Question

g find the 2 components of vector b = 2i + j - 3k, one parallel to a = 3i - j and another one perpendicular to a

Answer 1

Answer:

The components of $\vec{b}$ parallel and perpendicular to $\vec {a}$ are $\vec {b}_{\parallel} = \frac{3}{2}\,i-\frac{1}{2}\,j$ and $\vec b _{\perp} = \frac{1}{2}\,i+\frac{3}{2}\,j-3\,k$, respectively.

Step-by-step explanation:

Let be $\vec b = 2\,i+j-3\,k$ and $\vec a = 3\,i-j$, the component of $\vec b$ parallel to $\vec a$ is calculated by the following expression:

$\vec b_{\parallel} = (\vec b \bullet \hat{a}) \cdot \hat{a}$

Where $\hat{a}$ is the unit vector of $\vec a$, dimensionless and $\bullet$ is the operator of scalar product.

The unit vector of $\vec a$ is:

$\hat{a} = \frac{\vec {a}}{\|\vec a\|}$

Where $\|\vec {a}\|$ is the norm of $\vec a$, whose value is determined by Pythagorean Theorem.

The component of $\vec{b}$ parallel to $\vec {a}$ is:

$\|\vec {a}\| = \sqrt{3^{2}+(-1)^{2}+0^{2}}$

$\|\vec {a}\| = \sqrt{10}$

$\hat{a} = \frac{1}{\sqrt{10}} \cdot (3\,i-j)$

$\hat{a} = \frac{3}{\sqrt{10}}\,i -\frac{1}{\sqrt{10}} \,j$

$\vec{b}\bullet \hat{a} = (2)\cdot \left(\frac{3}{\sqrt{10}} \right)+(1)\cdot \left(-\frac{1}{\sqrt{10}} \right)+(-3)\cdot \left(0\right)$

$\vec b \bullet \hat{a} = \frac{5}{\sqrt{10}}$

$\vec b_{\parallel} = \frac{5}{\sqrt{10}}\cdot \left(\frac{3}{\sqrt{10}}\,i-\frac{1}{\sqrt{10}}\,j \right)$

$\vec {b}_{\parallel} = \frac{3}{2}\,i-\frac{1}{2}\,j$

Now, the component of $\vec {b}$ perpendicular to $\vec{a}$ is found by vector subtraction:

$\vec{b}_{\perp} = \vec {b}-\vec {b}_{\parallel}$

If $\vec b = 2\,i+j-3\,k$ and $\vec {b}_{\parallel} = \frac{3}{2}\,i-\frac{1}{2}\,j$, then:

$\vec{b}_{\perp} = (2\,i+j-3\,k)-\left(\frac{3}{2}\,i-\frac{1}{2}\,j \right)$

$\vec b _{\perp} = \frac{1}{2}\,i+\frac{3}{2}\,j-3\,k$