How do you find a unit vector that is oppositely directed to \[v=\left( 1,3,-4 \right)\]?

Hint: In this problem, we have to find the unit vector that is oppositely directed to \[v=\left( 1,3,-4 \right)\].. We can first see that, the given is in oppositely directed, where we can multiply a negative sign to the given vector. We can then write the formula for the unit vector and expand it, we can change it to oppositely directed and we can calculate the remaining part to get the answer for the unit vector.

Complete step by step solution:
We know that the given vector is,
\[v=\left( 1,3,-4 \right)\]
We can now write the opposite vector of the above vector by changing its sign, we get
\[-v=-<1,3,-4>\]…….. (1)
We can now write the formula for the opposite unit vector.
\[\dfrac{1}{\left| -\overset{\to }{\mathop{v}}\, \right|}\times \left( -\overset{\to }{\mathop{v}}\, \right)\] ………. (2)
Where,
\[\left| \overset{\to }{\mathop{v}}\, \right|=\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}\]
We can now substitute the vector (1) in the formula (2), we get
\[\Rightarrow \dfrac{1}{\sqrt{{{1}^{2}}+{{3}^{2}}+{{\left( -4 \right)}^{2}}}}\left( -<1,3,-4> \right)\]
We can now simplify the above step, we get
\[\Rightarrow -\dfrac{1}{26}<1,3,-4>\]
Therefore, a unit vector that is oppositely directed to \[v=\left( 1,3,-4 \right)\] is \[-\dfrac{1}{26}<1,3,-4>\].

Note: Students make mistakes while writing the magnitude which is in the formula, where the magnitude is the square root of the sum of the square of the given values. We should also remember that the unit vector is the magnitude multiplied by the vector. We should also note that, the given is in oppositely directed, where we can multiply a negative sign to the given vector.