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Q2:
Find the direction cosines of a line which makes equal angles with the coordinate axes.

Solution :

Step 1: Establishing the Geometric Relationship

Let the line make an angle $\alpha$ with each of the $x$, $y$, and $z$ axes. Consequently, the direction angles are $\alpha = \beta = \gamma = \alpha$.

The direction cosines of a line are defined as $l = \cos \alpha$, $m = \cos \beta$, and $n = \cos \gamma$. [Per the definition of direction cosines in 3D Euclidean space].

Step 2: Applying the Fundamental Property of Direction Cosines

For any line, the sum of the squares of its direction cosines is unity [By the fundamental identity for direction cosines]:

$l^2 + m^2 + n^2 = 1$

Substituting the condition of equal angles ($\cos \alpha = \cos \beta = \cos \gamma$):

$\cos^2 \alpha + \cos^2 \alpha + \cos^2 \alpha = 1$

$3\cos^2 \alpha = 1$

Step 3: Algebraic Derivation of Values

Solving for $\cos \alpha$:

$\cos^2 \alpha = \frac{1}{3}$

$\cos \alpha = \pm \frac{1}{\sqrt{3}}$

Since $\cos \alpha = \cos \beta = \cos \gamma$, the direction cosines are:

$(l, m, n) = \left( \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}} \right)$

X Y Z Vector (1,1,1)

Step 4: Interpretation of Results

The set of direction cosines identifies the orientation of the line in space. Given the symmetry of the problem relative to the axes, two cases emerge based on the direction of the vector:

Orientation Direction Cosines $(l, m, n)$
Case 1 $\left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right)$
Case 2 $\left( -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}} \right)$

Final Solution: The direction cosines are $\pm \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right)$.



CBSE Solutions for Class 12 Mathematics Three Dimensional Geometry


Chapters in CBSE - Class 12 Mathematics


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