given the vectors u= (3,5), v= (-1,2) and w= (10,2). determine a1 and a2, and write the new coordinates of the resulting vector w, such that w= a1 u+ a2 v
Accepted Solution
A:
To express vector w as a linear combination of u and v in the form $$(w = a_1u + a_2v),$$ we need to find the values of a_1 and a_2 that satisfy this equation. The equations for this can be set up as follows:
$$\[w = a_1u + a_2v\]$$
Given vectors:
$$\[u = (3, 5)\]$$
$$\[v = (-1, 2)\]$$
$$\[w = (10, 2)\]$$
Let's set up the equations using the components of the vectors:
$$\[10 = a_1 \times 3 + a_2 \times (-1)\]$$ (for the x-component)
$$\[2 = a_1 \times 5 + a_2 \times 2\]$$ (for the y-component)
Now, we can solve this system of equations for a_1 and a_2:
Solving for a_1:
$$\[10 = 3a_1 - a_2\]$$
Solving for a_2:
$$\[2 = 5a_1 + 2a_2\]$$
Now, let's solve for a_1 and a_2:
We'll use the substitution or elimination method to solve this system of linear equations.
Multiply the second equation by 3 to eliminate a_2:
$$\[6 = 15a_1 + 6a_2\]$$
Now, subtract the first equation from the modified second equation to eliminate a_2:
$$\[6 - 10 = 15a_1 + 6a_2 - (3a_1 - a_2)\]$$
$$\[a_1 = -\frac{4}{12} = -\frac{1}{3}\]$$
Now, substitute a_1 back into the first equation to solve for a_2:
$$\[10 = 3\left(-\frac{1}{3}\right) - a_2\]$$
$$\[a_2 = -\frac{7}{3}\]$$
So, $$\(a_1 = -\frac{1}{3}\)$$ and $$\(a_2 = -\frac{7}{3}\).$$
Now, let's find the new coordinates of w using these values:
$$\[w = a_1u + a_2v\]$$
$$\[w = \left(-\frac{1}{3}\right)u - \left(\frac{7}{3}\right)v\]$$
$$\[w = \left(-\frac{1}{3}\right)(3, 5) - \left(\frac{7}{3}\right)(-1, 2)\]$$
$$\[w = \left(-1, -\frac{5}{3}\right) - \left(-\frac{7}{3}, \frac{14}{3}\right)\]$$
$$\[w = \left(-1 + \frac{7}{3}, -\frac{5}{3} - \frac{14}{3}\right)\]$$
$$\[w = \left(-\frac{4}{3}, -\frac{19}{3}\right)\]$$
So, the new coordinates of w in the form $$\(w = a_1u + a_2v\)$$ are $$\(\left(-\frac{4}{3}, -\frac{19}{3}\right)\).$$