Question

Find the vector \(z,\) given that \(\mathbf{u}=\langle 1,2,3\rangle\) \(\mathbf{v}=\langle 2,2,-1\rangle,\) and \(\mathbf{w}=\langle 4,0,-4\rangle\) \(2 \mathbf{z}-3 \mathbf{u}=\mathbf{w}\)

Short answer

The vector \(z\) is \(z = \langle \frac{7}{2},\frac{6}{2},- \frac{1}{2}\rangle\).

Step-by-step solution

Rearranging the equation for z

First, let's rearrange the given equation for \(z\). Starting with \(2z - 3u = w\), we add \(3u\) on both sides of the equation to isolate \(2z\). This gives us \(2z = 3u + w\). Next, we divide each side of the equation by 2 to solve for \(z\). This gives us \(z = (3u + w)/2\).

Substituting the component vectors for u, w into the equation

Now, we substitute the given vectors \(u = \langle 1,2,3 \rangle\) and \(w = \langle 4,0,-4 \rangle\) into the equation. This gives us \(z = (3 \langle 1,2,3 \rangle + \langle 4,0,-4 \rangle)/2\).

Solving for z

We perform the vector addition and scalar multiplication to solve for \(z\). This results in \(z = \langle \frac{7}{2},\frac{6}{2},- \frac{1}{2}\rangle\).