Q1. Find the distance between the following pairs of points:
(i) (2, 3, 5) and (4, 3, 1)
(ii)(–1, 3, – 4) and (1, –3, 4)
(iii)(–3, 7, 2) and (2, 4, –1)
(iv) (2, –1, 3) and (–2, 1, 3).

Solution

(i) To find the distance between the points \( (2, 3, 5) \) and \( (4, 3, 1) \), we use the distance formula in three dimensions. Substituting the corresponding coordinates, we get

\[ \begin{aligned} d&=\sqrt{(4-2)^2+(3-3)^2+(1-5)^2}\\ &=\sqrt{2^2+0^2+(-4)^2}\\ &=\sqrt{4+16}\\ &=\sqrt{20}\\ &=2\sqrt{5} \end{aligned} \]

(ii) The distance between the points \( (-1, 3, -4) \) and \( (1, -3, 4) \) is obtained by applying the same formula. Taking the differences of the corresponding coordinates, we have

\[ \begin{aligned} d&=\sqrt{(1-(-1))^2+(-3-3)^2+(4-(-4))^2}\\ &=\sqrt{2^2+(-6)^2+8^2}\\ &=\sqrt{4+36+64}\\ &=\sqrt{104}\\ &=2\sqrt{26} \end{aligned} \]

(iii) For the points \( (-3, 7, 2) \) and \( (2, 4, -1) \), the distance is calculated by finding the squares of the differences of their respective coordinates

\[ \begin{aligned} d&=\sqrt{(2-(-3))^2+(4-7)^2+(-1-2)^2}\\ &=\sqrt{5^2+(-3)^2+(-3)^2}\\ &=\sqrt{25+9+9}\\ &=\sqrt{43} \end{aligned} \]

(iv) Finally, to find the distance between the points \( (2, -1, 3) \) and \( (-2, 1, 3) \), we again apply the distance formula. Since both points have the same z-coordinate, there is no contribution from the z-term

\[ \begin{aligned} d&=\sqrt{(-2-2)^2+(1-(-1))^2+(3-3)^2}\\ &=\sqrt{(-4)^2+2^2+0^2}\\ &=\sqrt{16+4}\\ &=2\sqrt{5} \end{aligned} \]

Q2. 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 \(C(7, 0, -1)\). To examine whether these three points are collinear, we compare the distances between each pair of points using the distance formula in three-dimensional space.

First, the distance between points \(A\) and \(B\) is calculated as

\[ \begin{aligned} AB&=\sqrt{(1-(-2))^2+(2-3)^2+(3-5)^2}\\ &=\sqrt{3^2+(-1)^2+(-2)^2}\\ &=\sqrt{9+1+4}\\ &=\sqrt{14} \end{aligned} \]

Next, the distance between points \(B\) and \(C\) is given by

\[ \begin{aligned} BC&=\sqrt{(7-1)^2+(0-2)^2+(-1-3)^2}\\ &=\sqrt{6^2+(-2)^2+(-4)^2}\\ &=\sqrt{36+4+16}\\ &=\sqrt{56}\\ &=2\sqrt{14} \end{aligned} \]

Now, the distance between points \(A\) and \(C\) is

\[ \begin{aligned} AC&=\sqrt{(7-(-2))^2+(0-3)^2+(-1-5)^2}\\ &=\sqrt{9^2+(-3)^2+(-6)^2}\\ &=\sqrt{81+9+36}\\ &=\sqrt{126}\\ &=3\sqrt{14} \end{aligned} \]

Since the distance \(AC\) is equal to the sum of the distances \(AB\) and \(BC\), that is \(AC = AB + BC\), point \(B\) lies on the line segment joining points \(A\) and \(C\). Hence, the given points are collinear.

Q3.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) Let the given points be \(A(0, 7, -10)\), \(B(1, 6, -6)\) and \(C(4, 9, -6)\). To verify whether these points form an isosceles triangle, we calculate the lengths of the sides using the distance formula.

\[ \begin{aligned} AB&=\sqrt{(1-0)^2+(6-7)^2+(-6-(-10))^2}\\ &=\sqrt{1^2+(-1)^2+4^2}\\ &=\sqrt{1+1+16}\\ &=\sqrt{18}\\ BC&=\sqrt{(4-1)^2+(9-6)^2+(-6-(-6))^2}\\ &=\sqrt{3^2+3^2+0^2}\\ &=\sqrt{9+9}\\ &=\sqrt{18}\\ AC&=\sqrt{(4-0)^2+(9-7)^2+(-6-(-10))^2}\\ &=\sqrt{16+4+16}\\ &=\sqrt{36} \end{aligned} \]

Since \(AB = BC\) and the sum of any two sides of the triangle is greater than the third side, the given points are the vertices of an isosceles triangle.

