19.  Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes \overrightarrow{r}.\left ( \widehat{i}-\widehat{j}+2\widehat{k} \right )=5 and \overrightarrow{r}.\left ( 3\widehat{i}+\widehat{j}+\widehat{k} \right )=6.

Answers (1)
P Pankaj Sanodiya


A point through which line passes

\vec a=\hat i+2\hat j+3\hat k

two plane

 \overrightarrow{r}.\left ( \widehat{i}-\widehat{j}+2\widehat{k} \right )=5  And

\overrightarrow{r}.\left ( 3\widehat{i}+\widehat{j}+\widehat{k} \right )=6 

it can be seen that normals of the planes are

\vec n_1=\hat i-\hat j+2\hat k

\vec n_2=3\hat i+\hat j+\hat k
since the line is parallel to both planes, its parallel vector will be perpendicular to normals of both planes.

So, a vector perpendicular to both these normal vector is

\vec d=\vec n_1\times\vec n_2

\vec d=\begin{vmatrix} \hat i &\hat j & \hat k\\ 1 &-1 &2 \\ 3& 1 & 1 \end{vmatrix}=\hat i(-1-2)-\hat j(1-6)+\hat k(1+3)

\vec d=-3\hat i+5\hat j+4\hat k

Now a line which passes through \vec a and parallels to \vec d is 

L=\vec a+\lambda\vec d

So the required line is

L=\vec a+\lambda\vec d

L=\hat i+2\hat j+3\hat k+\lambda(-3\hat i+5\hat j+4\hat k)

L=(1-3\lambda)\hat i+(2+5\lambda)\hat j+(3+4\lambda)\hat k