Question

1/√2, 1/√3, 1/√5 form the direction cosines of a line.
Is the statement true? Give reason for your answer, either with a short proof or a counterexample.

Answer 1

Answer on Question #78513 – Math – Analytic Geometry

Question

1/√2, 1/√3, 1/√5 form the direction cosines of a line.

Is the statement true? Give reason for your answer, either with a short proof or a counterexample.

Solution

The direction cosine of a line is defined as the cosine of the angles between the positive directed lines and the coordinate axes. If $\alpha$, $\beta$ and $\gamma$ are the three angles between the directed line segment and the coordinate axes, then these three angles are considered as direction angles. The cosine of these directed angles, $\cos \alpha$, $\cos \beta$, and $\cos \gamma$ are termed as direction cosines of the line with general notation $l$, $m$, and $n$, respectively. That is,

$$l = \cos \alpha = \frac{x}{\sqrt{x^2 + y^2 + z^2}};$$ $$m = \cos \beta = \frac{y}{\sqrt{x^2 + y^2 + z^2}};$$ $$n = \cos \gamma = \frac{z}{\sqrt{x^2 + y^2 + z^2}}.$$

The relation between the direction cosines is as follows:

$$l^2 + m^2 + n^2 = \frac{x^2}{x^2 + y^2 + z^2} + \frac{y^2}{x^2 + y^2 + z^2} + \frac{z^2}{x^2 + y^2 + z^2} = 1.$$

So we have to check the following conditions:

a) $l, m, n \leq 1$ (because cosine is always $\leq 1$);

b) $l^2 + m^2 + n^2 = 1$.

We can see that condition a) is satisfied:

$$\frac{1}{\sqrt{2}} < 1; \quad \frac{1}{\sqrt{3}} < 1; \quad \frac{1}{\sqrt{5}} < 1.$$

But condition b) is not satisfied:

$$\left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{3}}\right)^2 + \left(\frac{1}{\sqrt{5}}\right)^2 = \frac{1}{2} + \frac{1}{3} + \frac{1}{5} = \frac{31}{30} \neq 1.$$

Therefore, this statement is not true.

**Answer**: this statement is not true.

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