QUESTION 1
Let p . M ( x , y , z ) p.M\left(x,y,z\right) p . M ( x , y , z )
Then, P M → ∣ ∣ n → , n → = ( 2 , 1 , 3 ) − specified by condition \overrightarrow{PM}||\overrightarrow{n}, \,\,\,\overrightarrow{n}=\left(2,1,3\right)-\text{specified by condition} PM ∣∣ n , n = ( 2 , 1 , 3 ) − specified by condition
{ p . P ( 2 , 0 , − 1 ) p . M ( x , y , z ) → P M → ( x − 2 , y , z + 1 ) P M → ∣ ∣ n → → x − 2 2 = y 1 = z + 1 3 = t { x − 2 = 2 t y = t z + 1 = 3 t ⟶ { x = 2 + 2 t y = t z = − 1 + 3 t \left\{\begin{array}{l}
p.P\left(2,0,-1\right)\\[0.3cm]
p.M\left(x,y,z\right)
\end{array}\right.\rightarrow\overrightarrow{PM}\left(x-2,y,z+1\right)\\[0.3cm]
\overrightarrow{PM}||\overrightarrow{n}\rightarrow\frac{x-2}{2}=\frac{y}{1}=\frac{z+1}{3}=t\\[0.3cm]
\left\{\begin{array}{l}
x-2=2t\\[0.3cm]
y=t\\[0.3cm]
z+1=3t
\end{array}\right.\longrightarrow
\left\{\begin{array}{l}
x=2+2t\\[0.3cm]
y=t\\[0.3cm]
z=-1+3t
\end{array}\right. { p . P ( 2 , 0 , − 1 ) p . M ( x , y , z ) → PM ( x − 2 , y , z + 1 ) PM ∣∣ n → 2 x − 2 = 1 y = 3 z + 1 = t ⎩ ⎨ ⎧ x − 2 = 2 t y = t z + 1 = 3 t ⟶ ⎩ ⎨ ⎧ x = 2 + 2 t y = t z = − 1 + 3 t
ANSWER
{ x = 2 + 2 t y = t z = − 1 + 3 t \left\{\begin{array}{l}
x=2+2t\\[0.3cm]
y=t\\[0.3cm]
z=-1+3t
\end{array}\right. ⎩ ⎨ ⎧ x = 2 + 2 t y = t z = − 1 + 3 t
QUESRION 2
As we know, the parametric equation of the straight line has the form
{ x = x 0 + a t y = y 0 + b t z = z 0 + c t , where { p . M ( x 0 , y 0 , z 0 ) − any point on the line n → = ( a , b , c ) − direction vector \left\{\begin{array}{l}
x=x_0+at\\[0.3cm]
y=y_0+bt\\[0.3cm]
z=z_0+ct
\end{array}\right.,\,\,\,\text{where}
\left\{\begin{array}{l}
p.M\left(x_0,y_0,z_0\right)-\text{any point on the line}\\[0.3cm]
\overrightarrow{n}=\left(a,b,c\right)-\text{direction vector}
\end{array}\right. ⎩ ⎨ ⎧ x = x 0 + a t y = y 0 + b t z = z 0 + c t , where ⎩ ⎨ ⎧ p . M ( x 0 , y 0 , z 0 ) − any point on the line n = ( a , b , c ) − direction vector
In our case, since by the problem statement we know two points p . A = ( 1 , 2 , − 3 ) p.A=\left(1, 2, - 3\right) p . A = ( 1 , 2 , − 3 ) p . B = ( 7 , 2 , − 4 ) p.B=\left(7, 2, - 4\right) p . B = ( 7 , 2 , − 4 )
n → = A B → = ( 7 − 1 , 2 − 2 , − 4 − ( − 3 ) ) n → = ( 6 , 0 , − 1 ) \overrightarrow{n}=\overrightarrow{AB}=\left(7-1,2-2,-4-\left(-3\right)\right)\\[0.3cm]
