Chapter 11: Introduction to Three Dimensional Geometry

Ans: (i) $(2, 3, 5)$ and $(4, 3, 1)$:

We use the distance formula between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$:

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

$d = \sqrt{(4 - 2)^2 + (3 - 3)^2 + (1 - 5)^2} = \sqrt{(2)^2 + (0)^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$

(ii) $(-3, 7, 2)$ and $(2, 4, -1)$:

$d = \sqrt{(2 + 3)^2 + (4 - 7)^2 + (-1 - 2)^2} = \sqrt{(5)^2 + (-3)^2 + (-3)^2} = \sqrt{25 + 9 + 9} = \sqrt{43}$

(iii) $(-1, 3, -4)$ and $(1, -3, 4)$:

$d = \sqrt{(1 + 1)^2 + (-3 - 3)^2 + (4 + 4)^2} = \sqrt{(2)^2 + (-6)^2 + (8)^2} = \sqrt{4 + 36 + 64} = \sqrt{104} = 2\sqrt{26}$

(iv) $(2, -1, 3)$ and $(-2, 1, 3)$:

$d = \sqrt{(-2 - 2)^2 + (1 + 1)^2 + (3 - 3)^2} = \sqrt{(-4)^2 + (2)^2 + (0)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$

Ans: To prove collinearity, we check if the distances between the points satisfy $AB + BC = AC$. 

- Distance between $A(-2, 3, 5)$ and $B(1, 2, 3)$:

$AB = \sqrt{(1 + 2)^2 + (2 - 3)^2 + (3 - 5)^2} = \sqrt{(3)^2 + (-1)^2 + (-2)^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$

- Distance between $B(1, 2, 3)$ and $C(7, 0, -1)$:

$BC = \sqrt{(7 - 1)^2 + (0 - 2)^2 + (-1 - 3)^2} = \sqrt{(6)^2 + (-2)^2 + (-4)^2} = \sqrt{36 + 4 + 16} = \sqrt{56}$

- Distance between $A(-2, 3, 5)$ and $C(7, 0, -1)$:

$AC = \sqrt{(7 + 2)^2 + (0 - 3)^2 + (-1 - 5)^2} = \sqrt{(9)^2 + (-3)^2 + (-6)^2} = \sqrt{81 + 9 + 36} = \sqrt{126}$

Since $AB + BC \neq AC$, the points are not collinear.

Ans:

We check if two sides of the triangle are equal:

- Distance between $(0, 7, -10)$ and $(1, 6, -6)$:

$d = \sqrt{(1 - 0)^2 + (6 - 7)^2 + (-6 + 10)^2} = \sqrt{1 + 1 + 16} = \sqrt{18}$

- Distance between $(1, 6, -6)$ and $(4, 9, -6)$:

$d = \sqrt{(4 - 1)^2 + (9 - 6)^2 + (-6 + 6)^2} = \sqrt{9 + 9 + 0} = \sqrt{18}$

- Distance between $(0, 7, -10)$ and $(4, 9, -6)$:

$d = \sqrt{(4 - 0)^2 + (9 - 7)^2 + (-6 + 10)^2} = \sqrt{16 + 4 + 16} = \sqrt{36}$

Since two sides are equal, the triangle is isosceles.

Ans: To verify if the triangle is right-angled, we check if the square of the longest side equals the sum of the squares of the other two sides.

- Distance between $(0, 7, 10)$ and $(-1, 6, 6)$:

$d_1 = \sqrt{(-1 - 0)^2 + (6 - 7)^2 + (6 - 10)^2} = \sqrt{(-1)^2 + (-1)^2 + (-4)^2} = \sqrt{1 + 1 + 16} = \sqrt{18}$

- Distance between $(-1, 6, 6)$ and $(4, 9, 6)$:

$d_2 = \sqrt{(4 + 1)^2 + (9 - 6)^2 + (6 - 6)^2} = \sqrt{(5)^2 + (3)^2 + (0)^2} = \sqrt{25 + 9 + 0} = \sqrt{34}$

- Distance between $(0, 7, 10)$ and $(4, 9, 6)$:

$d_3 = \sqrt{(4 - 0)^2 + (9 - 7)^2 + (6 - 10)^2} = \sqrt{(4)^2 + (2)^2 + (-4)^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$

Now check if the Pythagorean theorem holds: 

$d_1^2 + d_2^2 = 18 + 34 = 52$

$d_3^2 = 6^2 = 36$

Since $ d_1^2 + d_2^2 \neq d_3^2 $, the triangle is not right-angled based on this configuration.

Ans: To verify if the points form a parallelogram, we need to show that the diagonals bisect each other, i.e., the midpoints of both diagonals should coincide.

- Midpoint of diagonal $(-1, 2, 1)$ and $(4, -7, 8)$:

$M_1 = \left(\frac{-1 + 4}{2}, \frac{2 - 7}{2}, \frac{1 + 8}{2}\right) = \left(\frac{3}{2}, \frac{-5}{2}, \frac{9}{2}\right)$

- Midpoint of diagonal $(1, -2, 5)$ and $(2, -3, 4)$:

$M_2 = \left(\frac{1 + 2}{2}, \frac{-2 - 3}{2}, \frac{5 + 4}{2}\right) = \left(\frac{3}{2}, \frac{-5}{2}, \frac{9}{2}\right)$

Since the midpoints are the same, the diagonals bisect each other, proving that the points form a parallelogram.

Ans: Let the coordinates of the point $P(x, y, z)$ be equidistant from the points $(1, 2, 3)$ and $(3, 2, -1)$. Then,

$\text{Distance from } P(x, y, z) \text{ to } (1, 2, 3) = \text{Distance from } P(x, y, z) \text{ to } (3, 2, -1)$

Using the distance formula:

$\sqrt{(x - 1)^2 + (y - 2)^2 + (z - 3)^2} = \sqrt{(x - 3)^2 + (y - 2)^2 + (z + 1)^2}$

Squaring both sides:

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

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:

$x^2 - 2x + 1 + y^2 - 4y + 4 + z^2 - 6z + 9 = x^2 - 6x + 9 + y^2 - 4y + 4 + z^2 + 2z + 1$

Cancel the common terms:

$-2x + 1 - 6z + 9 = -6x + 9 + 2z + 1$

Simplifying further:

$-2x - 6z + 10 = -6x + 2z + 10$

Bringing all terms to one side:

$4x - 8z = 0$

Thus, the equation of the set of points is:

$x = 2z$

Ans:This is the equation of an ellipse with foci at points $A(4, 0, 0)$ and $B(-4, 0, 0)$, and the sum of the distances from these points is equal to the constant 10. The standard equation of an ellipse with foci at $(a, 0, 0)$ and $(-a, 0, 0)$, and the sum of the distances equal to $2c$, is:

$2a = 10, \quad \Rightarrow a = 5, \quad b^2 = a^2 - f^2$

Here, the distance between the foci is 8, so $f = 4$. Thus:

$b^2 = 5^2 - 4^2 = 25 - 16 = 9, \quad b = 3$

Therefore, the equation of the ellipse is:

$\frac{x^2}{25} + \frac{y^2}{9} = 1$