NCERT Class 11 Mathematics — Introduction to Three Dimensional Geometry, Exercise 11.2. All 5 questions solved.

The plane distance formula extends to space with one extra term:

$$\mathrm{PQ} = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$$

Everything in this exercise follows from it:

• Collinear points: the longest of the three distances equals the sum of the other two.
• Isosceles triangle: two sides equal. Right-angled: Pythagoras holds for some labelling of the sides.
• Parallelogram: opposite sides equal in length.

Key insight. Compare squared distances wherever possible. In question 3(ii) the sides are $\sqrt{18}$, $\sqrt{18}$ and $6$, and the Pythagoras check $18 + 18 = 36$ is exact arithmetic — whereas $4.24 + 4.24$ against $6$ tells you nothing. Squaring removes the radicals and turns every test into whole numbers.

Question 1

Find the distance between the following pairs of points:

(i) $(2, 3, 5)$ and $(4, 3, 1)$    (ii) $(-3, 7, 2)$ and $(2, 4, -1)$ (iii) $(-1, 3, -4)$ and $(1, -3, 4)$    (iv) $(2, -1, 3)$ and $(-2, 1, 3)$

Solution.

(i) Differences $2$, $0$, $-4$:

$$\sqrt{4 + 0 + 16} = \sqrt{20} = 2\sqrt5$$

(ii) Differences $5$, $-3$, $-3$:

$$\sqrt{25 + 9 + 9} = \sqrt{43}$$

(iii) Differences $2$, $-6$, $8$:

$$\sqrt{4 + 36 + 64} = \sqrt{104} = 2\sqrt{26}$$

(iv) Differences $-4$, $2$, $0$:

$$\sqrt{16 + 4 + 0} = \sqrt{20} = 2\sqrt5$$

Question 2

Show that the points $(-2, 3, 5)$, $(1, 2, 3)$ and $(7, 0, -1)$ are collinear.

Solution. Label them $\mathrm A$, $\mathrm B$, $\mathrm C$ and compute all three distances.

$$\mathrm{AB} = \sqrt{3^2 + (-1)^2 + (-2)^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$$

$$\mathrm{BC} = \sqrt{6^2 + (-2)^2 + (-4)^2} = \sqrt{36 + 4 + 16} = \sqrt{56} = 2\sqrt{14}$$

$$\mathrm{AC} = \sqrt{9^2 + (-3)^2 + (-6)^2} = \sqrt{81 + 9 + 36} = \sqrt{126} = 3\sqrt{14}$$

Now

$$\mathrm{AB} + \mathrm{BC} = \sqrt{14} + 2\sqrt{14} = 3\sqrt{14} = \mathrm{AC}$$

Three points with one distance equal to the sum of the other two must lie on a single line — no triangle can have a side equal to the sum of the other two. Hence the points are collinear, with $\mathrm B$ lying between $\mathrm A$ and $\mathrm C$. $\blacksquare$

Question 3

Verify the following:

(i) $(0, 7, -10)$, $(1, 6, -6)$ and $(4, 9, -6)$ are the vertices of an isosceles triangle. (ii) $(0, 7, 10)$, $(-1, 6, 6)$ and $(-4, 9, 6)$ are the vertices of a right-angled triangle. (iii) $(-1, 2, 1)$, $(1, -2, 5)$, $(4, -7, 8)$ and $(2, -3, 4)$ are the vertices of a parallelogram.

Solution.

(i) With $\mathrm A(0,7,-10)$, $\mathrm B(1,6,-6)$, $\mathrm C(4,9,-6)$:

$$\mathrm{AB} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3\sqrt2$$ $$\mathrm{BC} = \sqrt{9 + 9 + 0} = \sqrt{18} = 3\sqrt2$$ $$\mathrm{AC} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$$

Two sides are equal, $\mathrm{AB} = \mathrm{BC} = 3\sqrt2$, so the triangle is isosceles. ✓

Worth noticing in passing: $\mathrm{AB}^2 + \mathrm{BC}^2 = 18 + 18 = 36 = \mathrm{AC}^2$, so this triangle is right-angled at $\mathrm B$ as well — it is an isosceles right-angled triangle. The question asks only for isosceles, and that is established.

