To solve the problem, we are given that a vector \( \mathbf{a} \) makes equal angles with all three coordinate axes and has a magnitude of \( 5\sqrt{3} \) units. We need to find the vector \( \mathbf{a} \).
1. Direction Cosines of the Vector:
If a vector makes equal angles with the x-, y-, and z-axes, then its direction cosines are equal. Let the common direction cosine be \( l \).
Since the sum of the squares of direction cosines equals 1:
\[
l^2 + l^2 + l^2 = 1 \Rightarrow 3l^2 = 1 \Rightarrow l^2 = \frac{1}{3} \Rightarrow l = \frac{1}{\sqrt{3}}
\]
2. Unit Vector in Direction of \( \mathbf{a} \):
The unit vector making equal angles with the axes is:
\[
\hat{\mathbf{a}} = \left\langle \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right\rangle
\]
3. Multiply by Magnitude:
We now multiply the unit vector by the magnitude \( 5\sqrt{3} \):
\[
\mathbf{a} = 5\sqrt{3} \cdot \left\langle \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right\rangle = \left\langle 5, 5, 5 \right\rangle
\]
Final Answer:
The vector \( \mathbf{a} \) is \( \boxed{\langle 5, 5, 5 \rangle} \).
A racing track is built around an elliptical ground whose equation is given by \[ 9x^2 + 16y^2 = 144 \] The width of the track is \(3\) m as shown. Based on the given information answer the following: 
(i) Express \(y\) as a function of \(x\) from the given equation of ellipse.
(ii) Integrate the function obtained in (i) with respect to \(x\).
(iii)(a) Find the area of the region enclosed within the elliptical ground excluding the track using integration.
OR
(iii)(b) Write the coordinates of the points \(P\) and \(Q\) where the outer edge of the track cuts \(x\)-axis and \(y\)-axis in first quadrant and find the area of triangle formed by points \(P,O,Q\).