
$\Rightarrow11\lambda=-11$
$\Rightarrow\lambda=-1$
So, the coordinates of the point are
$\Rightarrow\vec{\text{r}}\cdot(-9\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}})=-9-3-2$
$\Rightarrow\vec{\text{r}}\cdot\Big[-\big(9\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}\big)\Big]=-14$
Or $(3+2\lambda,-1-2\lambda,-1+\lambda)$
$(3+2\lambda,-1-2\lambda,-1+\lambda)$
$=(3-2,-1+2,-1-1)$
$=(1,1,-2)$

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