Question

How do you find the two unit vectors orthogonal to$a = \left( { - 1,2,1} \right){\text{ and }}b = \left( { - 4,0, - 5} \right)$?

Answer 1

Hint: Here we just need to convert these position vector in the form of vector and then we can find the cross product of the two vectors $a{\text{ and }}b$ as the cross product of two vectors means to find the vector that is perpendicular to both the vectors.

Complete step by step solution:
Whenever we are given two vectors and we need to find the vector that is orthogonal or perpendicular to both the vectors then we just need to find the cross product of the vector product of the two vectors.
When we get the cross product of these two vectors then we just need to convert that vector into the unit vector by dividing the vector with its magnitude.
So here we are given two vectors represented by $a = \left( { - 1,2,1} \right){\text{ and }}b = \left( { - 4,0, - 5} \right)$
These are the position vectors so we can write it also as:
$\overrightarrow a = - \widehat i + 2\widehat j + \widehat k$
$\overrightarrow b = - 4\widehat i - 5\widehat k$
Now we can find their cross or vector product:
$\left| {\begin{array}{*{20}{c}}
  {\widehat i}&{\widehat j}&{\widehat k} \\
  { - 1}&2&1 \\
  { - 4}&0&{ - 5}
\end{array}} \right|$
Hence we get the result as:
$\left( { - 10 - 0} \right)\widehat i - \left( {5 - \left( { - 4} \right)} \right)\widehat j + \left( {0 - \left( { - 8} \right)} \right)\widehat k$
$ - 10\widehat i - 9\widehat j + 8\widehat k$
So we can write it in the form of position vector as $\left( { - 10, - 9,8} \right)$
Now we have got just the vector that is orthogonal to both the given vectors but we need to find the vector that has the magnitude as $1$ because unit vector means a vector whose magnitude is one.
Hence we can convert this vector into a unit vector by the formula:
$\widehat a = \dfrac{{\overrightarrow a }}{{\left| {\overrightarrow a } \right|}}$
So here we have got the cross product therefore we just need to divide it just by its magnitude and we will get:
$\left| {\overrightarrow a \times \overrightarrow b } \right| = \sqrt {{{\left( { - 10} \right)}^2} + {{\left( { - 9} \right)}^2} + {8^2}} = \sqrt {245} $
And we can write this square root as:
$\sqrt {245} = \sqrt {\left( 5 \right)\left( 7 \right)\left( 7 \right)} = 7\sqrt 5 $
Hence we can write the unit vector as:
Required unit vectors can be both of plus as well as negative sign as it changes only the direction but orthogonality remains.

So required unit vectors$ = \pm \left( {\dfrac{{ - 10}}{{7\sqrt 5 }},\dfrac{{ - 9}}{{7\sqrt 5 }},\dfrac{8}{{7\sqrt 5 }}} \right)$

Note:
Here the student must know that when we are told to find the orthogonality of one vector with the other two vectors then we just need to find the cross product of the two vectors.