Given three vectors $\vec{u} = i + 2j - k$, $\vec{v} = 2i - j + 2k$, and $\vec{w} = 3i - 4j - 5k$. (1) Find the coordinates of the vector $\vec{u} + \vec{v} + \vec{w}$. Show that $\vec{w}$ is orthogonal to $\vec{u}$ and $\vec{v}$. (2) Show that $\vec{w} = \vec{u} \times \vec{v}$. Find a unit vector that is orthogonal to both $\vec{u}$ and $\vec{v}$.

1. Problem Description

Given three vectors $\vec{u} = i + 2j - k$, $\vec{v} = 2i - j + 2k$, and $\vec{w} = 3i - 4j - 5k$.

(1) Find the coordinates of the vector $\vec{u} + \vec{v} + \vec{w}$. Show that $\vec{w}$ is orthogonal to $\vec{u}$ and $\vec{v}$.

(2) Show that $\vec{w} = \vec{u} \times \vec{v}$. Find a unit vector that is orthogonal to both $\vec{u}$ and $\vec{v}$.

2. Solution Steps

(1) We first find the vector $\vec{u} + \vec{v} + \vec{w}$.

$\vec{u} + \vec{v} + \vec{w} = (i + 2j - k) + (2i - j + 2k) + (3i - 4j - 5k) = (1+2+3)i + (2-1-4)j + (-1+2-5)k = 6i - 3j - 4k$.

So, the coordinates of $\vec{u} + \vec{v} + \vec{w}$ are $(6, -3, -4)$.

Next, we show that $\vec{w}$ is orthogonal to $\vec{u}$ and $\vec{v}$. Two vectors are orthogonal if their dot product is zero.

$\vec{u} \cdot \vec{w} = (1)(3) + (2)(-4) + (-1)(-5) = 3 - 8 + 5 = 0$.

$\vec{v} \cdot \vec{w} = (2)(3) + (-1)(-4) + (2)(-5) = 6 + 4 - 10 = 0$.

Since $\vec{u} \cdot \vec{w} = 0$ and $\vec{v} \cdot \vec{w} = 0$, $\vec{w}$ is orthogonal to $\vec{u}$ and $\vec{v}$.

(2) We want to show that $\vec{w} = \vec{u} \times \vec{v}$.

$\vec{u} \times \vec{v} = \begin{vmatrix} i & j & k \\ 1 & 2 & -1 \\ 2 & -1 & 2 \end{vmatrix} = i(2(2) - (-1)(-1)) - j(1(2) - (-1)(2)) + k(1(-1) - 2(2)) = i(4 - 1) - j(2 + 2) + k(-1 - 4) = 3i - 4j - 5k = \vec{w}$.

Thus, $\vec{w} = \vec{u} \times \vec{v}$.

We want to find a unit vector that is orthogonal to both $\vec{u}$ and $\vec{v}$. The cross product $\vec{u} \times \vec{v}$ is orthogonal to both $\vec{u}$ and $\vec{v}$, so we need to find a unit vector in the direction of $\vec{w}$.

The magnitude of $\vec{w}$ is $||\vec{w}|| = \sqrt{3^2 + (-4)^2 + (-5)^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2}$.

The unit vector $\hat{w}$ is given by $\hat{w} = \frac{\vec{w}}{||\vec{w}||} = \frac{3i - 4j - 5k}{5\sqrt{2}} = \frac{3}{5\sqrt{2}}i - \frac{4}{5\sqrt{2}}j - \frac{5}{5\sqrt{2}}k = \frac{3\sqrt{2}}{10}i - \frac{4\sqrt{2}}{10}j - \frac{\sqrt{2}}{2}k$.

So, the unit vector is $(\frac{3\sqrt{2}}{10}, -\frac{4\sqrt{2}}{10}, -\frac{\sqrt{2}}{2})$.

3. Final Answer

(1) The coordinates of $\vec{u} + \vec{v} + \vec{w}$ are $(6, -3, -4)$. $\vec{w}$ is orthogonal to $\vec{u}$ and $\vec{v}$.

(2) $\vec{w} = \vec{u} \times \vec{v}$. The unit vector orthogonal to $\vec{u}$ and $\vec{v}$ is $(\frac{3\sqrt{2}}{10}, -\frac{4\sqrt{2}}{10}, -\frac{\sqrt{2}}{2})$.