Question

In Problems 19–24, decompose v into two vectors $v_1$and $v_2$, where $v_1$ is parallel to w and $v_2$ is orthogonal to w.

$$\begin{gathered} v=-3i+2j \\ w=2i+j \end{gathered}$$

Short answer

$$\begin{gathered} v_1=-1.6i-0.8j \\ {v}_2=-1.4i+2.8j \end{gathered}$$

Step-by-step solution

Step 1. Concept Used 

The decomposition of $v$into $v_1$and $v_2$, where $v_1$is parallel to $w$and $v_2$is orthogonal to $w$, is

role="math" localid="1646826707527" $$\begin{gathered} v_1=\frac{v·w}{{\left|w\right|}^2}w \\ {v}_2=v-v_1 \end{gathered}$$

Step 2. Find the dot product and w

For the vectors $v=-3i+2j$and $w=2i+j$the dot product is given as

$$\begin{gathered} v·w=(-3i+2j)·(2i+j) \\ v·w=(-3)·2+2·1 \\ v·w=-6+2 \\ v·w=-4 \end{gathered}$$

The magnitude of vector $w=2i+j$is given as

$$\begin{gathered} \left|w\right|=\sqrt{2^2+1^2} \\ \left|w\right|=\sqrt{4+1} \\ \left|w\right|=\sqrt{5} \end{gathered}$$

Step 3. Find v1 and v2

The vector $v_1$which is parallel to $w$is given by the formula

$v_1=\frac{v·w}{{\left|w\right|}^2}w$

On substituting the calculated values we get

$$\begin{gathered} v_1=\frac{-4}{{\left(\sqrt{5}\right)}^2}(2i+j) \\ {v}_1=\frac{-4}{5}(2i+j) \\ {v}_1=-0.8(2i+j) \\ {v}_1=-1.6i-0.8j \end{gathered}$$

and the vector $v_2$orthogonal to $w$is found as

$$\begin{gathered} v_2=v-v_1 \\ {v}_2=-3i+2j-(-1.6i-0.8j) \\ {v}_2=-3i+2j+1.6i+0.8j \\ {v}_2=(-3+1.6)i+(2+0.8)j \\ {v}_2=1.4i+2.8j \end{gathered}$$