Question

1. find the equation for the plane with the given description. (express the plane in the form: ( ax + by + cz = d ).) (a) passes through ( (1, 2, 3) ) and is parallel to the plane ( 2x + 3y + 4z = 5 ). (b) passes through ( (2, 3, 4) ) and is parallel to the yz-plane. (c) passes through ( (3, 4, 5) ) and has normal vector ( hat{i} + hat{j} ). (d) contains the point ( (4, 5, 6) ) and the line ( vec{r}(t) = langle 1, 0, -1

angle + t langle 1, 1, 1
angle ).

Explanation:

Part (a)

Step1: Recall parallel plane normal vectors

Parallel planes have the same normal vector. The plane \(2x + 3y + 4z = 5\) has normal vector \(\langle 2, 3, 4
angle\), so our plane also has normal vector \(\langle 2, 3, 4
angle\).

Step2: Use point - normal form

The point - normal form of a plane is \(A(x - x_0)+B(y - y_0)+C(z - z_0)=0\), where \((x_0,y_0,z_0)=(1,2,3)\) and \(\langle A,B,C
angle=\langle 2,3,4
angle\).
Substitute the values: \(2(x - 1)+3(y - 2)+4(z - 3)=0\)
Expand: \(2x-2 + 3y-6 + 4z-12 = 0\)
Simplify: \(2x+3y + 4z=2 + 6+12=20\)

Step1: Determine normal vector of yz - plane

The yz - plane has equation \(x = 0\), so its normal vector is \(\langle 1,0,0
angle\). A plane parallel to the yz - plane also has normal vector \(\langle 1,0,0
angle\).

Step2: Use point - normal form

The plane passes through \((2,3,4)\). The point - normal form is \(1(x - 2)+0(y - 3)+0(z - 4)=0\)
Simplify: \(x-2=0\), or \(x = 2\) (in the form \(Ax+By + Cz=D\), this is \(1x+0y + 0z=2\))

Step1: Identify normal vector and point

The normal vector \(\vec{n}=\langle 1,1,0
angle\) (since \(\hat{i}+\hat{j}=\langle 1,1,0
angle\)) and the point is \((3,4,5)\).

Step2: Use point - normal form

The point - normal form is \(1(x - 3)+1(y - 4)+0(z - 5)=0\)
Expand: \(x-3+y - 4=0\)
Simplify: \(x + y-7=0\), or \(x + y=7\) (in the form \(Ax+By + Cz=D\), \(1x+1y+0z = 7\))

Answer:

\(2x + 3y+4z = 20\)

Part (b)