We are given the following information:

• $b = (2 - 9)$
• $c = (1x)$
• $a = 7b + 7c$
• The magnitude of $a$ is $\sqrt{3577}$

Step 1: Compute $7b + 7c$

First, multiply both $b$ and $c$ by 7: $7b = 7(2 - 9) = (14 - 63)$ $7c = 7(1x) = (7x)$

Thus, $a$ is: $a = (14 - 63) + (7x) = (14 + 7, -63 + 7x)$ $a = (21, 7x - 63)$

Step 2: Find the magnitude of $a$

The magnitude of $a$ is given by: $|a| = \sqrt{(21)^2 + (7x - 63)^2}$ We are also told that the magnitude of $a$ is $\sqrt{3577}$. Therefore: $\sqrt{(21)^2 + (7x - 63)^2} = \sqrt{3577}$

Square both sides to eliminate the square roots: $(21)^2 + (7x - 63)^2 = 3577$ $441 + (7x - 63)^2 = 3577$ $(7x - 63)^2 = 3577 - 441$ $(7x - 63)^2 = 3136$

Step 3: Solve for $x$

Take the square root of both sides: $7x - 63 = \pm \sqrt{3136}$ $7x - 63 = \pm 56$

Now, solve for $x$:

Case 1: $7x - 63 = 56$

$7x = 56 + 63$ $7x = 119$ $x = \frac{119}{7} = 17$

Case 2: $7x - 63 = -56$

$7x = -56 + 63$ $7x = 7$ $x = \frac{7}{7} = 1$

Conclusion:

The possible values of $x$ are $x = 17$ and $x = 1$.

Would you like further details or clarifications?

Here are 5 related questions you might find interesting:

1. How can you interpret the direction of vector $a$ geometrically?
2. What happens if the magnitude of $a$ is changed to a different value?
3. How would the solution change if $b$ and $c$ were in three dimensions?
4. Can you derive a general formula for vectors of this type?
5. What if the components of $b$ and $c$ were parametric functions of time?

Tip: Always double-check your calculations when working with square roots and solving quadratic equations.