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 \( \boxed{\langle 5, 5, 5 \rangle} \).
Determine whether each of the following relations are reflexive, symmetric, and transitive.
Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.