Find the equation of plane containing line described by Please help me in this really easy task
Find the equation of plane containing line described by
$x+3y-2z=1$, $2x-y+2z=3$, containing point $(1,1,3)$
 A: Any eq.n of the plane containing that line is $L_1+\lambda L_2=0$.
Hence general eq.n of the plane is $(1+2\lambda)x+(3-\lambda)y+(-2+2\lambda)z=1+3\lambda$.
Since it passes through $(1,1,3)$, putting it in above eq.n we get $\lambda=0.75$.
Hence the plane is $2.5x+2.25y-0.5z=3.25$ i.e. $10x+9y-2z=13$
A: So, plane $$p=>(x+3y−2z-1) + \lambda \times (2x−y+2z-3)=0$$
ie. $p=P1+\lambda P2;$
As plane passes through $(1,1,3)$ it satisfy the eq.
ie. $(-3)+ \lambda(4) =0$;
$\lambda=3/4;$
Eq. $$p=>10x+9y−2z=13$$
A: Hint: If $\pi_1$ is the plane given by $x+3y-2z-1=0$ and $\pi_2$ is the other one, write the equation of the pencil of planes $\mathcal{F}$ generated by $\pi_1$ and $\pi_2$. Then find the plane in $\mathcal{F}$ passing through $(1,1,3)$.
A: The normals to the planes are $n_1 = (1,3,-2)$, $n_2 = (2,-1,2)$, hence the direction of the line is given by $d = n_1 \times n_2 = (4, -6,-7)$.
The line has the form $x_0+t d$ for $t \in \mathbb{R}$, and lies on both planes. Hence $\langle n_1, x_0 \rangle = 1$, $\langle n_2, x_0 \rangle = 3$. Solving these equations shows $x_0 \in \{(\frac{10-4t}{7},\frac{6t-1}{7},t)\}$. Picking $t=-1$ gives $x_0 = (2,-1,-1)$.
The other point is $p_0 = (1,1,3)$, hence a normal to the required plane is given by $\hat{n} = d \times (p_0-x_0) = (-10, -9 , 2)$. We have $\langle \hat{n}, x_0 \rangle = -13$, hence the required plane is $-10x -9 y +2 z = -13$.