\boxed{\overrightarrow{n}=\left(6,0,-1\right)} n = A B = ( 7 − 1 , 2 − 2 , − 4 − ( − 3 ) ) n = ( 6 , 0 , − 1 )
As an "unknown" p . M ( x 0 , y 0 , z 0 ) p.M\left(x_0,y_0,z_0\right) p . M ( x 0 , y 0 , z 0 ) p . A = ( 1 , 2 , − 3 ) p.A=\left(1,2,-3\right) p . A = ( 1 , 2 , − 3 )
Then,
{ x = 1 + 6 t y = 2 z = − 3 − t \left\{\begin{array}{l}
x=1+6t\\[0.3cm]
y=2\\[0.3cm]
z=-3-t
\end{array}\right. ⎩ ⎨ ⎧ x = 1 + 6 t y = 2 z = − 3 − t
ANSWER
{ x = 1 + 6 t y = 2 z = − 3 − t \left\{\begin{array}{l}
x=1+6t\\[0.3cm]
y=2\\[0.3cm]
z=-3-t
\end{array}\right. ⎩ ⎨ ⎧ x = 1 + 6 t y = 2 z = − 3 − t
QUESTION 3
From the beginning, we find any point p . M ( x 0 , y 0 , z 0 ) p.M\left(x_0,y_0,z_0\right) p . M ( x 0 , y 0 , z 0 )
{ − 5 x + y − 2 z = 3 → y = 3 + 5 x + 2 z 2 x − 3 y + 5 z = − 7 { y = 3 + 5 x + 2 z 2 x − 3 ⋅ ( 3 + 5 x + 2 z ) + 5 z = − 7 { y = 3 + 5 x + 2 z 2 x − 9 − 15 x − 6 z + 5 z = − 7 { y = 3 + 5 x + 2 z − 13 x − z = 2 → z = − 2 − 13 x { y = 3 + 5 x + 2 ⋅ ( − 2 − 13 x ) z = − 2 − 13 x { y = − 1 − 21 x z = − 2 − 13 x x = C o n s t \left\{\begin{array}{l}
- 5x + y - 2z =3\to y=3+5x+2z\\[0.3cm]
2x - 3y + 5z =-7
\end{array}\right.\\[0.3cm]
\left\{\begin{array}{l}
y=3+5x+2z\\[0.3cm]
2x - 3\cdot\left(3+5x+2z\right) + 5z =-7
\end{array}\right.\\[0.3cm]
\left\{\begin{array}{l}
y=3+5x+2z\\[0.3cm]
2x - 9-15x-6z + 5z =-7
\end{array}\right.\\[0.3cm]
\left\{\begin{array}{l}
y=3+5x+2z\\[0.3cm]
-13x-z =2\to z=-2-13x
\end{array}\right.\\[0.3cm]
\left\{\begin{array}{l}
y=3+5x+2\cdot\left(-2-13x\right)\\[0.3cm]
z=-2-13x
\end{array}\right.\\[0.3cm]
\left\{\begin{array}{l}
y=-1-21x\\[0.3cm]
z=-2-13x\\[0.3cm]
x=Const
\end{array}\right.\\[0.3cm] { − 5 x + y − 2 z = 3 → y = 3 + 5 x + 2 z 2 x − 3 y + 5 z = − 7 { y = 3 + 5 x + 2 z 2 x − 3 ⋅ ( 3 + 5 x + 2 z ) + 5 z = − 7 { y = 3 + 5 x + 2 z 2 x − 9 − 15 x − 6 z + 5 z = − 7 { y = 3 + 5 x + 2 z − 13 x − z = 2 → z = − 2 − 13 x { y = 3 + 5 x + 2 ⋅ ( − 2 − 13 x ) z = − 2 − 13 x ⎩ ⎨ ⎧ y = − 1 − 21 x z = − 2 − 13 x x = C o n s t
Let x = 0 ⟶ p . M ( 0 , − 1 , − 2 ) x=0\longrightarrow \boxed{p.M\left(0,-1,-2\right)} x = 0 ⟶ p . M ( 0 , − 1 , − 2 )
As we know, from the canonical equation of the plane we can find the coordinates of the normal vector as follows
