# What is the unit vector that is orthogonal to the plane containing  ( - 5 i + 4 j - 5 k)  and  (4 i + 4 j + 2 k) ?

Jan 23, 2016

There are two steps: (1) find the cross product of the vectors, (2) normalise the resultant vector. In this case, the answer is:
$\left(\frac{28}{46.7} i - \frac{10}{46.7} j - \frac{36}{46.7} k\right)$

#### Explanation:

The cross product of two vectors yields a vector that is orthogonal (at right angles) to both.

The cross product of two vectors (ai$+ b$j$+ c$k) and (pi$+ q$j$+ r$k) is given by $\left(b \cdot r - c \cdot q\right) i + \left(c \cdot p - a \cdot r\right) j + \left(a \cdot q - b \cdot p\right) k$

First step is to find the cross product:

(−5i+4j−5k) xx (4i+4j+2k) = ((4*2)-(4*-5)i + ((-5*4)-(-5*2))j + ((-5*4)-(4*4))k =((8-(-20))i+(-20-(-10)j+((-20)-16)k)=(28i-10j-36k)

This vector is orthogonal to both the original vectors, but it is not a unit vector. To make it a unit vector we need to normalise it: divide each of its components by the length of the vector.

$l = \sqrt{{28}^{2} + {\left(- 10\right)}^{2} + {\left(- 36\right)}^{2}} = 46.7$ units

The unit vector orthogonal to the original vectors is:

$\left(\frac{28}{46.7} i - \frac{10}{46.7} j - \frac{36}{46.7} k\right)$

This is one unit vector that is orthogonal to both the original vectors, but there is another - the one in the exact opposite direction. Simply changing the sign of each of the components yields a second vector orthogonal to the original vectors.

$\left(- \frac{28}{46.7} i + \frac{10}{46.7} j + \frac{36}{46.7} k\right)$

(but it's the first vector that you should offer as the answer on a test or assignment!)