Question

2. (5 points) (a) Find a vector perpendicular to the plane through the points A(0, -2,0), B(4,1, -2) and C(5,3,1). (b) Find an equation of the plane through the points A, B, and C. (b) Find the area of the triangle ABC.

Answer 1

a)

let assume that

n

the equation of the plane is,

$\dpi{100} \begin{vmatrix} x-x_1 & y-y_1 & z-z_1\\ x_2-x_1 & y_2-y_1 & z_2-z_1\\ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{vmatrix}=0$

1-0 Y-(-2) 2-0 → 4-0 1-(-2) -2-0 15-03-(-2) 1-0 0

7 9 +2 4 3 این اے -2 = 0 1

$\Rightarrow x(3+10)-(y+2)(4+10)+z(20-15)=0$

$\Rightarrow 13x-14y-28+5z=0$

$\Rightarrow 13x-14y+5z=28$

b)

$13x-14y+5z=28$

the vector of plane is,

$\vec{a}=(13,-14,5)$

now the area of triangle ABC,

$A=\frac{1}{2}|\vec{a}|$

$A=\frac{1}{2}\sqrt{13^2+(-14)^2+5^2}$

$A=\frac{\sqrt{390}}{2}$

$A=9.8742$