NCERT Class 11 Mathematics — Introduction to Three Dimensional Geometry, Miscellaneous Exercise on Chapter 11. All 4 questions solved.

Three tools, each the plane version with a third coordinate added.

Midpoint of $\mathrm P(x_1,y_1,z_1)$ and $\mathrm Q(x_2,y_2,z_2)$:

$$\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2},\ \frac{z_1+z_2}{2}\right)$$

Centroid of a triangle with vertices $\mathrm A$, $\mathrm B$, $\mathrm C$:

$$\left(\frac{x_1+x_2+x_3}{3},\ \frac{y_1+y_2+y_3}{3},\ \frac{z_1+z_2+z_3}{3}\right)$$

Distance as in Exercise 11.2.

Key insight. In a parallelogram $\mathrm{ABCD}$ the two diagonals $\mathrm{AC}$ and $\mathrm{BD}$ bisect each other, so they share a midpoint. Setting the two midpoint expressions equal gives $\mathrm D = \mathrm A – \mathrm B + \mathrm C$ immediately, coordinate by coordinate — which is question 1 in a single line.

Question 1

Three vertices of a parallelogram $\mathrm{ABCD}$ are $\mathrm A(3, -1, 2)$, $\mathrm B(1, 2, -4)$ and $\mathrm C(-1, 1, 2)$. Find the coordinates of the fourth vertex.

Solution. In parallelogram $\mathrm{ABCD}$ the vertices are named in order, so the diagonals are $\mathrm{AC}$ and $\mathrm{BD}$, and they bisect each other. Writing $\mathrm D = (x, y, z)$ and equating the two midpoints:

$$\text{midpoint of } \mathrm{AC} = \left(\frac{3 + (-1)}{2},\ \frac{-1 + 1}{2},\ \frac{2 + 2}{2}\right) = (1,\ 0,\ 2)$$

$$\text{midpoint of } \mathrm{BD} = \left(\frac{1 + x}{2},\ \frac{2 + y}{2},\ \frac{-4 + z}{2}\right)$$

Equating coordinate by coordinate:

$$\frac{1+x}{2} = 1 \ \Longrightarrow\ x = 1$$ $$\frac{2+y}{2} = 0 \ \Longrightarrow\ y = -2$$ $$\frac{-4+z}{2} = 2 \ \Longrightarrow\ z = 8$$

$$\mathrm D = (1, -2, 8)$$

Equivalently $\mathrm D = \mathrm A – \mathrm B + \mathrm C = (3-1-1,\ -1-2+1,\ 2+4+2)$, which gives the same triple.

Question 2

Find the lengths of the medians of the triangle with vertices $\mathrm A(0, 0, 6)$, $\mathrm B(0, 4, 0)$ and $\mathrm C(6, 0, 0)$.

Solution. Each median joins a vertex to the midpoint of the opposite side, so find the three midpoints first.

$$\mathrm D = \text{midpoint of BC} = \left(3,\ 2,\ 0\right)$$ $$\mathrm E = \text{midpoint of AC} = \left(3,\ 0,\ 3\right)$$ $$\mathrm F = \text{midpoint of AB} = \left(0,\ 2,\ 3\right)$$

Now the three lengths.

Median AD, from $(0,0,6)$ to $(3,2,0)$:

$$\sqrt{9 + 4 + 36} = \sqrt{49} = 7$$

Median BE, from $(0,4,0)$ to $(3,0,3)$:

$$\sqrt{9 + 16 + 9} = \sqrt{34}$$

Median CF, from $(6,0,0)$ to $(0,2,3)$:

$$\sqrt{36 + 4 + 9} = \sqrt{49} = 7$$

Two medians are equal, which is consistent with $\mathrm{AB} = \mathrm{CB}$ — both equal $\sqrt{52}$ — so the triangle is isosceles about $\mathrm B$.

Question 3

If the origin is the centroid of the triangle $\mathrm{PQR}$ with vertices $\mathrm P(2a, 2, 6)$, $\mathrm Q(-4, 3b, -10)$ and $\mathrm R(8, 14, 2c)$, then find the values of $a$, $b$ and $c$.

