Find the vector equation of the plane passing through three point with position vectors i + j -2 k ,2 i - j + k and i +2 j + k . Also, find coordinates of the point of intersection of this plane and the line r =3 i - j - k + (2 i -2 j + k ).

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 a = i + j -2 k and is normal to the vector n given by n = AB AC Clearly, AB = OB - OA =(2 i - j + k )-( i + j -2 k ) = i -2 j +3 k AC = OC - OA =( i +2 j + k )-( i + j -2 k ) =0 i + j +3 k n = AB AC = i & j & k 1&-2&3 0&1&3 =-9 i -3 j + k The vector equation of the required plane is r n = a n r (-9 i -3 j + k )=( i + j -2 k ) (-9 i -3 j + k ) r (9 i +3 j - k )=14 To find the point of intersection of the plane The given equation of the line is r =(3 i - j - k )+ (2 i -2j+ k ) r =(3+2 ) i +(-1-2 ) j +(-1+ ) k The coordinates of any point on this line are in the form of (3+2 ) i +(-1-2 ) j +(-1+ ) k Since this point lies on the plane r (9 i +3 j - k )=14, [(3+2 ) i +(-1-2 ) j +(-1+ ) k ] (9 i +3 j - k )=14 27+18 -3-6 +1- =14 11 =-11 =-1 So, the coordinates of the point are r (-9 i -3 j + k )=-9-3-2 r [- (9 i +3 j - k ) ]=-14 Or (3+2 ,-1-2 ,-1+ ) (3+2 ,-1-2 ,-1+ ) =(3-2,-1+2,-1-1) =(1,1,-2)

Find the vector equation of the plane passing through three point…

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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 $(3+2\lambda,-1-2\lambda,-1+\lambda)$
$(3+2\lambda,-1-2\lambda,-1+\lambda)$
$=(3-2,-1+2,-1-1)$
$=(1,1,-2)$