A standard Class 12 problem on line direction cosines, not tied to a specific exam paper.
Problem
The equation of a line is $5x-3=15y+7=3-10z$. Write the direction cosines of the line and the coordinates of a point through which it passes.
Key insight. The three expressions aren’t yet in the standard symmetric form $\dfrac{x-x_1}{l}=\dfrac{y-y_1}{m}=\dfrac{z-z_1}{n}$ — each has a different coefficient multiplying the variable. Factoring each coefficient out front and then dividing everything by their LCM converts all three into the same “per unit change” scale, revealing the true direction ratios.
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Approach
Factor each of the three expressions so the variable has coefficient $1$, then divide the whole equation by the LCM of the three original coefficients so all three fractions share the same denominator scale. The resulting numbers in the denominators are the direction ratios; normalising them by their magnitude gives the direction cosines, and the point used to zero out each numerator gives a point on the line.
Solution
Step 1 — Factor each coefficient out
$$5x-3 = 5\left(x-\frac{3}{5}\right), \qquad 15y+7 = 15\left(y+\frac{7}{15}\right), \qquad 3-10z = -10\left(z-\frac{3}{10}\right)$$
So the equation becomes:
$$5\left(x-\frac{3}{5}\right) = 15\left(y+\frac{7}{15}\right) = -10\left(z-\frac{3}{10}\right)$$
Step 2 — Divide by the LCM of the coefficients
The LCM of $5,15,10$ is $30$. Dividing each part by $30$:
$$\frac{5\left(x-\frac{3}{5}\right)}{30} = \frac{15\left(y+\frac{7}{15}\right)}{30} = \frac{-10\left(z-\frac{3}{10}\right)}{30}$$
$$\frac{x-\frac{3}{5}}{6} = \frac{y+\frac{7}{15}}{2} = \frac{z-\frac{3}{10}}{-3}$$
Step 3 — Read off the direction ratios and the point
This is now in standard symmetric form, with direction ratios $6, 2, -3$ and a point $\left(\dfrac{3}{5},-\dfrac{7}{15},\dfrac{3}{10}\right)$ on the line.
Step 4 — Normalise to get direction cosines
$$\sqrt{6^2+2^2+(-3)^2} = \sqrt{36+4+9} = \sqrt{49} = 7$$
$$\text{Direction cosines} = \left(\frac{6}{7},\frac{2}{7},-\frac{3}{7}\right)$$
Answer
Direction cosines: $\left(\dfrac{6}{7},\dfrac{2}{7},-\dfrac{3}{7}\right)$, passing through $\left(\dfrac{3}{5},-\dfrac{7}{15},\dfrac{3}{10}\right)$
Common mistakes
- Using the LCM of the wrong numbers. The LCM needed is of the original coefficients of $x$, $y$, $z$ (here $5,15,10$), not of any intermediate numbers that appear while factoring.
- Forgetting to normalise the direction ratios into direction cosines. $6,2,-3$ are direction ratios — they only become direction cosines after dividing each by the magnitude $\sqrt{6^2+2^2+(-3)^2}=7$.
Practise next
- Find the direction cosines and a point on the line $3x-1=6y+2=2-4z$, using the same factor-and-normalise approach.
Show answer
Direction cosines $\left(\tfrac{4}{\sqrt{29}},\tfrac{2}{\sqrt{29}},-\tfrac{3}{\sqrt{29}}\right)$, through $\left(\tfrac13,-\tfrac13,\tfrac12\right)$.
Factor each part so the variable has coefficient $1$: $3x-1=3\left(x-\tfrac13\right)$, $6y+2=6\left(y+\tfrac13\right)$, $2-4z=-4\left(z-\tfrac12\right)$.
Dividing through puts the line in symmetric form with direction ratios $\left(\tfrac13,\tfrac16,-\tfrac14\right)$, or $(4,2,-3)$ after scaling by $12$.
Since $|(4,2,-3)|=\sqrt{29}$, normalising gives the direction cosines above. Forgetting to factor out the coefficient first is the usual slip — the ratios are not $(3,6,-4)$.
Doubts are answered by Shiwam, usually within a day. Ask about this question specifically — a general question about the chapter is better asked in class.