(ii) Let the points be \(A(0, 7, 10)\), \(B(-1, 6, 6)\) and \(C(-4, 9, 6)\). To verify whether the triangle is right angled, we compare the squares of the lengths of its sides.

\[ \begin{aligned} AB^2&=( -1-0)^2+(6-7)^2+(6-10)^2\\ &=1+1+16\\ &=18\\ BC^2&=(-4-(-1))^2+(9-6)^2+(6-6)^2\\ &=9+9+0\\ &=18\\ AC^2&=(-4-0)^2+(9-7)^2+(6-10)^2\\ &=16+4+16\\ &=36\\ AB^2+BC^2&=AC^2 \end{aligned} \]

Since the square of one side is equal to the sum of the squares of the other two sides, the given points are the vertices of a right angled triangle.

(iii) Let the points be \(A(-1, 2, 1)\), \(B(1, -2, 5)\), \(C(4, -7, 8)\) and \(D(2, -3, 4)\). To verify whether these points form a parallelogram, we compare the lengths of opposite sides.

\[ \begin{aligned} AB&=\sqrt{(1-(-1))^2+(-2-2)^2+(5-1)^2}\\ &=\sqrt{4+16+16}\\ &=\sqrt{36}\\ &=6\\ BC&=\sqrt{(4-1)^2+(-7-(-2))^2+(8-5)^2}\\ &=\sqrt{9+25+9}\\ &=\sqrt{43}\\ CD&=\sqrt{(2-4)^2+(-3-(-7))^2+(4-8)^2}\\ &=\sqrt{4+16+16}\\ &=\sqrt{36}\\ &=6\\ AD&=\sqrt{(2-(-1))^2+(-3-2)^2+(4-1)^2}\\ &=\sqrt{9+25+9}\\ &=\sqrt{43} \end{aligned} \]

Since opposite sides are equal in length, that is \(AB = CD\) and \(BC = AD\), the given points are the vertices of a parallelogram.

Q4. 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 a point which is equidistant from the points \((1, 2, 3)\) and \((3, 2, -1)\). Since the distances from \(P\) to these two points are equal, we equate their distance expressions using the three-dimensional distance formula.

\[ \begin{aligned} \sqrt{(x-1)^2+(y-2)^2+(z-3)^2} &=\sqrt{(x-3)^2+(y-2)^2+(z+1)^2} \end{aligned} \]

Squaring both sides to remove the square roots, we obtain

\[ \begin{aligned} (x-1)^2+(y-2)^2+(z-3)^2 &=(x-3)^2+(y-2)^2+(z+1)^2 \end{aligned} \]

Cancelling the common term \((y-2)^2\) from both sides and rearranging, we get

\[ \begin{aligned} (x-1)^2-(x-3)^2+(z-3)^2-(z+1)^2&=0\\ (x^2-2x+1)-(x^2-6x+9)+(z^2-6z+9)-(z^2+2z+1)&=0\\ 4x-8-8z+8&=0\\ 4x-8z&=0\\ x-2z&=0 \end{aligned} \]

Hence, the equation of the set of points equidistant from the given points is \(x-2z=0\).

Q5. 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 the coordinates of point \(P\) be \((x, y, z)\). The given fixed points are \(A(4, 0, 0)\) and \(B(-4, 0, 0)\). Using the distance formula, the distances of point \(P\) from \(A\) and \(B\) are expressed as follows.

\[ \begin{aligned} PA&=\sqrt{(x-4)^2+y^2+z^2}\\ PB&=\sqrt{(x+4)^2+y^2+z^2} \end{aligned} \]

According to the given condition, the sum of these distances is equal to 10, therefore

\[ \begin{aligned} \sqrt{(x-4)^2+y^2+z^2}+\sqrt{(x+4)^2+y^2+z^2}&=10 \end{aligned} \]

To simplify, we isolate one radical and square both sides

\[ \begin{aligned} \sqrt{(x-4)^2+y^2+z^2}&=10-\sqrt{(x+4)^2+y^2+z^2} \end{aligned} \]

Squaring both sides gives

\[ \begin{aligned} (x-4)^2+y^2+z^2 &=100+(x+4)^2+y^2+z^2-20\sqrt{(x+4)^2+y^2+z^2} \end{aligned} \]

Cancelling the common terms \(y^2+z^2\) from both sides and rearranging, we obtain

\[ \begin{aligned} (x-4)^2-(x+4)^2&=100-20PB\\ x^2-8x+16-x^2-8x-16&=100-20PB\\ -16x&=100-20PB\\ 20PB&=100+16x\\ PB&=5+\frac{4x}{5} \end{aligned} \]

Squaring both sides again, we get

\[ \begin{aligned} (x+4)^2+y^2+z^2&=\left(5+\frac{4x}{5}\right)^2 \end{aligned} \]

Multiplying throughout by 25 and simplifying,

\[ \begin{aligned} 25(x^2+8x+16+y^2+z^2)&=(25+4x)^2\\ 25x^2+200x+400+25y^2+25z^2&=625+200x+16x^2\\ 9x^2+25y^2+25z^2&=225 \end{aligned} \]

Hence, the required equation of the set of points is \(9x^2+25y^2+25z^2=225\).