Uestion 2 the points a(-2, 3,-1), b(0, 5, 2) and c(-1, -2, 1) lies on the same plane. (a) find the vector equation of the plane. (b) find the cartesian of the plane

Answer with explanation:

Equation of Plane having Direction cosines A, B and C passing through points, p, q and r is

⇒A (x-p)+B(y-q)+C(z-r)=0

The plane passes through the points A(-2, 3,-1), B(0, 5, 2) and C(-1, -2, 1).

→A(x+2)+B(y-3)+C(z+1)=0----------(1)

→A(0+2)+B(5-3)+C(2+1)=0

2 A +2 B+3 C=0

→A(-1+2)+B(-2-3)+C(1+1)=0

A -5 B+2 C=0

Substituting the value of A , B and C in equation (1)

⇒19 K(x+2)-K(y-3)-12 K(z+1)=0

⇒19 x +38 -y +3-12 z-12=0

⇒19 x -y -12 z +29=0, is the required equation of Plane in Cartesian form.

⇒(19 i -j -12 k)(xi +y j+z k)+29=0 ,is the required  vector equation of the plane.

step-by-step explanation:

their are no angles. sorry cant .

answer: solve 5\cdot 2^x=2405⋅2  

x

=2405, dot, 2, start superscript, x, end superscript, equals, 240.

solution

to solve for xxx, we must first isolate the exponential part. to do this, divide both sides by 555 as shown below. we do not multiply the 555 and the 222 as this goes against the order of operations!

\begin{aligned} 5\cdot 2^x& =240 2^x& =48 \\ \end{aligned}  

5⋅2  

x

2  

x

​    

=240

=48

​

step-by-step explanation: