Ex 11.1 - Question 1

Question: A point is on the $x$-axis. What are its $y$-coordinate and $z$-coordinates?

Solution:
In a three-dimensional Cartesian coordinate system, any point in space is represented by an ordered triplet $(x, y, z)$.
By definition, the $x$-axis is the line of intersection of the $XY$-plane and $XZ$-plane. For any point lying directly on the $x$-axis, its perpendicular distances from the $XZ$-plane and $XY$-plane are both zero.
Therefore, the $y$-coordinate and $z$-coordinate of a point on the $x$-axis are both equal to $0$.
Hence, a point on the $x$-axis is of the form $(x, 0, 0)$.

Ex 11.1 - Question 2

Question: A point is in the $XZ$-plane. What can you say about its $y$-coordinate?

Solution:
The $XZ$-plane is formed by the $x$-axis and the $z$-axis.
Any point lying within the $XZ$-plane has a perpendicular distance of zero from this plane itself. Since the $y$-coordinate represents the perpendicular distance of a point from the $XZ$-plane (measured along the $y$-axis), a point lying strictly in the $XZ$-plane must have a $y$-coordinate of zero.
Hence, for any point in the $XZ$-plane, its $y$-coordinate is $0$, and its coordinates are of the form $(x, 0, z)$.

Ex 11.1 - Question 3

Question: Name the octants in which the following points lie: $(1, 2, 3), (4, -2, 3), (4, -2, -5), (4, 2, -5), (-4, 2, -5), (-4, 2, 5), (-3, -1, 6), (-2, -4, -7)$.

Solution:
The eight octants in three-dimensional space are determined by the signs of the coordinates $(x, y, z)$ as follows:

• Octant I: $(+, +, +)$
• Octant II: $(-, +, +)$
• Octant III: $(-, -, +)$
• Octant IV: $(+, -, +)$
• Octant V: $(+, +, -)$
• Octant VI: $(-, +, -)$
• Octant VII: $(-, -, -)$
• Octant VIII: $(+, -, -)$

1. $(1, 2, 3) \rightarrow (+, +, +) \Rightarrow$ Octant I
2. $(4, -2, 3) \rightarrow (+, -, +) \Rightarrow$ Octant IV
3. $(4, -2, -5) \rightarrow (+, -, -) \Rightarrow$ Octant VIII
4. $(4, 2, -5) \rightarrow (+, +, -) \Rightarrow$ Octant V
5. $(-4, 2, -5) \rightarrow (-, +, -) \Rightarrow$ Octant VI
6. $(-4, 2, 5) \rightarrow (-, +, +) \Rightarrow$ Octant II
7. $(-3, -1, 6) \rightarrow (-, -, +) \Rightarrow$ Octant III
8. $(-2, -4, -7) \rightarrow (-, -, -) \Rightarrow$ Octant VII

Ex 11.1 - Question 4

Question: Fill in the blanks: (i) The $x$-axis and $y$-axis taken together determine a plane known as $\underline{\hspace{2cm}}$. (ii) The coordinates of points in the $XY$-plane are of the form $\underline{\hspace{2cm}}$. (iii) Coordinate planes divide the space into $\underline{\hspace{2cm}}$ octants.

Solution:
(i) The plane determined by the $x$-axis and $y$-axis has the equation $z = 0$ and is known as the $XY$-plane.
(ii) Since any point lying in the $XY$-plane has its $z$-coordinate equal to $0$, the coordinates of points in the $XY$-plane are of the form $(x, y, 0)$.
(iii) The three coordinate planes ($XY$, $YZ$, and $ZX$ planes) intersect at right angles and divide the three-dimensional space into eight (8) octants.

