NCERT Class 12 Mathematics — Three Dimensional Geometry, Exercise 11.1. All 5 questions solved.

The whole exercise rests on one distinction. Direction cosines are the cosines of the angles a line makes with the positive $x$, $y$ and $z$ axes; they are written $l, m, n$ and always satisfy

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

Direction ratios are any three numbers proportional to them, written $a, b, c$. Direction ratios are not unique — $(2, 4, 6)$ and $(1, 2, 3)$ describe the same direction — but direction cosines are unique up to an overall sign, and are recovered by dividing by the magnitude:

$$l = \frac{a}{\sqrt{a^2+b^2+c^2}}, \qquad m = \frac{b}{\sqrt{a^2+b^2+c^2}}, \qquad n = \frac{c}{\sqrt{a^2+b^2+c^2}}$$

Key insight. $l^2 + m^2 + n^2 = 1$ is not just a fact to memorise — it is the check that catches almost every arithmetic slip in this chapter. Any triple you write down as direction cosines should square-and-sum to $1$ before you move on. It is what identifies the misprint in question 5’s printed answer.

Question 1

If a line makes angles $90^\circ$, $135^\circ$, $45^\circ$ with the $x$, $y$ and $z$ axes respectively, find its direction cosines.

Solution. Direction cosines are the cosines of those very angles — nothing to compute beyond evaluating them:

$$l = \cos 90^\circ = 0, \qquad m = \cos 135^\circ = -\frac{1}{\sqrt2}, \qquad n = \cos 45^\circ = \frac{1}{\sqrt2}$$

Check: $0 + \tfrac12 + \tfrac12 = 1$ ✓.

Question 2

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

Solution. Equal angles means $l = m = n$. Substituting into the identity:

$$3l^2 = 1 \quad\Longrightarrow\quad l = \pm\frac{1}{\sqrt3}$$

Both signs are genuine answers. A line has two opposite directions, and reversing the direction negates all three cosines together — which is why the $\pm$ applies to the whole triple, not to each component independently.

Question 3

If a line has the direction ratios $-18$, $12$, $-4$, then what are its direction cosines?

Solution. Divide each ratio by the magnitude of the triple:

$$\sqrt{(-18)^2 + 12^2 + (-4)^2} = \sqrt{324 + 144 + 16} = \sqrt{484} = 22$$

$$\frac{-18}{22},\ \frac{12}{22},\ \frac{-4}{22}$$

Reducing each fraction:

Question 4

Show that the points $(2, 3, 4)$, $(-1, -2, 1)$, $(5, 8, 7)$ are collinear.

Solution. Three points are collinear when the lines joining them have proportional direction ratios and share a common point.

Direction ratios of $AB$, where $A(2,3,4)$ and $B(-1,-2,1)$:

$$-1-2,\ -2-3,\ 1-4 \quad\text{i.e.}\quad -3,\ -5,\ -3$$

Direction ratios of $BC$, where $C(5,8,7)$:

$$5-(-1),\ 8-(-2),\ 7-1 \quad\text{i.e.}\quad 6,\ 10,\ 6$$

Now $6 = -2(-3)$, $10 = -2(-5)$ and $6 = -2(-3)$, so $BC = -2 \times AB$ as direction ratios: the two lines are parallel. Since $B$ lies on both, they are the same line, and $A$, $B$, $C$ are collinear.

The second half of the argument matters. Parallel direction ratios alone would only show $AB \parallel BC$; the shared point $B$ is what forces them onto one line.

Question 5

Find the direction cosines of the sides of the triangle whose vertices are $(3, 5, -4)$, $(-1, 1, 2)$ and $(-5, -5, -2)$.

Solution. Call the vertices $A(3, 5, -4)$, $B(-1, 1, 2)$ and $C(-5, -5, -2)$, and take each side in turn.

Side $AB$. Direction ratios $-1-3,\ 1-5,\ 2+4$, that is $-4, -4, 6$. The magnitude is $\sqrt{16+16+36} = \sqrt{68} = 2\sqrt{17}$, so the direction cosines are

$$\frac{-4}{2\sqrt{17}},\ \frac{-4}{2\sqrt{17}},\ \frac{6}{2\sqrt{17}} \quad=\quad \frac{-2}{\sqrt{17}},\ \frac{-2}{\sqrt{17}},\ \frac{3}{\sqrt{17}}$$

Side $BC$. Direction ratios $-5+1,\ -5-1,\ -2-2$, that is $-4, -6, -4$. The magnitude is $\sqrt{16+36+16} = \sqrt{68} = 2\sqrt{17}$ again:

$$\frac{-2}{\sqrt{17}},\ \frac{-3}{\sqrt{17}},\ \frac{-2}{\sqrt{17}}$$

Side $CA$. Direction ratios $3+5,\ 5+5,\ -4+2$, that is $8, 10, -2$. The magnitude is $\sqrt{64+100+4} = \sqrt{168} = 2\sqrt{42}$:

$$\frac{4}{\sqrt{42}},\ \frac{5}{\sqrt{42}},\ \frac{-1}{\sqrt{42}}$$

Each triple squares and sums to $1$: $\tfrac{4+4+9}{17} = 1$, $\tfrac{4+9+4}{17} = 1$ and $\tfrac{16+25+1}{42} = 1$.

A note on the printed answer. The NCERT key gives the third direction cosine of $AB$ as $\tfrac{3}{17}$, without the square root, while printing $\tfrac{-2}{\sqrt{17}}$ for the other two components of the same triple. That is a typographical slip. With $\tfrac{3}{17}$ the sum of squares comes to $\tfrac{4}{17} + \tfrac{4}{17} + \tfrac{9}{289}$, which is not $1$; with $\tfrac{3}{\sqrt{17}}$ it is exactly $1$. The value above is the correct one.

Common mistakes

• Dividing direction ratios by the wrong magnitude. The divisor is $\sqrt{a^2+b^2+c^2}$ for the triple as a whole, not the largest component or the distance between a different pair of points.
• Treating direction ratios as if they were unique. $(-4, -4, 6)$ and $(-2, -2, 3)$ are the same direction; both are valid direction ratios and both give the same direction cosines.
• Giving only one sign in question 2. A line has two directions. NCERT prints $\pm$ for this reason, and marks are lost for dropping it.
• Concluding collinearity from proportional ratios alone. Two parallel lines are not the same line. You must also point to a shared vertex, as in question 4.
• Not checking $l^2 + m^2 + n^2 = 1$. It costs five seconds and catches everything, including the misprint above.

Practise next

• Exercise 11.2 — the equations of lines built from these direction ratios, the angle between two lines, and shortest distances.
• Exercise 10.2 — the vector algebra underlying direction ratios; a direction ratio triple is just a vector along the line.