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Q3:
If a line has the direction ratios –18, 12, – 4, then what are its direction cosines ?

Solution :

Step 1: Definition of Direction Ratios and Cosines

Given the direction ratios $(a, b, c)$ of a line as $-18, 12, -4$, the direction cosines $(l, m, n)$ are defined by the normalized components of the direction vector divided by its magnitude $r$.

The relations are defined as: $l = \frac{a}{r}, \quad m = \frac{b}{r}, \quad n = \frac{c}{r}$ where $r = \sqrt{a^2 + b^2 + c^2}$ [Per the definition of direction cosines in 3D Euclidean space].

Step 2: Calculation of the Magnitude ($r$)

Substitute the given values $a = -18, b = 12, c = -4$ into the magnitude formula:

$r = \sqrt{(-18)^2 + (12)^2 + (-4)^2}$

$r = \sqrt{324 + 144 + 16}$

$r = \sqrt{484}$

$r = 22$ [By the property of square roots where $\sqrt{484} = 22$].

Step 3: Determining the Direction Cosines

Since the direction cosines can be both positive and negative (depending on the direction of the line segment), we consider $r = \pm 22$.

For $r = 22$:

  • $l = \frac{-18}{22} = -\frac{9}{11}$
  • $m = \frac{12}{22} = \frac{6}{11}$
  • $n = \frac{-4}{22} = -\frac{2}{11}$

For $r = -22$:

  • $l = \frac{-18}{-22} = \frac{9}{11}$
  • $m = \frac{12}{-22} = -\frac{6}{11}$
  • $n = \frac{-4}{-22} = \frac{2}{11}$

Visual Representation of Direction Ratios

x: -18 y: 12 z: -4 Origin (0,0,0)

Verification via Identity

We verify using the identity $l^2 + m^2 + n^2 = 1$ [Fundamental property of direction cosines]:

$(-\frac{9}{11})^2 + (\frac{6}{11})^2 + (-\frac{2}{11})^2 = \frac{81}{121} + \frac{36}{121} + \frac{4}{121} = \frac{121}{121} = 1$.

Final Solution: The direction cosines are $\pm \left( -\frac{9}{11}, \frac{6}{11}, -\frac{2}{11} \right)$ or simply $\left( -\frac{9}{11}, \frac{6}{11}, -\frac{2}{11} \right)$ and $\left( \frac{9}{11}, -\frac{6}{11}, \frac{2}{11} \right)$.



CBSE Solutions for Class 11 Mathematics Introduction to Three Dimensional Geometry


Chapters in CBSE - Class 11 Mathematics


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