Question

Find the value or values of m for which m (i + j + k) is a unit vector.

Answer 1

Hint: To solve this problem, we will use the basic definition of the magnitude of vector (ai + bj + ck) is $\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}$. In this case, we have the vector as m (i + j + k), thus, the magnitude would be $\sqrt{{{m}^{2}}+{{m}^{2}}+{{m}^{2}}}=\sqrt{3{{m}^{2}}}$. We would then equate this to 1 (since, the magnitude of the unit vector is 1).

Complete step-by-step answer:
To solve this problem, we start with the definition of unit vectors. A unit vector is one which has a magnitude of 1. The formula for magnitude of any vector (ai + bj + ck) is $\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}$. To give some examples of unit vectors, j is a unit vector (since, magnitude is given by $\sqrt{1}=1$). Another example would be $\dfrac{1}{\sqrt{2}}i+\dfrac{1}{\sqrt{2}}j$. Here, again would magnitude would be $\sqrt{{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}+{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}}=1$. Now, coming back to the problem in hand, we have the vector as m (i + j + k). Thus, the magnitude is given by –
$\sqrt{{{m}^{2}}+{{m}^{2}}+{{m}^{2}}}=\sqrt{3{{m}^{2}}}$
Now, the magnitude is given by 1. Thus, we have,
$\sqrt{3{{m}^{2}}}=1$
Now, we have,
$\left| m \right|\sqrt{3}=1$
Here, |m| represents the absolute value of m. That is, if m is negative, we take the additive inverse of it. But, if m is positive, we take as it is. For example, |-3| = 3 and |3| = 3. Hence, in this case,
|m| = $\dfrac{1}{\sqrt{3}}$
Thus, m = $\pm \dfrac{1}{\sqrt{3}}$ (Since, absolute values of either of these values would give $\left| m \right|=\dfrac{1}{\sqrt{3}}$.)
Hence, the values of m are $\pm \dfrac{1}{\sqrt{3}}$ for which m (i + j + k) is a unit vector.

Note: While solving algebraic equations involving the square term of a variable, it is important to include the negative solutions. Further, in general, one should be aware of the related terms to unit vectors like null vectors (magnitude of 0). Similarly, to solve problems involving these vectors (like null vectors) we equate the magnitude to 0 (since, magnitude of null vector is 0).