Solution. The centroid is the average of the three vertices, and it is given to be $(0, 0, 0)$. That gives one equation per coordinate.

$x$:

$$\frac{2a + (-4) + 8}{3} = 0 \quad\Longrightarrow\quad 2a + 4 = 0 \quad\Longrightarrow\quad a = -2$$

$y$:

$$\frac{2 + 3b + 14}{3} = 0 \quad\Longrightarrow\quad 3b + 16 = 0 \quad\Longrightarrow\quad b = -\frac{16}{3}$$

$z$:

$$\frac{6 + (-10) + 2c}{3} = 0 \quad\Longrightarrow\quad 2c – 4 = 0 \quad\Longrightarrow\quad c = 2$$

The three coordinates are independent, so the three equations decouple completely — no simultaneous solving is needed.

Question 4

If $\mathrm A$ and $\mathrm B$ are the points $(3, 4, 5)$ and $(-1, 3, -7)$ respectively, find the equation of the set of points $\mathrm P$ such that $\mathrm{PA}^2 + \mathrm{PB}^2 = k^2$, where $k$ is a constant.

Solution. With $\mathrm P(x, y, z)$:

$$\mathrm{PA}^2 = (x-3)^2 + (y-4)^2 + (z-5)^2$$ $$\mathrm{PB}^2 = (x+1)^2 + (y-3)^2 + (z+7)^2$$

Adding, and expanding each square:

$$\left(x^2 – 6x + 9\right) + \left(y^2 – 8y + 16\right) + \left(z^2 – 10z + 25\right)$$ $$+ \left(x^2 + 2x + 1\right) + \left(y^2 – 6y + 9\right) + \left(z^2 + 14z + 49\right) = k^2$$

Collecting like terms:

$$2x^2 + 2y^2 + 2z^2 – 4x – 14y + 4z + 109 = k^2$$

Moving the constant across and dividing by $2$:

$$x^2 + y^2 + z^2 – 2x – 7y + 2z = \frac{k^2 – 109}{2}$$

The equation describes a sphere — completing the square would put it in the form $(x-1)^2 + \left(y – \tfrac72\right)^2 + (z+1)^2 = \text{constant}$, centred at the midpoint-related point $\left(1, \tfrac72, -1\right)$, which is precisely the midpoint of $\mathrm{AB}$.

Common mistakes

• Question 1, taking $\mathrm{AB}$ and $\mathrm{CD}$ as the diagonals. In $\mathrm{ABCD}$ the vertices are in cyclic order, so the sides are $\mathrm{AB}$, $\mathrm{BC}$, $\mathrm{CD}$, $\mathrm{DA}$ and the diagonals are $\mathrm{AC}$ and $\mathrm{BD}$. Using the wrong pair gives a different — and wrong — fourth vertex.
• Question 2, joining vertices to vertices. A median runs to the midpoint of the opposite side. Computing $\mathrm{AB}$, $\mathrm{BC}$, $\mathrm{CA}$ gives the sides, not the medians.
• Question 3, treating the unknowns as entangled. The centroid condition is three separate scalar equations, one per axis, each in a single unknown.
• Question 3, forgetting the coefficients. The $x$-coordinate of $\mathrm P$ is $2a$, not $a$, so the equation gives $2a = -4$ and $a = -2$. The same applies to $3b$ and $2c$.
• Question 4, dropping the factor of $2$. Both $\mathrm{PA}^2$ and $\mathrm{PB}^2$ contribute $x^2$, $y^2$ and $z^2$, so the sum carries $2x^2$ and so on. Dividing by $2$ at the end is what puts the sphere in standard shape.
• Expecting a plane in question 4. Equidistance (Exercise 11.2 question 4) gives a plane because the squared terms cancel; a constant sum of squares keeps them and gives a sphere.

Practise next

• Exercise 12.1 — the next chapter, Limits and Derivatives, which opens a new topic.
• Exercise 11.2 — worth revising alongside question 2, since every median length is a distance-formula computation.