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Question: Find a unit vector in the direction of vector \(\overrightarrow a = 2\widehat i + 3\widehat j + 6\wi...

Find a unit vector in the direction of vector a=2i^+3j^+6k^\overrightarrow a = 2\widehat i + 3\widehat j + 6\widehat k.

Explanation

Solution

We will calculate a unit vector in the direction of the given vector by using the formula: a^=aa\widehat a = \dfrac{{\overrightarrow a }}{{\left| {\overrightarrow a } \right|}} where a^\widehat a is the unit vector in direction ofa\overrightarrow a and a\left| {\overrightarrow a } \right|is the magnitude of a\overrightarrow a . We will calculate the magnitude of a\overrightarrow a by the formula: a\left| {\overrightarrow a } \right|= x2+y2+z2\sqrt {{x^2} + {y^2} + {z^2}} when a=xi^+yj^+zk^\overrightarrow a = x\widehat i + y\widehat j + z\widehat k.

Complete step-by-step answer:
We are given a vector a=2i^+3j^+6k^\overrightarrow a = 2\widehat i + 3\widehat j + 6\widehat k.
We are required to find a unit vector in the direction of a=2i^+3j^+6k^\overrightarrow a = 2\widehat i + 3\widehat j + 6\widehat k.
We will first calculate the magnitude of a=2i^+3j^+6k^\overrightarrow a = 2\widehat i + 3\widehat j + 6\widehat kby the formula: a\left| {\overrightarrow a } \right|= x2+y2+z2\sqrt {{x^2} + {y^2} + {z^2}}
Here, x = 2, y = 3 and z = 6, substituting them in the formula of the magnitude, we get
a=22+32+62 a=4+9+36 a=49=7  \Rightarrow \left| {\overrightarrow a } \right| = \sqrt {{2^2} + {3^2} + {6^2}} \\\ \Rightarrow \left| {\overrightarrow a } \right| = \sqrt {4 + 9 + 36} \\\ \Rightarrow \left| {\overrightarrow a } \right| = \sqrt {49} = 7 \\\
Now, the relation of the unit vector is given by: a^=aa\widehat a = \dfrac{{\overrightarrow a }}{{\left| {\overrightarrow a } \right|}} where a^\widehat a is the unit vector in direction ofa\overrightarrow a and a\left| {\overrightarrow a } \right|is the magnitude of a\overrightarrow a .
Substituting the values, we get
a^=aa a^=2i^+3j^+6k^7  \Rightarrow \widehat a = \dfrac{{\overrightarrow a }}{{\left| {\overrightarrow a } \right|}} \\\ \Rightarrow \widehat a = \dfrac{{2\widehat i + 3\widehat j + 6\widehat k}}{7} \\\
We can write this equation as:
a^=27i^+37j^+67k^\Rightarrow \widehat a = \dfrac{2}{7}\widehat i + \dfrac{3}{7}\widehat j + \dfrac{6}{7}\widehat k
Therefore, the required unit vector in the direction of a=2i^+3j^+6k^\overrightarrow a = 2\widehat i + 3\widehat j + 6\widehat kis found to be: a^=27i^+37j^+67k^\widehat a = \dfrac{2}{7}\widehat i + \dfrac{3}{7}\widehat j + \dfrac{6}{7}\widehat k

Note: In this question, you may get confused in the formula used and the calculation of the magnitude of vector a. Put the correct values of the vector and its magnitude to calculate the unit vector. You can only write the condensed form as there is no compulsion to write it as a^=27i^+37j^+67k^\widehat a = \dfrac{2}{7}\widehat i + \dfrac{3}{7}\widehat j + \dfrac{6}{7}\widehat k. This is just for the simplicity of the vector components.