l : a x + b y + c z + d = 0 → n l → = ( a , b , c ) l : ax+by+cz+d=0\to\overrightarrow{n_l}=\left(a,b,c\right) l : a x + b y + cz + d = 0 → n l = ( a , b , c ) ( more information : https://en.wikipedia.org/wiki/Normal_(geometry) )
In our case,
{ l 1 : − 5 x + y − 2 z = 3 → n l 1 → = ( − 5 , 1 , − 2 ) l 2 : 2 x − 3 y + 5 z = − 7 → n l 2 → = ( 2 , − 3 , 5 ) \left\{\begin{array}{l}
l_1 :-5x + y - 2z =3\to\overrightarrow{n_{l_1}}=\left(-5,1,-2\right)\\[0.3cm]
l_2 :2x - 3y + 5z =-7\to\overrightarrow{n_{l_2}}=\left(2,-3,5\right)
\end{array}\right. ⎩ ⎨ ⎧ l 1 : − 5 x + y − 2 z = 3 → n l 1 = ( − 5 , 1 , − 2 ) l 2 : 2 x − 3 y + 5 z = − 7 → n l 2 = ( 2 , − 3 , 5 )
Then,
n → = n l 1 → × n l 2 → − direction vector \overrightarrow{n}=\overrightarrow{n_{l_1}}\times\overrightarrow{n_{l_2}}-\text{direction vector}\\[0.3cm] n = n l 1 × n l 2 − direction vector
n → = ∣ i → j → k → − 5 1 − 2 2 − 3 5 ∣ = = i → ⋅ ( 5 − 6 ) − j → ⋅ ( − 25 + 4 ) + k → ⋅ ( 15 − 2 ) = = − 1 ⋅ i → + 21 ⋅ j → + 13 ⋅ k → n → = ( − 1 , 21 , 13 ) { p . M ( 0 , − 1 , − 2 ) n → = ( − 1 , 21 , 13 ) ⟶ { x = − t y = − 1 + 21 t z = − 2 + 13 t \overrightarrow{n}=\left|\begin{array}{ccc}
\overrightarrow{i}&\overrightarrow{j}&\overrightarrow{k}\\[0.3cm]
-5&1&-2\\[0.3cm]
2&-3&5
\end{array}\right|=\\[0.3cm]
=\overrightarrow{i}\cdot\left(5-6\right)-\overrightarrow{j}\cdot\left(-25+4\right)+\overrightarrow{k}\cdot\left(15-2\right)=\\[0.3cm]
=-1\cdot\overrightarrow{i}+21\cdot\overrightarrow{j}+13\cdot\overrightarrow{k}\\[0.3cm]
\boxed{\overrightarrow{n}=\left(-1,21,13\right)}\\[0.3cm]
\left\{\begin{array}{l}
p.M\left(0,-1,-2\right)\\[0.3cm]
\overrightarrow{n}=\left(-1,21,13\right)
\end{array}\right.\longrightarrow
\left\{\begin{array}{l}
x=-t\\[0.3cm]
y=-1+21t\\[0.3cm]
z=-2+13t
\end{array}\right. n = ∣ ∣ i − 5 2 j 1 − 3 k − 2 5 ∣ ∣ = = i ⋅ ( 5 − 6 ) − j ⋅ ( − 25 + 4 ) + k ⋅ ( 15 − 2 ) = = − 1 ⋅ i + 21 ⋅ j + 13 ⋅ k n = ( − 1 , 21 , 13 ) ⎩ ⎨ ⎧ p . M ( 0 , − 1 , − 2 ) n = ( − 1 , 21 , 13 ) ⟶ ⎩ ⎨ ⎧ x = − t y = − 1 + 21 t z = − 2 + 13 t
ANSWER
{ x = − t y = − 1 + 21 t z = − 2 + 13 t \left\{\begin{array}{l}
x=-t\\[0.3cm]
y=-1+21t\\[0.3cm]
z=-2+13t
\end{array}\right. ⎩ ⎨ ⎧ x = − t y = − 1 + 21 t z = − 2 + 13 t
#339914 on Dec 2023
Comments