Riddle Time!! Answer only if you know
Professor Devonshire was losing his marbles, Dr. Wallaby thought, staring at the equations filling the board in Dr. Devonshire’s office. What kind of addition was this?
1+1 = 6
2+1 = 6
1+6 = 6
2+6 = 6
6+6 = 6
1+3 = 8
7+9 = ?
I wonder what goes in the blank?
Answers
Answer:
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.
=
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
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.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.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).
Answer:
yes bro
your answer is correct
