Q-5
4
Days
1
1
Observe the railway schedule given below and write the answers to the following
questions.
Jamnagar Bandra Humsafar Express
Sr.
Time of
Name of station
Time of Waiting time Distance
No
arrival departure (Minutes) (Kilometers)
1 Jamnagar
20:00
0
2 Rajkot Junction 21:10 21:12
2
82
3 Surendranagar 23:26 23:28
2
198
Ahmedabad
4
01:55 02:00
5
328
Junction
Vadodara
5
03:57 04:03
5
428
Junction
6 Surat Junction 06:12 06:17
5
556
7 Bandra Junction 10:20
808
1. Time taken to reach Surendranagar from Jamnagar railway
1
2
2
2
2
station
2. Distance between Rajkot and Ahmedabad
3. Reached Bandra terminal in
days from Jamnagar.
2
Answers
Explanation:
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
2Find 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
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.