please help me guys....Instructions
1. Research the above two problems and provide accurate
information to these two young people. You may use books,
pamphlets from health clinics or the Internet to find this
information
2. You are also required to provide the names and telephone
numbers of local resources in your community that offer sexual
and reproductive health services.
3. Make a poster for display to generate awareness in your school
that
shows the problems and your advice
(10 x 2 = 20
the local resources
(40
is well planned and designed
(10
Answers
Answer:
Find the vector form as well as Cartesian form of the equation of the plane passing through the three points (2, 2, -1), (3, 4, 2) and (7, 0, 6).
Solution:-
Our plane passes through the three points A(2, 2, -1), B(3, 4, 2) and C(7, 0, 6). So we have the three following vectors that lie on our plane.
\vec{b_1}=\vec{AB}=\left < 1,\ 2,\ 3\right >
b
1
=
AB
=⟨1, 2, 3⟩
\vec{b_2}=\vec{BC}=\left < 4,\ -4,\ 4\right > =4\left < 1,\ -1,\ 1\right >
b
2
=
BC
=⟨4, −4, 4⟩=4⟨1, −1, 1⟩
\vec{b_3}=\vec{AC}=\left < 5,\ -2,\ 7\right >
b
3
=
AC
=⟨5, −2, 7⟩
To find a vector \vec{n}
n
normal to our plane, let us find the cross product of any two vectors from the above. I'm taking \vec{b_1}
b
1
and \vec{b_2}.
b
2
.
[On taking \vec{b_2}
b
2
we can ignore that 4 in it.]
So,
\longrightarrow \vec{n}=\vec{b_1}\times\vec{b_2}⟶
n
=
b
1
×
b
2
\begin{gathered}\longrightarrow\vec{n}=\left|\begin{array}{ccc}\hat i&\hat j&\hat k\\1&2&3\\1&-1&1\end{array}\right|\end{gathered}
⟶
n
=
∣
∣
∣
∣
∣
∣
∣
i
^
1
1
j
^
2
−1
k
^
3
1
∣
∣
∣
∣
∣
∣
∣
\longrightarrow\vec{n}=\left < 5,\ 2,\ -3\right >⟶
n
=⟨5, 2, −3⟩
Let (x, y, z) be a point on our plane such that the vector \left < x-7,\ y,\ z-6\right >⟨x−7, y, z−6⟩ lies on our plane but is perpendicular to \vec{n},
n
, thus,
\longrightarrow\left < x-7,\ y,\ z-6\right > \cdot\left < 5,\ 2,\ -3\right > =0⟶⟨x−7, y, z−6⟩⋅⟨5, 2, −3⟩=0
\longrightarrow 5(x-7)+2y-3(z-6)=0⟶5(x−7)+2y−3(z−6)=0
\longrightarrow\underline{\underline{5x+2y-3z-17=0}}⟶
5x+2y−3z−17=0
This is the Cartesian form of the equation of our plane.
\longrightarrow 5x+2y-3z-17=0⟶5x+2y−3z−17=0
\longrightarrow \left < x,\ y,\ z\right > \cdot\left < 5,\ 2,\ -3\right > -17=0⟶⟨x, y, z⟩⋅⟨5, 2, −3⟩−17=0
Let \vec{r}=\left < x,\ y,\ z\right > .
r
=⟨x, y, z⟩. Then,
\longrightarrow\underline{\underline{\vec{r}\cdot\left < 5,\ 2,\ -3\right > -17=0}}⟶
r
⋅⟨5, 2, −3⟩−17=0
This is the vector form of the equation of our plane.