Let find a plane S containing points A(3,0,0) ,B(0,2,0),C(0,0,1):
∣ x − x a y − y a z − z a x b − x a y b − y a z b − z a x c − x a y c − y a z c − z a ∣ \begin{vmatrix}
x-x_a & y-y_a & z-z_a\\
x_b-x_a & y_b-y_a & z_b-z_a\\
x_c-x_a & y_c-y_a & z_c-z_a\\
\end{vmatrix} ∣ ∣ x − x a x b − x a x c − x a y − y a y b − y a y c − y a z − z a z b − z a z c − z a ∣ ∣ = 0 = 0 = 0
∣ x − 3 y − 0 z − 0 0 − 0 2 − 0 0 − 0 0 − 0 0 − 0 1 − 0 ∣ \begin{vmatrix}
x-3 & y-0 & z-0\\
0-0 & 2-0 & 0-0\\
0-0 & 0-0 & 1-0\\
\end{vmatrix} ∣ ∣ x − 3 0 − 0 0 − 0 y − 0 2 − 0 0 − 0 z − 0 0 − 0 1 − 0 ∣ ∣ = 0 =0 = 0
Plane S equation:
2 ∗ x + 3 ∗ y + 6 ∗ z − 6 = 0 2*x+3*y+6*z-6=0 2 ∗ x + 3 ∗ y + 6 ∗ z − 6 = 0
Let find a height OH of a cone. OH is perpendicular to the plane S and passes through point O(0,0,0)
x − 0 2 = y − 0 3 = z − 0 6 \dfrac{x-0}{2}=\dfrac{y-0}{3}=\dfrac{z-0}{6} 2 x − 0 = 3 y − 0 = 6 z − 0
Let us find the equation of the straight line passing through the origin and point A
As a result, the parametric equation of the straight line D was obtained
x = 3 ∗ t x=3*t x = 3 ∗ t
Find the angle between the straight lines A and D
ϕ = 73.381 ° \phi = 73.381° ϕ = 73.381°
Length of line OH is 0.85:
t g ( ϕ ) = H A 0.85 tg(\phi)=\dfrac{HA}{0.85} t g ( ϕ ) = 0.85 H A
H A = 2.78 HA = 2.78 H A = 2.78
Equation of cone
x 2 2.7 8 2 + y 2 2.7 8 2 − z 2 0.8 5 2 = 0 \dfrac{x^2}{2.78^2}+\dfrac{y^2}{2.78^2}-\dfrac{z^2}{0.85^2}=0 2.7 8 2 x 2 + 2.7 8 2 y 2 − 0.8 5 2 z 2 = 0
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