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.

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

Short answer

$$\begin{gathered} v_1=-0.2i-0.4j \\ {v}_2=1.2i-0.6j \end{gathered}$$

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 localid="1646827028342" $v_2$is orthogonal to $w$, is

localid="1646827499665" $$\begin{gathered} v_1=\frac{v·w}{{\left|w\right|}^2}w \\ {v}_2=v-v_1 \end{gathered}$$

Step 2. Find the dot product and w

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

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

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

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

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

$$\begin{gathered} v_1=\frac{1}{{\left(\sqrt{5}\right)}^2}(-i-2j) \\ {v}_1=\frac{1}{5}(-i-2j) \\ {v}_1=0.2(-i-2j) \\ {v}_1=-0.2i-0.4j \end{gathered}$$

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

$$\begin{gathered} v_2=v-v_1 \\ {v}_2=i-j-(-0.2i-0.4j) \\ {v}_2=i-j+0.2i+0.4j \\ {v}_2=(1+0.2)i+(-1+0.4)j \\ {v}_2=1.2i-0.6j \end{gathered}$$