Q.Find the vector equation of the plane passing through the points $\hat i + \hat j - 2\hat k$, $\hat i + 2\hat j + \hat k$, $2\hat i - \hat j + \hat k$. Hence find the cartesian equation of the plane.

Find two vectors in the plane from the three given points, take their cross product for the normal, then write $\bar r\cdot\bar n=\bar a\cdot\bar n$.

Points: $A(1,1,-2)$, $B(1,2,1)$, $C(2,-1,1)$ (from position vectors $\hat i+\hat j-2\hat k$, $\hat i+2\hat j+\hat k$, $2\hat i-\hat j+\hat k$).

Step 1: Two vectors lying in the plane:

$\vec{AB}=B-A=(0,1,3), \qquad \vec{AC}=C-A=(1,-2,3)$

Step 2: Normal vector $\bar n=\vec{AB}\times\vec{AC}$:

$\bar n=\begin{vmatrix}\hat i&\hat j&\hat k\\0&1&3\\1&-2&3\end{vmatrix}=\hat i(1\cdot3-3(-2))-\hat j(0\cdot3-3\cdot1)+\hat k(0(-2)-1\cdot1)$

$=\hat i(3+6)-\hat j(0-3)+\hat k(0-1)=9\hat i+3\hat j-\hat k$

Step 3: $\bar n\cdot\bar a$ using $A(1,1,-2)$: …