NCERT Class 12 Mathematics — Three Dimensional Geometry, Miscellaneous Exercise on Chapter 11. All 5 questions solved.

Five questions, and between them they cover everything the chapter contains: the angle between two lines, the equation of a line, a perpendicularity condition solved for an unknown, the shortest distance between skew lines, and a line constructed perpendicular to two others.

Key insight. Question 5 introduces the one idea not drilled in Exercise 11.2: a line perpendicular to two given directions must be along their cross product. That is what the cross product is for — it produces the unique direction (up to scale) orthogonal to both inputs. Once you see that, the question is a cross product and a substitution.

Question 1

Find the angle between the lines whose direction ratios are $a$, $b$, $c$ and $b-c$, $c-a$, $a-b$.

Solution. Take the dot product of the two direction-ratio triples and expand:

$$a(b-c) + b(c-a) + c(a-b) = ab – ac + bc – ab + ac – bc$$

Every term cancels against another, so the dot product is $0$ regardless of what $a$, $b$ and $c$ are.

The cancellation is not a coincidence: each product appears once positive and once negative, because the second triple is built by cycling the letters.

Question 2

Find the equation of a line parallel to the $x$-axis and passing through the origin.

Solution. A line parallel to the $x$-axis has direction ratios $(1, 0, 0)$, and it passes through $(0, 0, 0)$. Substituting into the standard form:

$$\frac{x-0}{1} = \frac{y-0}{0} = \frac{z-0}{0}$$

The zero denominators look alarming but are standard notation in this chapter: they are read as $y = 0$ and $z = 0$, which together with $x$ free is exactly the $x$-axis. Nothing is being divided by zero — the symmetric form is shorthand for the parametric equations $x = \lambda$, $y = 0$, $z = 0$.

Question 3

If the lines $\dfrac{x-1}{-3} = \dfrac{y-2}{2k} = \dfrac{z-3}{2}$ and $\dfrac{x-1}{3k} = \dfrac{y-1}{1} = \dfrac{z-6}{-5}$ are perpendicular, find the value of $k$.

Solution. Both equations are already in standard form, so the direction ratios read off directly as $(-3, 2k, 2)$ and $(3k, 1, -5)$. Perpendicularity means the dot product vanishes:

$$(-3)(3k) + (2k)(1) + (2)(-5) = 0$$

$$-9k + 2k – 10 = 0 \quad\Longrightarrow\quad -7k = 10$$

Question 4

Find the shortest distance between lines $\vec r = 6\hat i + 2\hat j + 2\hat k + \lambda\left(\hat i – 2\hat j + 2\hat k\right)$ and $\vec r = -4\hat i – \hat k + \mu\left(3\hat i – 2\hat j – 2\hat k\right)$.

Solution. The directions $(1, -2, 2)$ and $(3, -2, -2)$ are not proportional, so the lines are skew:

$$\vec a_2 – \vec a_1 = (-4 – 6,\ 0 – 2,\ -1 – 2) = (-10, -2, -3)$$

$$\vec b_1 \times \vec b_2 = \begin{vmatrix}\hat i & \hat j & \hat k \\ 1 & -2 & 2 \\ 3 & -2 & -2\end{vmatrix} = (4+4)\hat i – (-2-6)\hat j + (-2+6)\hat k = (8, 8, 4)$$

$$\left|\vec b_1 \times \vec b_2\right| = \sqrt{64 + 64 + 16} = \sqrt{144} = 12$$

$$\left(\vec a_2 – \vec a_1\right) \cdot \left(\vec b_1 \times \vec b_2\right) = -80 – 16 – 12 = -108$$

$$d = \frac{|-108|}{12}$$

Note that the second line’s point is $(-4, 0, -1)$: the $\hat j$ term is absent from $-4\hat i – \hat k$, which means its coefficient is zero, not that it should be ignored.

Question 5

Find the vector equation of the line passing through the point $(1, 2, -4)$ and perpendicular to the two lines $\dfrac{x-8}{3} = \dfrac{y+19}{-16} = \dfrac{z-10}{7}$ and $\dfrac{x-15}{3} = \dfrac{y-29}{8} = \dfrac{z-5}{-5}$.

Solution. The required line must be perpendicular to both directions $(3, -16, 7)$ and $(3, 8, -5)$, so its direction is their cross product:

$$\begin{vmatrix}\hat i & \hat j & \hat k \\ 3 & -16 & 7 \\ 3 & 8 & -5\end{vmatrix}$$

Expanding:

$$\hat i\big[(-16)(-5) – (7)(8)\big] – \hat j\big[(3)(-5) – (7)(3)\big] + \hat k\big[(3)(8) – (-16)(3)\big]$$

$$= \hat i(80 – 56) – \hat j(-15 – 21) + \hat k(24 + 48) = 24\hat i + 36\hat j + 72\hat k$$

Every component is divisible by $12$, so $(2, 3, 6)$ is an equally valid direction — simpler, and the one NCERT prints. Any scalar multiple describes the same line.

Common mistakes

• Panicking at a zero denominator in question 2. $\tfrac{y}{0}$ is standard symmetric-form notation meaning $y = 0$. Writing the answer as $y = z = 0$ is equally correct, but the symmetric form is what NCERT expects.
• Missing a zero coefficient. In question 4 the point $-4\hat i – \hat k$ has $y$-coordinate $0$; reading it as a two-component vector loses a term in $\vec a_2 – \vec a_1$.
• Not simplifying the cross product in question 5. $(24, 36, 72)$ is correct but should be reduced to $(2, 3, 6)$; the marker expects the simplest direction ratios.
• Trying to find the perpendicular direction by solving simultaneous equations. It works, but the cross product does it in one step and with fewer chances to slip.
• Sign errors in the $\hat j$ cofactor. Both questions 4 and 5 turn on it. The middle term of a $3 \times 3$ expansion is subtracted.

Practise next

• Exercise 11.2 — the fifteen questions that drill each of these formulas individually.
• Exercise 10.4 — the cross product used in questions 4 and 5.