Three Dimensional Geometry
Exercise 11.2 Solutions
Distance Formula in 3D:
$d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2}$
$d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2}$
Q1
1. Find the distance between the following pairs of points:
(i) (2, 3, 5) and (4, 3, 1)
Calc
$\sqrt{(4-2)^2 + (3-3)^2 + (1-5)^2} = \sqrt{2^2 + 0^2 + (-4)^2} = \sqrt{4+16} = \sqrt{20}$
$2\sqrt{5}$ units
(ii) (-3, 7, 2) and (2, 4, -1)
Calc
$\sqrt{(2 – (-3))^2 + (4-7)^2 + (-1-2)^2} = \sqrt{5^2 + (-3)^2 + (-3)^2} = \sqrt{25+9+9}$
$\sqrt{43}$ units
(iii) (-1, 3, -4) and (1, -3, 4)
Calc
$\sqrt{(1 – (-1))^2 + (-3-3)^2 + (4 – (-4))^2} = \sqrt{2^2 + (-6)^2 + 8^2} = \sqrt{4+36+64}$
$\sqrt{104} = 2\sqrt{26}$ units
(iv) (2, -1, 3) and (-2, 1, 3)
Calc
$\sqrt{(-2-2)^2 + (1 – (-1))^2 + (3-3)^2} = \sqrt{(-4)^2 + 2^2 + 0} = \sqrt{16+4}$
$\sqrt{20} = 2\sqrt{5}$ units
Q2
2. Show that the points (-2, 3, 5), (1, 2, 3) and (7, 0, -1) are collinear.
Points
Let $A(-2,3,5)$, $B(1,2,3)$, $C(7,0,-1)$.
Dist AB
$\sqrt{(1+2)^2 + (2-3)^2 + (3-5)^2} = \sqrt{9+1+4} = \sqrt{14}$.
Dist BC
$\sqrt{(7-1)^2 + (0-2)^2 + (-1-3)^2} = \sqrt{36+4+16} = \sqrt{56} = 2\sqrt{14}$.
Dist AC
$\sqrt{(7+2)^2 + (0-3)^2 + (-1-5)^2} = \sqrt{81+9+36} = \sqrt{126} = 3\sqrt{14}$.
Check
$AB + BC = \sqrt{14} + 2\sqrt{14} = 3\sqrt{14} = AC$.
Since $AB + BC = AC$, points are collinear.
Q3
3. Verify the following:
(i) Isosceles Triangle: (0, 7, -10), (1, 6, -6), (4, 9, -6)
AB
$\sqrt{(1-0)^2 + (6-7)^2 + (-6+10)^2} = \sqrt{1+1+16} = \sqrt{18}$.
BC
$\sqrt{(4-1)^2 + (9-6)^2 + (-6+6)^2} = \sqrt{9+9+0} = \sqrt{18}$.
AC
$\sqrt{(4-0)^2 + (9-7)^2 + (-6+10)^2} = \sqrt{16+4+16} = \sqrt{36} = 6$.
Since $AB = BC$, it is an isosceles triangle.
(ii) Right Angled Triangle: (0, 7, 10), (-1, 6, 6), (-4, 9, 6)
AB²
$(-1-0)^2 + (6-7)^2 + (6-10)^2 = 1 + 1 + 16 = 18$.
BC²
$(-4+1)^2 + (9-6)^2 + (6-6)^2 = 9 + 9 + 0 = 18$.
AC²
$(-4-0)^2 + (9-7)^2 + (6-10)^2 = 16 + 4 + 16 = 36$.
Since $AB^2 + BC^2 = 18 + 18 = 36 = AC^2$, it is a right angled triangle.
(iii) Parallelogram: (-1, 2, 1), (1, -2, 5), (4, -7, 8), (2, -3, 4)
AB
$\sqrt{(1+1)^2 + (-2-2)^2 + (5-1)^2} = \sqrt{4+16+16} = 6$.
CD
$\sqrt{(2-4)^2 + (-3+7)^2 + (4-8)^2} = \sqrt{4+16+16} = 6$.
BC
$\sqrt{(4-1)^2 + (-7+2)^2 + (8-5)^2} = \sqrt{9+25+9} = \sqrt{43}$.
DA
$\sqrt{(-1-2)^2 + (2+3)^2 + (1-4)^2} = \sqrt{9+25+9} = \sqrt{43}$.
Opposite sides are equal ($AB=CD, BC=DA$). Hence, parallelogram.
Q4
4. Find the equation of set of points equidistant from (1, 2, 3) and (3, 2, -1).
Let
Let point be $P(x, y, z)$. Given $PA = PB \Rightarrow PA^2 = PB^2$.
Equation
$(x-1)^2 + (y-2)^2 + (z-3)^2 = (x-3)^2 + (y-2)^2 + (z+1)^2$
Expand
$x^2-2x+1 + z^2-6z+9 = x^2-6x+9 + z^2+2z+1$ (Cancelled y-terms).
Simplify
$-2x – 6z + 10 = -6x + 2z + 10$
Result
$4x – 8z = 0 \Rightarrow x – 2z = 0$
$x – 2z = 0$
Q5
5. Find equation of points P sum of whose distances from A(4,0,0) and B(-4,0,0) is 10.
Given
$PA + PB = 10 \Rightarrow \sqrt{(x-4)^2+y^2+z^2} + \sqrt{(x+4)^2+y^2+z^2} = 10$.
Rearrange
$\sqrt{(x-4)^2+y^2+z^2} = 10 – \sqrt{(x+4)^2+y^2+z^2}$.
Square
$(x-4)^2+y^2+z^2 = 100 + (x+4)^2+y^2+z^2 – 20\sqrt{(x+4)^2+y^2+z^2}$.
Simplify
$x^2-8x+16 = 100 + x^2+8x+16 – 20\sqrt{\dots} \Rightarrow -16x – 100 = -20\sqrt{\dots}$.
Divide
Divide by -4: $4x + 25 = 5\sqrt{(x+4)^2+y^2+z^2}$.
Square
$(4x+25)^2 = 25[(x+4)^2+y^2+z^2]$
Expand
$16x^2 + 200x + 625 = 25[x^2+8x+16+y^2+z^2]$
$16x^2 + 200x + 625 = 25x^2 + 200x + 400 + 25y^2 + 25z^2$.
$16x^2 + 200x + 625 = 25x^2 + 200x + 400 + 25y^2 + 25z^2$.
$9x^2 + 25y^2 + 25z^2 – 225 = 0$