Ex 11.2 - Question 1

Question: 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: The distance $d$ between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by the distance formula: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$$
(i) Distance between $(2, 3, 5)$ and $(4, 3, 1)$: $$d = \sqrt{(4 - 2)^2 + (3 - 3)^2 + (1 - 5)^2}$$ $$d = \sqrt{(2)^2 + (0)^2 + (-4)^2} = \sqrt{4 + 0 + 16} = \sqrt{20} = 2\sqrt{5}$$
(ii) Distance between $(-3, 7, 2)$ and $(2, 4, -1)$: $$d = \sqrt{(2 - (-3))^2 + (4 - 7)^2 + (-1 - 2)^2}$$ $$d = \sqrt{(5)^2 + (-3)^2 + (-3)^2} = \sqrt{25 + 9 + 9} = \sqrt{43}$$
(iii) Distance between $(-1, 3, -4)$ and $(1, -3, 4)$: $$d = \sqrt{(1 - (-1))^2 + (-3 - 3)^2 + (4 - (-4))^2}$$ $$d = \sqrt{(2)^2 + (-6)^2 + (8)^2} = \sqrt{4 + 36 + 64} = \sqrt{104} = 2\sqrt{26}$$
(iv) Distance between $(2, -1, 3)$ and $(-2, 1, 3)$: $$d = \sqrt{(-2 - 2)^2 + (1 - (-1))^2 + (3 - 3)^2}$$ $$d = \sqrt{(-4)^2 + (2)^2 + (0)^2} = \sqrt{16 + 4 + 0} = \sqrt{20} = 2\sqrt{5}$$

Ex 11.2 - Question 2

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

Solution: Let the given points be $A(-2, 3, 5)$, $B(1, 2, 3)$, and $C(7, 0, -1)$. We compute the distances between each pair of points: 1. Distance $AB$: $$AB = \sqrt{(1 - (-2))^2 + (2 - 3)^2 + (3 - 5)^2} = \sqrt{(3)^2 + (-1)^2 + (-2)^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$$ 2. Distance $BC$: $$BC = \sqrt{(7 - 1)^2 + (0 - 2)^2 + (-1 - 3)^2} = \sqrt{(6)^2 + (-2)^2 + (-4)^2} = \sqrt{36 + 4 + 16} = \sqrt{56} = 2\sqrt{14}$$ 3. Distance $AC$: $$AC = \sqrt{(7 - (-2))^2 + (0 - 3)^2 + (-1 - 5)^2} = \sqrt{(9)^2 + (-3)^2 + (-6)^2} = \sqrt{81 + 9 + 36} = \sqrt{126} = 3\sqrt{14}$$ Since $AB + BC = \sqrt{14} + 2\sqrt{14} = 3\sqrt{14} = AC$, the sum of the lengths of two line segments is equal to the length of the third segment. Therefore, the points $A$, $B$, and $C$ are collinear.

Ex 11.2 - Question 3

Question: 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) Verification for vertices of an isosceles triangle: Let $A(0, 7, -10)$, $B(1, 6, -6)$, and $C(4, 9, -6)$ be the given points. $$AB = \sqrt{(1 - 0)^2 + (6 - 7)^2 + (-6 - (-10))^2} = \sqrt{1^2 + (-1)^2 + 4^2} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3\sqrt{2}$$ $$BC = \sqrt{(4 - 1)^2 + (9 - 6)^2 + (-6 - (-6))^2} = \sqrt{3^2 + 3^2 + 0^2} = \sqrt{9 + 9 + 0} = \sqrt{18} = 3\sqrt{2}$$ $$CA = \sqrt{(0 - 4)^2 + (7 - 9)^2 + (-10 - (-6))^2} = \sqrt{(-4)^2 + (-2)^2 + (-4)^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$$ Since $AB = BC = 3\sqrt{2} \neq CA$, two sides are equal in length. Hence, $\triangle ABC$ is an isosceles triangle.
(ii) Verification for vertices of a right-angled triangle: Let $P(0, 7, 10)$, $Q(-1, 6, 6)$, and $R(-4, 9, 6)$ be the given points. $$PQ^2 = (-1 - 0)^2 + (6 - 7)^2 + (6 - 10)^2 = 1 + 1 + 16 = 18$$ $$QR^2 = (-4 - (-1))^2 + (9 - 6)^2 + (6 - 6)^2 = (-3)^2 + 3^2 + 0 = 9 + 9 = 18$$ $$PR^2 = (-4 - 0)^2 + (9 - 7)^2 + (6 - 10)^2 = (-4)^2 + 2^2 + (-4)^2 = 16 + 4 + 16 = 36$$ We observe that $PQ^2 + QR^2 = 18 + 18 = 36 = PR^2$. By the converse of Pythagoras theorem, $\triangle PQR$ is a right-angled triangle (right-angled at $Q$).
(iii) Verification for vertices of a parallelogram: Let $A(-1, 2, 1)$, $B(1, -2, 5)$, $C(4, -7, 8)$, and $D(2, -3, 4)$ be the given points. We find the lengths of all four sides: $$AB = \sqrt{(1 - (-1))^2 + (-2 - 2)^2 + (5 - 1)^2} = \sqrt{2^2 + (-4)^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$$ $$CD = \sqrt{(2 - 4)^2 + (-3 - (-7))^2 + (4 - 8)^2} = \sqrt{(-2)^2 + 4^2 + (-4)^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$$ $$BC = \sqrt{(4 - 1)^2 + (-7 - (-2))^2 + (8 - 5)^2} = \sqrt{3^2 + (-5)^2 + 3^2} = \sqrt{9 + 25 + 9} = \sqrt{43}$$ $$DA = \sqrt{(-1 - 2)^2 + (2 - (-3))^2 + (1 - 4)^2} = \sqrt{(-3)^2 + 5^2 + (-3)^2} = \sqrt{9 + 25 + 9} = \sqrt{43}$$ Since opposite sides are equal ($AB = CD = 6$ and $BC = DA = \sqrt{43}$), the quadrilateral $ABCD$ is a parallelogram.

