Question

The direction cosines $\ell, m, n$ of the line $\frac{x+2}{2}=\frac{2y-5}{3} ; z=-1$ are

• A. $\ell = \pm \frac{1}{5},\ m = 0,\ n = \pm \frac{2}{5}$
• B. $\ell = \pm \frac{3}{5},\ m = \pm \frac{4}{5},\ n = 0$
• C. $\ell = \pm \frac{4}{5},\ m = \pm \frac{3}{5},\ n = 0$
• D. $\ell = \pm \frac{1}{3},\ m = \pm \frac{1}{3},\ n = \pm \frac{1}{3}$

Hint:

Notice that $z = -1$ is a fixed constant, which means the line runs completely horizontal to the $XY$-plane.
Since the line does not move along the $Z$-axis, its third direction cosine $n$ must be 0. This immediately eliminates option (D) and leaves you with an easy choice between options with a standard 3-4-5 right triangle component ratio!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks us to determine the direction cosines ($\ell, m, n$) for a line in three-dimensional space whose algebraic equations are given.

Step 2: Key Formula or Approach:
First, convert the equation of the line into standard symmetrical form:
$$\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$$ This allows us to identify the direction ratios ($a, b, c$). Once direction ratios are found, direction cosines are determined using the normalization ratios:
$$\ell = \frac{\pm a}{\sqrt{a^2+b^2+c^2}}, \quad m = \frac{\pm b}{\sqrt{a^2+b^2+c^2}}, \quad n = \frac{\pm c}{\sqrt{a^2+b^2+c^2}}$$

Step 3: Detailed Explanation:
Let's modify the given equations to isolate $x, y,$ and $z$ with coefficients of 1:
The given equation contains $\frac{x+2}{2}$ and $2y-5$. Let's refactor the $y$-expression:
$$2y - 5 = 2\left(y - \frac{5}{2}\right) = \frac{y - \frac{5}{2}}{\frac{1}{2}}$$ So the equation becomes:
$$\frac{x+2}{2} = \frac{y-\frac{5}{2}}{\frac{3}{2}}$$ The constraint $z = -1$ can be expressed as $z - (-1) = 0$, meaning the component along the $Z$-axis does not change relative to parameter changes, so its direction ratio is $c = 0$.
Thus, our primary direction ratios are:
$$a = 2, \quad b = \frac{3}{2}, \quad c = 0$$ Multiply the components by 2 to clear fractions and establish integer direction ratios:
$$a' = 4, \quad b' = 3, \quad c' = 0$$ Calculate the normalizing square-root magnitude:
$$\sqrt{(a')^2 + (b')^2 + (c')^2} = \sqrt{4^2 + 3^2 + 0^2} = \sqrt{16 + 9 + 0} = \sqrt{25} = 5$$ Now, calculate the final direction cosines:
$$\ell = \pm \frac{4}{5}, \quad m = \pm \frac{3}{3}, \quad n = \frac{0}{5} = 0$$

Step 4: Final Answer:
The direction cosines are $\ell = \pm \frac{4}{5},\ m = \pm \frac{3}{5},\ n = 0$, which matches option (C).