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

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

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}}\`

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}\| = √(10)`

`\hat{a} = (1)/(√(10)) \cdot (3\,i-j)`

`\hat{a} = (3)/(√(10))\,i -(1)/(√(10)) \,j`

`\vec{b}\bullet \hat{a} = (2)\cdot \left((3)/(√(10)) \right)+(1)\cdot \left(-(1)/(√(10)) \right)+(-3)\cdot \left(0\right)`

`\vec b \bullet \hat{a} = (5)/(√(10))`

`\vec b_(\parallel) = (5)/(√(10))\cdot \left((3)/(√(10))\,i-(1)/(√(10))\,j \right)`

`\vec {b}_(\parallel) = (3)/(2)\,i-(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) = (3)/(2)\,i-(1)/(2)\,j`, then:

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

`\vec b _(\perp) = (1)/(2)\,i+(3)/(2)\,j-3\,k`