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Question: How would you find the unit vector along the line joining point \((2,4,4)\) to point \((-3,2,2)\)?...

How would you find the unit vector along the line joining point (2,4,4)(2,4,4) to point (−3,2,2)(-3,2,2)?

Explanation

Solution

A vector is an object that has both a magnitude and a direction. In this we need to find the unit vector of the line joining points. Unit vector refers to the normal vector space of length 1. First we need to represent points in vector form. Then with given points we have form position vector and then we substitute in unit vector formula. Unit vector is the ratio of the vector to the magnitude of the vector.

Formula used: Unit vector in the direction of AB→=1∣AB→∣×AB→\overrightarrow {{\text{AB}}} = \dfrac{1}{{\left| {\overrightarrow {{\text{AB}}} } \right|}} \times \overrightarrow {{\text{AB}}}
∣AB→∣=(i→)2+(j→)2+(k→)2\left| {\overrightarrow {{\text{AB}}} } \right| = \sqrt {{{(\overrightarrow i )}^2} + {{(\overrightarrow j )}^2} + {{(\overrightarrow k )}^2}}

Complete step-by-step solution:
Let us consider points A(2,4,4) and B(-3,2,2)
Now convert the points into position vector
OA→=2i→+4j→+4k→\overrightarrow {{\text{OA}}} = 2\overrightarrow i + 4\overrightarrow j + 4\overrightarrow k
OB→=−3i→+2j→+2k→\overrightarrow {{\text{OB}}} = - 3\overrightarrow i + 2\overrightarrow j + 2\overrightarrow k
Now we need to find the director of AB→\overrightarrow {{\text{AB}}} is
AB → = (−3i→+2j→+2k→)−(2i→+4j→+4k→)\overrightarrow {{\text{AB }}} {\text{ = (}} - 3\overrightarrow i + 2\overrightarrow j + 2\overrightarrow k ) - (2\overrightarrow i + 4\overrightarrow j + 4\overrightarrow k )
Now we need to subtract the position vectors.
AB → = (−3i→−2i→+2j→−4j→+2k→−4k→)\overrightarrow {{\text{AB }}} {\text{ = (}} - 3\overrightarrow i - 2\overrightarrow i + 2\overrightarrow j - 4\overrightarrow j + 2\overrightarrow k - 4\overrightarrow k )
AB → = (−5i→−2j→−2k→)\overrightarrow {{\text{AB }}} {\text{ = (}} - 5\overrightarrow i - 2\overrightarrow j - 2\overrightarrow k )
Now applying unit formula mentioned in formula used, we get
Unit vector in the direction of AB→=1∣−5i→−2j→−2k→∣×(−5i→−2j→−2k→)\overrightarrow {{\text{AB}}} = \dfrac{1}{{\left| { - 5\overrightarrow i - 2\overrightarrow j - 2\overrightarrow k } \right|}} \times {\text{(}} - 5\overrightarrow i - 2\overrightarrow j - 2\overrightarrow k )
By applying vector multiplication, we get
∣AB→∣=25+4+4\left| {\overrightarrow {{\text{AB}}} } \right| = \sqrt {25 + 4 + 4}
∣AB→∣=33\left| {\overrightarrow {{\text{AB}}} } \right| = \sqrt {33}
Now, the unit vector of direction AB→\overrightarrow {{\text{AB}}} is
Unit Vector in the direction of AB→=133×(−5i→−2j→−2k→){\text{Unit Vector in the direction of }}\overrightarrow {{\text{AB}}} = \dfrac{1}{{\sqrt {33} }} \times {\text{(}} - 5\overrightarrow i - 2\overrightarrow j - 2\overrightarrow k )
AB→=−533i→−233j→−233k→\overrightarrow {{\text{AB}}} = - \dfrac{5}{{\sqrt {33} }}\overrightarrow i - \dfrac{2}{{\sqrt {33} }}\overrightarrow j - \dfrac{2}{{\sqrt {33} }}\overrightarrow k

Thus we obtain the unit vector along the line joining point (2,4,4) to point (-3,2,2) is
AB→=−533i→−233j→−233k→\overrightarrow {{\text{AB}}} = - \dfrac{5}{{\sqrt {33} }}\overrightarrow i - \dfrac{2}{{\sqrt {33} }}\overrightarrow j - \dfrac{2}{{\sqrt {33} }}\overrightarrow k

Note: In this problem we also have vector subtraction and multiplication. Vector subtraction is the process of taking vector difference and is the inverse operation to vector addition. Here comes dot product of vector multiplication. The dot product between a unit vector and itself is also simple to compute. In this case the angle is zero and 1. Given that the vectors are all of length one, the dot products are i×i=j×j=k×k=1i \times i = j \times j = k \times k = 1. Then we solve by square and square root methods and basic mathematical calculation.