Question

Line $\mathop r\limits^ \to {\text{ }} = \left( {{\text{ }}i{\text{ }} - {\text{ }}j{\text{ }} + {\text{ }}k{\text{ }}} \right){\text{ }} + {\text{ }}t{\text{ }}\left( {{\text{ }}2i{\text{ }} - {\text{ }}j{\text{ }} + {\text{ }}k{\text{ }}} \right)$ contained in a plane to which vector $\mathop n\limits^ \to {\text{ }} = {\text{ }}3i{\text{ }} - {\text{ }}2j{\text{ }} + {\text{ }}\lambda k$ is normal . Find the value of $\lambda$ . Also find the vector equation of the plane .

Answer 1

Hint : We have to find the value of $\lambda$ and the vector equation of the plane . We solve this question using the concept of three dimensional geometry and vector algebra and we also use the concept of two vectors perpendicular to each other . We use the formula of two vectors perpendicular to each other to find the value of $\lambda$ and then find the vector equation of the plane using the formula for the vector equations .

Complete step-by-step answer :
Given :
$\mathop r\limits^ \to {\text{ }} = {\text{ }}\left( {{\text{ }}i{\text{ }} - {\text{ }}j{\text{ }} + {\text{ }}k{\text{ }}} \right){\text{ }} + {\text{ }}t{\text{ }}\left( {{\text{ }}2i{\text{ }} - {\text{ }}j{\text{ }} + {\text{ }}k{\text{ }}} \right)$
$\mathop r\limits^ \to {\text{ }} = {\text{ }}\left( {{\text{ }}2t{\text{ }} + {\text{ }}1{\text{ }}} \right){\text{ }}i{\text{ }} - {\text{ }}\left( {{\text{ }}t{\text{ }} + {\text{ }}1{\text{ }}} \right){\text{ }}j{\text{ }} + {\text{ }}\left( {{\text{ }}t{\text{ }} + {\text{ }}1{\text{ }}} \right){\text{ }}k$
$\mathop n\limits^ \to {\text{ }} = {\text{ }}3i{\text{ }} - {\text{ }}2j{\text{ }} + {\text{ }}\lambda k$
As given , the line $\mathop r\limits^ \to {\text{ }} = {\text{ }}\left( {{\text{ }}i{\text{ }} - {\text{ }}j{\text{ }} + {\text{ }}k{\text{ }}} \right){\text{ }} + {\text{ }}t{\text{ }}\left( {{\text{ }}2i{\text{ }} - {\text{ }}j{\text{ }} + {\text{ }}k{\text{ }}} \right)$ is normal to $\mathop n\limits^ \to {\text{ }} = {\text{ }}3i{\text{ }} - {\text{ }}2j{\text{ }} + {\text{ }}\lambda k{\text{ }}.$
This means that $\mathop r\limits^ \to {\text{ }}$ and $\mathop n\limits^ \to {\text{ }}$ are perpendicular to each other .
We also know that the dot product to two vectors is zero if both the vectors are perpendicular .
So , we get
$\left( {{\text{ }}2i{\text{ }} - {\text{ }}j{\text{ }} + {\text{ }}k{\text{ }}} \right){\text{ }}.{\text{ }}\left( {{\text{ }}3i{\text{ }} - {\text{ }}2j{\text{ }} + {\text{ }}\lambda k{\text{ }}} \right){\text{ }} = {\text{ }}0$
As , we know that
$i{\text{ }}.{\text{ }}i{\text{ }} = {\text{ }}j{\text{ }}.{\text{ }}j{\text{ }} = {\text{ }}k{\text{ }}.{\text{ }}k{\text{ }} = {\text{ }}1$ Also ,
i . j = i . k = j . k = 0
By multiplication of terms , we get
$6{\text{ }} + {\text{ }}2{\text{ }} + {\text{ }}\lambda {\text{ }} = {\text{ }}0$
We get ,
$\lambda {\text{ }} = {\text{ }} - 8$
The points that lies on the plane are $\left( {{\text{ }}1{\text{ }},{\text{ }} - 1{\text{ }},{\text{ }}1{\text{ }}} \right)$
Using this points , we get values of points as :
$\left( {{\text{ }}x{\text{ }} - {\text{ }}1{\text{ }}} \right){\text{ }},{\text{ }}\left( {{\text{ }}y{\text{ }} + {\text{ }}1{\text{ }}} \right){\text{ }},{\text{ }}\left( {{\text{ }}z{\text{ }} - {\text{ }}1{\text{ }}} \right)$
Vector equation of plane is given as :
Putting these points in $\overrightarrow n$ , we get
$3{\text{ }}\left( {{\text{ }}x{\text{ }} - {\text{ }}1{\text{ }}} \right){\text{ }} - {\text{ }}2{\text{ }}\left( {{\text{ }}y{\text{ }} + {\text{ }}1{\text{ }}} \right){\text{ }} + {\text{ }}\lambda {\text{ }}\left( {{\text{ }}z{\text{ }} - {\text{ }}1{\text{ }}} \right){\text{ }} = {\text{ }}0$
putting the value of $\lambda$ , we get
$3{\text{ }}\left( {{\text{ }}x{\text{ }} - {\text{ }}1{\text{ }}} \right){\text{ }} - {\text{ }}2{\text{ }}\left( {{\text{ }}y{\text{ }} + {\text{ }}1{\text{ }}} \right){\text{ }} - {\text{ }}8{\text{ }}\left( {{\text{ }}z{\text{ }} - {\text{ }}1{\text{ }}} \right){\text{ }} = {\text{ }}0$
Expanding the terms , we get
$3{\text{ }}x{\text{ }} - {\text{ }}2{\text{ }}y{\text{ }} - {\text{ }}8{\text{ }}z{\text{ }} + {\text{ }}3{\text{ }} = {\text{ }}0$
Thus , the value of $\lambda$ is $- 8$ and the vector equation of the plane is $3{\text{ }}x{\text{ }} - {\text{ }}2{\text{ }}y{\text{ }} - {\text{ }}8{\text{ }}z{\text{ }} + {\text{ }}3{\text{ }} = {\text{ }}0{\text{ }}.$

Note : The scalar product of two given vectors $\mathop a\limits^ \to {\text{ }}$ and $\mathop b\limits^ \to {\text{ }}$ having angle $\theta$ between them is defined as :
$\mathop a\limits^ \to {\text{ }}.{\text{ }}\mathop b\limits^ \to {\text{ }} = {\text{ }}\left| a \right|{\text{ }}\left| b \right|{\text{ }}cos{\text{ }}\theta$ Also , when $\mathop a\limits^ \to {\text{ }}.{\text{ }}\mathop b\limits^ \to {\text{ }}$ is given , the angle ‘$\theta$’ between the vectors $\mathop a\limits^ \to {\text{ }}$ and $\mathop b\limits^ \to {\text{ }}$.