Ex 11.2 - Question 4

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

Solution: Let $P(x, y, z)$ be any point on the required set of points. Let the given points be $A(1, 2, 3)$ and $B(3, 2, -1)$. According to the given condition, the distance of $P$ from $A$ is equal to the distance of $P$ from $B$, i.e., $$PA = PB$$ Squaring on both sides, we get: $$PA^2 = PB^2$$ Using the distance formula: $$(x - 1)^2 + (y - 2)^2 + (z - 3)^2 = (x - 3)^2 + (y - 2)^2 + (z - (-1))^2$$ Expanding both sides: $$(x^2 - 2x + 1) + (y^2 - 4y + 4) + (z^2 - 6z + 9) = (x^2 - 6x + 9) + (y^2 - 4y + 4) + (z^2 + 2z + 1)$$ Simplifying and canceling common terms ($x^2, y^2, z^2, 4, 1$) from both sides: $$-2x - 6z + 9 = -6x + 2z + 9$$ Rearranging all terms to one side: $$-2x + 6x - 6z - 2z + 9 - 9 = 0$$ $$4x - 8z = 0$$ Dividing by $4$, we get the required equation: $$x - 2z = 0$$

Ex 11.2 - Question 5

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

Solution: Let $P(x, y, z)$ be any point in the set. The given points are $A(4, 0, 0)$ and $B(-4, 0, 0)$. According to the question, the sum of the distances from $P$ to $A$ and $P$ to $B$ is $10$: $$PA + PB = 10$$ $$PA = 10 - PB$$ Using the distance formula for $PA$ and $PB$: $$\sqrt{(x - 4)^2 + y^2 + z^2} = 10 - \sqrt{(x + 4)^2 + y^2 + z^2}$$ Squaring both sides: $$(x - 4)^2 + y^2 + z^2 = \left(10 - \sqrt{(x + 4)^2 + y^2 + z^2}\right)^2$$ $$x^2 - 8x + 16 + y^2 + z^2 = 100 - 20\sqrt{(x + 4)^2 + y^2 + z^2} + (x + 4)^2 + y^2 + z^2$$ Expanding $(x+4)^2$: $$x^2 - 8x + 16 + y^2 + z^2 = 100 - 20\sqrt{(x + 4)^2 + y^2 + z^2} + x^2 + 8x + 16 + y^2 + z^2$$ Canceling common terms ($x^2, 16, y^2, z^2$) from both sides: $$-8x = 100 - 20\sqrt{(x + 4)^2 + y^2 + z^2} + 8x$$ Rearranging to isolate the square root term: $$20\sqrt{(x + 4)^2 + y^2 + z^2} = 16x + 100$$ Dividing the entire equation by $4$: $$5\sqrt{(x + 4)^2 + y^2 + z^2} = 4x + 25$$ Squaring again on both sides: $$25 \left[(x + 4)^2 + y^2 + z^2\right] = (4x + 25)^2$$ $$25(x^2 + 8x + 16 + y^2 + z^2) = 16x^2 + 200x + 625$$ $$25x^2 + 200x + 400 + 25y^2 + 25z^2 = 16x^2 + 200x + 625$$ Subtracting $16x^2 + 200x$ from both sides: $$9x^2 + 25y^2 + 25z^2 + 400 = 625$$ $$9x^2 + 25y^2 + 25z^2 = 625 - 400$$ $$9x^2 + 25y^2 + 25z^2 = 225$$ Thus, the required equation is: $$9x^2 + 25y^2 + 25z^2 - 225 = 0$$

