Question

How do you find the unit vector along the line joining point $\left( {2,4,4} \right)$ to $\left( { - 3,2,2} \right)$?

Answer 1

A vector with magnitude 1 is called unit vector, and the unit vector between points P and Q is given by the formula $\overrightarrow {PQ} = \dfrac{{\overrightarrow {PQ} }}{{\overrightarrow {\left| {PQ} \right|} }}$, where$\left| {PQ} \right|$ is the magnitude of the vector $\overrightarrow {PQ}$, and is given by the formula $\left| {\overrightarrow {PQ} } \right| = \sqrt {{P^2} + {Q^2}}$,and $\overrightarrow {PQ}$ is the vector between the points P and Q.
In the given question we are given two points, whose unit vector is to be found, first we have to find the vector along the two points and find the magnitude and by substituting the values in the vector in the unit vector formula, we will find the required unit vector in the direction of the given two points.

Complete step-by-step solution:
Given points are $\left( {2,4,4} \right)$ and $\left( { - 3,2,2} \right)$, we have to find the unit vector along the line joining these points.
Let $P = \left( {2,4,4} \right)$, and
$Q = \left( { - 3,2,2} \right)$,
Now we will find the vector along the line joining points P and Q, we get,
$\Rightarrow \overrightarrow {PQ} = \left( { - 3 - 2} \right)i + \left( {2 - 4} \right)j + \left( {2 - 4} \right)k$,
By simplifying we get,
$\Rightarrow \overrightarrow {PQ} = - 5i - 2j - 2j$,
Now we have to find the magnitude of vector $\overrightarrow {PQ} = - 5i - 2j - 2k$,
$\Rightarrow \left| {\overrightarrow {PQ} } \right| = \sqrt {{{\left( { - 5} \right)}^2} + {{\left( { - 2} \right)}^2} + {{\left( { - 2} \right)}^2}}$,
By simplifying we get,
$\Rightarrow \left| {\overrightarrow {PQ} } \right| = \sqrt {25 + 4 + 4}$,
Now simplifying we get,
$\Rightarrow \left| {\overrightarrow {PQ} } \right| = \sqrt {25 + 8}$,
So now simplifying we get,
$\Rightarrow \left| {\overrightarrow {PQ} } \right| = \sqrt {33}$,
So, the magnitude of $\left| {\overrightarrow {PQ} } \right|$ is equal to $\sqrt {33}$,
Now using the unit vector formula in the direction P and Q i.e.,$\overrightarrow {PQ} = \dfrac{{\overrightarrow {PQ} }}{{\overrightarrow {\left| {PQ} \right|} }}$, where $\left| {PQ} \right|$ is the magnitude of the vector $\overrightarrow {PQ}$,
So by substituting the values in the formula we get,
Unit vector formula in the direction P and Q $\overrightarrow {PQ}$=$\dfrac{{ - 5i - 2j - 2k}}{{\sqrt {33} }}$.
So, required unit vector is $\dfrac{{ - 5}}{{\sqrt {33} }}i - \dfrac{2}{{\sqrt {33} j}} - \dfrac{2}{{\sqrt {33} }}k$.

Unit vector along the line joining point $\left( {2,4,4} \right)$ to $\left( { - 3,2,2} \right)$ is equal to $\dfrac{{ - 5}}{{\sqrt {33} }}i - \dfrac{2}{{\sqrt {33} j}} - \dfrac{2}{{\sqrt {33} }}k$.

Note: A unit vector has the magnitude of 1. If two vectors of equal magnitude pointing in opposite directions will have sum as zero and if the magnitude of two vectors are not equal then their sum can never be zero.
$\overrightarrow i$ represent unit vector in positive $x$-axis direction,
$\overrightarrow j$ represent unit vector in positive $y$-axis direction,
$\overrightarrow k$ represent unit vectors in positive $z$-axis direction.