Question

Find the equation of the plane

Answer 1

Dear Student:

The equation of the plane can be determinig by getting normal to the vector.

So, for finding normal vector we need two vectors which is in the plane.

By taking cross product we will get Normal vector.

See the attachment.

Answer 2

The plane passes through $(2,\ 0,\ 1)$ and contain the line,

$\longrightarrow\vec{r}=\left<1,\ 4,\ -1\right>+k\left<2,\ -3,\ 3\right>$

So the plane contain these three points -

• $(2,\ 0,\ 1)$

• $(1,\ 4,\ -1)$

• $(x,\ y,\ z)\quad[Let]$

Using these three points we can make the following two vectors. which lie along our plane.

• $\vec{b_1}=\left<1,\ -4,\ 2\right>$

• $\vec{b_2}=\left<x-2,\ y,\ z-1\right>$

And the direction ratio of the line also lies on our plane.

• $\vec{b_3}=\left<2,\ -3,\ 3\right>$

Taking cross product of $\vec{b_1}$ and $\vec{b_3},$

$\longrightarrow\vec{b_1}\times\vec{b_3}=\left|\begin{array}{ccc}\hat i&\hat j&\hat k\\1&-4&2\\2&-3&3\end{array}\right|$

$\longrightarrow\vec{b_1}\times\vec{b_3}=\left<-6,\ 1,\ 5\right>$

This vector is perpendicular to the plane or $\vec{b_2},$ since the three vectors are along the plane. So,

$\longrightarrow\vec{b_2}\cdot\left(\vec{b_1}\times\vec{b_3}\right)=0$

$\longrightarrow\left<x-2,\ y,\ z-1\right>\cdot\left<-6,\ 1,\ 5\right>=0$

$\longrightarrow-6(x-2)+y+5(z-1)=0$

$\longrightarrow\underline{\underline{6x-y-5z-7=0}}$