Miscellaneous Exercise - Question 1

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

Solution: Let the coordinates of the fourth vertex $D$ be $(x, y, z)$. We are given the vertices of parallelogram $ABCD$: - $A(3, -1, 2)$ - $B(1, 2, -4)$ - $C(-1, 1, 2)$ - $D(x, y, z)$ In a parallelogram, the diagonals bisect each other. Therefore, the midpoint of diagonal $AC$ is equal to the midpoint of diagonal $BD$. Midpoint of $AC$: $$\left(\frac{3 + (-1)}{2}, \frac{-1 + 1}{2}, \frac{2 + 2}{2}\right) = \left(\frac{2}{2}, \frac{0}{2}, \frac{4}{2}\right) = (1, 0, 2)$$ Midpoint of $BD$: $$\left(\frac{1 + x}{2}, \frac{2 + y}{2}, \frac{-4 + z}{2}\right)$$ Equating the coordinates of the midpoints: $$\frac{1 + x}{2} = 1 \implies 1 + x = 2 \implies x = 1$$ $$\frac{2 + y}{2} = 0 \implies 2 + y = 0 \implies y = -2$$ $$\frac{-4 + z}{2} = 2 \implies -4 + z = 4 \implies z = 8$$ Thus, the coordinates of the fourth vertex $D$ are $(1, -2, 8)$.

Miscellaneous Exercise - Question 2

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

Solution: Let the vertices of the triangle be $A(0, 0, 6)$, $B(0, 4, 0)$, and $C(6, 0, 0)$. Let the medians from vertices $A, B,$ and $C$ be $AD, BE,$ and $CF$ respectively, where $D, E, F$ are the midpoints of sides $BC, CA,$ and $AB$. 1. **Finding the midpoint $D$ of $BC$:** $$D = \left(\frac{0 + 6}{2}, \frac{4 + 0}{2}, \frac{0 + 0}{2}\right) = (3, 2, 0)$$ Length of median $AD$ (distance between $A(0, 0, 6)$ and $D(3, 2, 0)$): $$AD = \sqrt{(3 - 0)^2 + (2 - 0)^2 + (0 - 6)^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$$ 2. **Finding the midpoint $E$ of $AC$:** $$E = \left(\frac{0 + 6}{2}, \frac{0 + 0}{2}, \frac{6 + 0}{2}\right) = (3, 0, 3)$$ Length of median $BE$ (distance between $B(0, 4, 0)$ and $E(3, 0, 3)$): $$BE = \sqrt{(3 - 0)^2 + (0 - 4)^2 + (3 - 0)^2} = \sqrt{9 + 16 + 9} = \sqrt{34}$$ 3. **Finding the midpoint $F$ of $AB$:** $$F = \left(\frac{0 + 0}{2}, \frac{0 + 4}{2}, \frac{6 + 0}{2}\right) = (0, 2, 3)$$ Length of median $CF$ (distance between $C(6, 0, 0)$ and $F(0, 2, 3)$): $$CF = \sqrt{(0 - 6)^2 + (2 - 0)^2 + (3 - 0)^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7$$ Thus, the lengths of the medians are $7, \sqrt{34},$ and $7$.

Miscellaneous Exercise - Question 3

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

Solution: Let the coordinates of the vertices of triangle $PQR$ be given as:

• $P(2a, 2, 6)$
• $Q(-4, 3b, -10)$
• $R(8, 14, 2c)$

Miscellaneous Exercise - Question 4

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

Solution:

Let the coordinates of the moving point $P$ be $(x, y, z)$.

We are given the fixed points:

• $A(3, 4, 5)$
• $B(-1, 3, -7)$

Using the distance formula, the square of the distance from $P(x, y, z)$ to $A(3, 4, 5)$ is:

Similarly, the square of the distance from $P(x, y, z)$ to $B(-1, 3, -7)$ is:

We are given the condition $PA^2 + PB^2 = k^2$. Substituting our expressions for $PA^2$ and $PB^2$:

Combining like terms:

Dividing the entire equation by 2:

This is the required equation of the set of points $P$.