(ii) With $\mathrm A(0,7,10)$, $\mathrm B(-1,6,6)$, $\mathrm C(-4,9,6)$:

$$\mathrm{AB}^2 = 1 + 1 + 16 = 18$$ $$\mathrm{BC}^2 = 9 + 9 + 0 = 18$$ $$\mathrm{AC}^2 = 16 + 4 + 16 = 36$$

Checking Pythagoras:

$$\mathrm{AB}^2 + \mathrm{BC}^2 = 18 + 18 = 36 = \mathrm{AC}^2$$

So the triangle is right-angled, with the right angle at $\mathrm B$. ✓

(iii) With $\mathrm A(-1,2,1)$, $\mathrm B(1,-2,5)$, $\mathrm C(4,-7,8)$, $\mathrm D(2,-3,4)$, compute the four sides in order:

$$\mathrm{AB} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$$ $$\mathrm{BC} = \sqrt{9 + 25 + 9} = \sqrt{43}$$ $$\mathrm{CD} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$$ $$\mathrm{DA} = \sqrt{9 + 25 + 9} = \sqrt{43}$$

Opposite sides are equal — $\mathrm{AB} = \mathrm{CD}$ and $\mathrm{BC} = \mathrm{DA}$ — so $\mathrm{ABCD}$ is a parallelogram. ✓

The diagonals are $\mathrm{AC} = \sqrt{155}$ and $\mathrm{BD} = \sqrt3$, which are unequal, so it is not a rectangle. As a check, the parallelogram law holds: $155 + 3 = 158 = 2(36 + 43)$ ✓

Question 4

Find the equation of the set of points which are equidistant from the points $(1, 2, 3)$ and $(3, 2, -1)$.

Solution. Let the point be $\mathrm P(x, y, z)$. Equating the squared distances avoids radicals entirely:

$$(x-1)^2 + (y-2)^2 + (z-3)^2 = (x-3)^2 + (y-2)^2 + (z+1)^2$$

The $(y-2)^2$ terms are identical and cancel. Expanding the rest:

$$x^2 – 2x + 1 + z^2 – 6z + 9 = x^2 – 6x + 9 + z^2 + 2z + 1$$

The $x^2$ and $z^2$ terms cancel as well, leaving a linear equation:

$$-2x – 6z = -6x + 2z \quad\Longrightarrow\quad 4x – 8z = 0 \quad\Longrightarrow\quad x – 2z = 0$$

The set is a plane — the perpendicular bisector plane of the segment joining the two points. Note that $y$ has disappeared entirely, which says the plane is parallel to the $y$-axis.

Question 5

Find the equation of the set of points $\mathrm P$, the sum of whose distances from $\mathrm A(4, 0, 0)$ and $\mathrm B(-4, 0, 0)$ is equal to $10$.

Solution. With $\mathrm P(x, y, z)$, the condition is

$$\sqrt{(x-4)^2 + y^2 + z^2} + \sqrt{(x+4)^2 + y^2 + z^2} = 10$$

Isolate one radical and square:

$$\sqrt{(x-4)^2 + y^2 + z^2} = 10 – \sqrt{(x+4)^2 + y^2 + z^2}$$

$$(x-4)^2 + y^2 + z^2 = 100 – 20\sqrt{(x+4)^2 + y^2 + z^2} + (x+4)^2 + y^2 + z^2$$

Everything except the $x$ terms cancels:

$$x^2 – 8x = 100 – 20\sqrt{\cdots} + x^2 + 8x$$

$$20\sqrt{(x+4)^2 + y^2 + z^2} = 100 + 16x$$

Divide by $4$ and square again:

$$5\sqrt{(x+4)^2 + y^2 + z^2} = 25 + 4x$$

$$25\left[(x+4)^2 + y^2 + z^2\right] = 625 + 200x + 16x^2$$

$$25x^2 + 200x + 400 + 25y^2 + 25z^2 = 625 + 200x + 16x^2$$

The $200x$ terms cancel:

$$9x^2 + 25y^2 + 25z^2 = 225$$

$$9x^2 + 25y^2 + 25z^2 – 225 = 0$$

This is an ellipsoid of revolution — the three-dimensional analogue of the ellipse in Exercise 10.3 question 7, with the same defining property of a constant distance sum.

Common mistakes

• Forgetting the third term. The space distance formula has three squared differences; dropping the $z$ term silently returns the projection’s length instead.
• Question 2, testing collinearity with slopes. Slope is a plane notion. In space, compare the three distances and check that the largest equals the sum of the other two.
• Question 3(ii), checking Pythagoras against the wrong side. The hypotenuse is the longest side, so it is $\mathrm{AC}$ whose square must equal the sum. Test the largest against the other two, not an arbitrary pairing.
• Question 3(iii), taking the vertices out of order. $\mathrm{AB}$, $\mathrm{BC}$, $\mathrm{CD}$, $\mathrm{DA}$ are the sides; $\mathrm{AC}$ and $\mathrm{BD}$ are the diagonals. Pairing $\mathrm{AB}$ with $\mathrm{BC}$ as “opposite” gives a false negative.
• Questions 4 and 5, squaring both sides only once. Question 5 has two radicals, so one must be isolated and the squaring done twice. Squaring the sum directly leaves a cross term with a radical still in it.
• Question 4, expecting a curved surface. All the squared terms cancel, so the locus is a plane — which is what “equidistant from two points” always gives.

Practise next

• Miscellaneous Exercise on Chapter 11 — parallelograms, medians and centroids in space, all built on the same distance formula.
• Exercise 10.3 — worth revisiting alongside question 5, which is the ellipse’s defining property carried into three dimensions.