A lever OA 1m long is inclined at 60 degree with horizontal. A force of 200N is acting at A vertically downwards. The moment about O
Answers
Explanation:
Appropriate Question :-
The sine of the angle between the straight line
\begin{gathered}\rm \: \dfrac{x - 2}{3} = \dfrac{y - 3}{4} = \dfrac{z - 4}{5} \: and \: plane \: 2x - 2y + z - 5 = 0 \\ \end{gathered}
3
x−2
=
4
y−3
=
5
z−4
andplane2x−2y+z−5=0
is ______
\large\underline{\sf{Solution-}}
Solution−
Given equation of line is
\begin{gathered}\rm \: \dfrac{x - 2}{3} = \dfrac{y - 3}{4} = \dfrac{z - 4}{5} \\ \end{gathered}
3
x−2
=
4
y−3
=
5
z−4
So, direction ratios of the line be (3, 4, 5).
So, in vector form, direction ratio of the line is
\begin{gathered}\rm \: \vec{b} = 3\hat{i} + 4\hat{j} + 5\hat{k} \\ \end{gathered}
b
=3
i
^
+4
j
^
+5
k
^
Also, given equation of plane is
\begin{gathered}\rm \: 2x - 2y + z - 5 = 0 \\ \end{gathered}
2x−2y+z−5=0
So, normal to the surface of plane is (2, - 2, 1).
So, normal vector to the plane is
\begin{gathered}\rm \: \vec{n} = 2\hat{i} - 2\hat{j} + \hat{k} \\ \end{gathered}
n
=2
i
^
−2
j
^
+
k
^
We know, Angle between the line and plane is given by
\begin{gathered}\boxed{ \rm{ \:sin \theta \: = \: \dfrac{\vec{b} \: . \: \vec{n}}{ |\vec{b}| \: |\vec{n}| } \: }} \\ \end{gathered}
sinθ=
∣
b
∣∣
n
∣
b
.
n
So,
\begin{gathered}\rm \: \vec{b}.\vec{n} \\ \end{gathered}
b
.
n
\begin{gathered}\rm \: = \: (3\hat{i} + 4\hat{j} + 5\hat{k}).(2\hat{i} - 2\hat{j} + \hat{k}) \\ \end{gathered}
=(3
i
^
+4
j
^
+5
k
^
).(2
i
^
−2
j
^
+
k
^
)
\begin{gathered}\rm \: = \: 6 - 8 + 5 \\ \end{gathered}
=6−8+5
\begin{gathered}\rm \: = \: 3 \\ \end{gathered}
=3
Now,
\begin{gathered}\rm \: |\vec{b}| \\ \end{gathered}
∣
b
∣
\begin{gathered}\rm \: = \: |3\hat{i} + 4\hat{j} + 5\hat{k}| \\ \end{gathered}
=∣3
i
^
+4
j
^
+5
k
^
∣
\begin{gathered}\rm \: = \: \sqrt{ {3}^{2} + {4}^{2} + {5}^{2} } \\ \end{gathered}
=
3
2
+4
2
+5
2