Question

(5.1) Find the vector form of the equation of the plane that passes through the point P0 = (1, −2, 3)

and has normal vector ~n =< 3, 1, −1 >.

(5.2) Find an equation for the plane that contains the line x = −1 + 3t, y = 5 + 3t, z = 2 + t and is

parallel to the line of intersection of the planes x −2(y −1) + 3z = −1 and y = −2x −1 = 0.

Answer 1

Question 5.1 :

The points through which the plane passing is,

P0 = (1, −2, 3)

Normal vector to the plane is,

n =< 3, 1, −1 >

The vector equation of plane is given by,

$(\overrightarrow{r}-\overrightarrow{a}).\overrightarrow{n}=0$

$\overrightarrow{r}=x\overrightarrow{i}+y\overrightarrow{j}+z\overrightarrow{k}$ is the position vector for any point (x,y,z) in the plane.

$\overrightarrow{a}$ is the position vector of point through which plane passing $\overrightarrow{n}$ is the normal vector to the plane.

Position vector of point P0(1,-2,3) is,

$\overrightarrow{a}=\hat{i}-2\hat{j}+3\hat{k}\\\overrightarrow{n}=3\hat{i}+\hat{j}-\hat{k}$

the vector equation of plane is,

$(\overrightarrow{r}-\hat{i}-2\hat{j}+3\hat{k})).(3\hat{i}+\hat{j}-\hat{k})$

this can also be written as,

$[(x\hat{i}+y\hat{j}+z\hat{k})-(\hat{i}-2\hat{j}+3\hat{k})].(3\hat{i}+\hat{j}-\hat{k})\\=[(x-1)\hat{i}+(y+2)\hat{j}+(z-3)\hat{k}].(3\hat{i}+\hat{j}-\hat{k})$

by taking the dot product of above equation, we will obtain the Cartesian form of equation of plane.

2)

$x−2(y−1)+3z=−1\\ y = −2x −1\\ y=-2x-1\\ x+4x+4+3z=-1\\ y=-2x-1\\ z=-\dfrac{5}{3}x-\dfrac{5}{3}\\ <0,−1,−\frac{5}{3}>+t<1,−2,−\frac{5}{3}>\\ The\ plane\ contains\ the\ line\\ x = −1 + 3t, y = 5 + 3t, z = 2 + t\\ t=0: Point(-1, 5, 2)\\ t=-2: Point(-7, -1, 0)\\ -a+5b+2c+d=0\\ -7a-b+d=0\\ b=-7a+d\\ -a-35a+5d+2c+d=0\\ b=-7a+d\\ c=18a-3d\\ Substitute\\ a+14a-2d-30a+5d=0\\ d=5a\\ If\ a=1\\ d=5\\ b=-2\\ c=3\\ The\ equation\ for\ the\ plane\ is\\ x-2y+3z+5=0$