Answer:
The components of [tex]\vec{b}[/tex] parallel and perpendicular to [tex]\vec {a}[/tex] are [tex]\vec {b}_{\parallel} = \frac{3}{2}\,i-\frac{1}{2}\,j[/tex] and [tex]\vec b _{\perp} = \frac{1}{2}\,i+\frac{3}{2}\,j-3\,k[/tex], respectively.
Step-by-step explanation:
Let be [tex]\vec b = 2\,i+j-3\,k[/tex] and [tex]\vec a = 3\,i-j[/tex], the component of [tex]\vec b[/tex] parallel to [tex]\vec a[/tex] is calculated by the following expression:
[tex]\vec b_{\parallel} = (\vec b \bullet \hat{a}) \cdot \hat{a}[/tex]
Where [tex]\hat{a}[/tex] is the unit vector of [tex]\vec a[/tex], dimensionless and [tex]\bullet[/tex] is the operator of scalar product.
The unit vector of [tex]\vec a[/tex] is:
[tex]\hat{a} = \frac{\vec {a}}{\|\vec a\|}[/tex]
Where [tex]\|\vec {a}\|[/tex] is the norm of [tex]\vec a[/tex], whose value is determined by Pythagorean Theorem.
The component of [tex]\vec{b}[/tex] parallel to [tex]\vec {a}[/tex] is:
[tex]\|\vec {a}\| = \sqrt{3^{2}+(-1)^{2}+0^{2}}[/tex]
[tex]\|\vec {a}\| = \sqrt{10}[/tex]
[tex]\hat{a} = \frac{1}{\sqrt{10}} \cdot (3\,i-j)[/tex]
[tex]\hat{a} = \frac{3}{\sqrt{10}}\,i -\frac{1}{\sqrt{10}} \,j[/tex]
[tex]\vec{b}\bullet \hat{a} = (2)\cdot \left(\frac{3}{\sqrt{10}} \right)+(1)\cdot \left(-\frac{1}{\sqrt{10}} \right)+(-3)\cdot \left(0\right)[/tex]
[tex]\vec b \bullet \hat{a} = \frac{5}{\sqrt{10}}[/tex]
[tex]\vec b_{\parallel} = \frac{5}{\sqrt{10}}\cdot \left(\frac{3}{\sqrt{10}}\,i-\frac{1}{\sqrt{10}}\,j \right)[/tex]
[tex]\vec {b}_{\parallel} = \frac{3}{2}\,i-\frac{1}{2}\,j[/tex]
Now, the component of [tex]\vec {b}[/tex] perpendicular to [tex]\vec{a}[/tex] is found by vector subtraction:
[tex]\vec{b}_{\perp} = \vec {b}-\vec {b}_{\parallel}[/tex]
If [tex]\vec b = 2\,i+j-3\,k[/tex] and [tex]\vec {b}_{\parallel} = \frac{3}{2}\,i-\frac{1}{2}\,j[/tex], then:
[tex]\vec{b}_{\perp} = (2\,i+j-3\,k)-\left(\frac{3}{2}\,i-\frac{1}{2}\,j \right)[/tex]
[tex]\vec b _{\perp} = \frac{1}{2}\,i+\frac{3}{2}\,j-3\,k[/tex]
