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: Understand the goal.
We must find the direction cosines $\ell, m, n$ of the line $\dfrac{x+2}{2} = \dfrac{2y-5}{3};\ z = -1$.
Step 2: Make every variable have coefficient 1.
The $y$ part $2y - 5$ should look like $y$ minus a number. Factor out the 2: $2y - 5 = 2\left(y - \tfrac{5}{2}\right)$, so $\dfrac{2y-5}{3} = \dfrac{y - \frac{5}{2}}{3/2}$.
Step 3: Read off the direction ratios.
So $x$ has denominator 2, $y$ has denominator $\tfrac32$, and $z = -1$ is fixed so its ratio is 0. Direction ratios: $\left(2, \tfrac32, 0\right)$.
Step 4: Clear the fraction.
Multiply all by 2 to get whole numbers: $(4, 3, 0)$. Scaling does not change a line's direction.
Step 5: Find the magnitude.
\[ \sqrt{4^2 + 3^2 + 0^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]
Step 6: Divide to get cosines.
\[ \ell = \pm\frac{4}{5},\quad m = \pm\frac{3}{5},\quad n = 0 \] \[ \boxed{\ell = \pm\tfrac{4}{5},\ m = \pm\tfrac{3}{5},\ n = 0} \]