Concept:
The cross product of two vectors \(\vec{a}\) and \(\vec{b}\) gives a vector perpendicular to both.
A vector of magnitude \(\lambda\) in the direction of \(\vec{n}\) is: \[ \vec{v}=\lambda\hat{n} \] where \(\hat{n}\) is the unit vector in the direction of \(\vec{n}\).
Step 1: Find a vector perpendicular to both rods
Let: \[ \vec{a}=4\hat{i}-\hat{j}+3\hat{k} \] and \[ \vec{b}=-2\hat{i}+\hat{j}-2\hat{k} \] Using the cross product: \[ \vec{n}=\vec{a}\times\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -1 & 3 \\ -2 & 1 & -2 \end{vmatrix} \] \[ =\hat{i}[(-1)(-2)-3(1)] -\hat{j}[4(-2)-3(-2)] +\hat{k}[4(1)-(-1)(-2)] \] \[ =\hat{i}(2-3)-\hat{j}(-8+6)+\hat{k}(4-2) \] \[ \vec{n}=-\hat{i}+2\hat{j}+2\hat{k} \]
Step 2: Find the unit vector
The magnitude of \(\vec{n}\) is: \[ |\vec{n}|=\sqrt{(-1)^2+2^2+2^2} \] \[ =\sqrt{1+4+4}=3 \] Therefore, the unit vector is: \[ \hat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{-\hat{i}+2\hat{j}+2\hat{k}}{3} \]
Step 3: Find the vector representing the flag-post
The magnitude of the flag-post vector is \(5\) m. Therefore: \[ \vec{F}=5\hat{n} \] \[ \vec{F} = 5\left(\frac{-\hat{i}+2\hat{j}+2\hat{k}}{3}\right) \] \[ \vec{F} = -\frac{5}{3}\hat{i} +\frac{10}{3}\hat{j} +\frac{10}{3}\hat{k} \]
Final Answer:
Therefore, the required vector is: \[ \boxed{ \vec{F} = -\frac{5}{3}\hat{i} +\frac{10}{3}\hat{j} +\frac{10}{3}\hat{k} } \] The vector in the opposite direction, \[ \frac{5}{3}\hat{i} -\frac{10}{3}\hat{j} -\frac{10}{3}\hat{k}, \] is also perpendicular to both rods and has magnitude \(5\).
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.