# What is the unit vector that is normal to the plane containing ( i +k) and (i+2j+2k) ?

Jun 10, 2017

$\vec{n} = \frac{2}{3} i + \frac{1}{3} j - \frac{2}{3} k$

#### Explanation:

The vector we're looking for is $\vec{n} = a \vec{i} + b \vec{j} + c \vec{k}$ where $\vec{n} \cdot \left(i + k\right) = 0$ AND $\vec{n} \cdot \left(i + 2 j + 2 k\right) = 0$, since $\vec{n}$ is perpendicular to both of those vectors.

Using this fact, we can make a system of equations:

$\vec{n} \cdot \left(i + 0 j + k\right) = 0$

$\left(a i + b j + c k\right) \left(i + 0 j + k\right) = 0$

$a + c = 0$

$\vec{n} \cdot \left(i + 2 j + 2 k\right) = 0$

$\left(a i + b j + c k\right) \cdot \left(i + 2 j + 2 k\right) = 0$

$a + 2 b + 2 c = 0$

Now we have $a + c = 0$ and $a + 2 b + 2 c = 0$, so we can say that:

$a + c = a + 2 b + 2 c$

$0 = 2 b + c$

$\therefore a + c = 2 b + c$

$a = 2 b$

$\frac{a}{2} = b$

Now we know that $b = \frac{a}{2}$ and $c = - a$. Therefore, our vector is:

$a i + \frac{a}{2} j - a k$

Finally, we need to make this a unit vector, meaning we need to divide each coefficient of the vector by its magnitude. The magnitude is:

$| \vec{n} | = \sqrt{{a}^{2} + {\left(\frac{a}{2}\right)}^{2} + {\left(- a\right)}^{2}}$

$| \vec{n} | = \sqrt{\frac{9}{4} {a}^{2}}$

$| \vec{n} | = \frac{3}{2} a$

So our unit vector is:

$\vec{n} = \frac{a}{\frac{3}{2} a} i + \frac{\frac{a}{2}}{\frac{3}{2} a} j + \frac{- a}{\frac{3}{2} a} k$

$\vec{n} = \frac{2}{3} i + \frac{1}{3} j - \frac{2}{3} k$