GSEB Class 11 Maths Solutions Chapter 12 Introduction to three Dimensional Geometry Ex 12.3

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

Answer:
(i) The given points are \( (2, 3, 5) \) and \( (4, 3, 1) \).
The required distance is calculated using the distance formula:
\( \sqrt{(4-2)^2 + (3-3)^2 + (1-5)^2} \)
\( = \sqrt{2^2 + 0^2 + (-4)^2} \)
\( = \sqrt{4 + 0 + 16} \)
\( = \sqrt{20} \)
\( = 2\sqrt{5} \).
(ii) The given points are \( (-3, 7, 2) \) and \( (2, 4, -1) \).
The required distance is:
\( \sqrt{(2-(-3))^2 + (4-7)^2 + (-1-2)^2} \)
\( = \sqrt{(2+3)^2 + (-3)^2 + (-3)^2} \)
\( = \sqrt{5^2 + (-3)^2 + (-3)^2} \)
\( = \sqrt{25 + 9 + 9} \)
\( = \sqrt{43} \).
(iii) The given points are \( (-1, 3, -4) \) and \( (1, -3, 4) \).
The required distance is:
\( \sqrt{(1-(-1))^2 + (-3-3)^2 + (4-(-4))^2} \)
\( = \sqrt{(1+1)^2 + (-6)^2 + (4+4)^2} \)
\( = \sqrt{2^2 + (-6)^2 + 8^2} \)
\( = \sqrt{4 + 36 + 64} \)
\( = \sqrt{104} \)
\( = 2\sqrt{26} \).
(iv) The given points are \( (2, -1, 3) \) and \( (-2, 1, 3) \).
The required distance is:
\( \sqrt{(-2-2)^2 + (1-(-1))^2 + (3-3)^2} \)
\( = \sqrt{(-4)^2 + (1+1)^2 + 0^2} \)
\( = \sqrt{(-4)^2 + 2^2 + 0^2} \)
\( = \sqrt{16 + 4 + 0} \)
\( = \sqrt{20} \)
\( = 2\sqrt{5} \).
In simple words: To find the distance between two points, use the distance formula. This formula involves subtracting the coordinates, squaring the results, adding them up, and then taking the square root.

Exam Tip: Remember the distance formula in 3D: \( \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2} \). Be careful with negative signs when subtracting coordinates.

Question 2. Show that the points (- 2, 3, 5), (1, 2, 3) and (7,0,-1) are collinear.
Answer: Let the given points be \( A(-2, 3, 5) \), \( B(1, 2, 3) \), and \( C(7, 0, -1) \).
First, we calculate the distance between points A and B (AB):
\( AB = \sqrt{(-2-1)^2 + (3-2)^2 + (5-3)^2} \)
\( = \sqrt{(-3)^2 + 1^2 + 2^2} \)
\( = \sqrt{9 + 1 + 4} \)
\( = \sqrt{14} \)
Next, we determine the distance between points B and C (BC):
\( BC = \sqrt{(7-1)^2 + (0-2)^2 + (-1-3)^2} \)
\( = \sqrt{6^2 + (-2)^2 + (-4)^2} \)
\( = \sqrt{36 + 4 + 16} \)
\( = \sqrt{56} \)
\( = 2\sqrt{14} \)
Then, we find the distance between points A and C (AC):
\( AC = \sqrt{(-2-7)^2 + (3-0)^2 + (5-(-1))^2} \)
\( = \sqrt{(-9)^2 + 3^2 + (5+1)^2} \)
\( = \sqrt{81 + 9 + 6^2} \)
\( = \sqrt{81 + 9 + 36} \)
\( = \sqrt{126} \)
\( = 3\sqrt{14} \)
Now, we check if the sum of two distances equals the third distance. We see that \( AB + BC = \sqrt{14} + 2\sqrt{14} = 3\sqrt{14} \).
This sum is equal to AC.
Thus, since \( AB + BC = AC \), points A, B, and C lie on the same line, which means they are collinear.
In simple words: To show that three points are on the same line, find the distances between all three pairs of points. If the sum of the two shorter distances equals the longest distance, then the points are collinear.

Exam Tip: For collinearity, calculate all three pairwise distances. Remember that \( \sqrt{a} + \sqrt{b} = \sqrt{c} \) is the condition, not \( a+b=c \). Simplify square roots to easily check the sum.

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), В(1, – 2, 5), C(4, – 7, 8) and D(2, – 3, 4) are the vertices of a parallelogram.

Answer:
(i) Let \( A(0, 7, -10) \), \( B(1, 6, -6) \), and \( C(4, 9, -6) \) be the given points.
First, we calculate the length of side AB:
\( AB = \sqrt{(1-0)^2 + (6-7)^2 + (-6-(-10))^2} \)
\( = \sqrt{1^2 + (-1)^2 + (4)^2} \)
\( = \sqrt{1 + 1 + 16} \)
\( = \sqrt{18} \)
Next, we calculate the length of side BC:
\( BC = \sqrt{(4-1)^2 + (9-6)^2 + (-6-(-6))^2} \)
\( = \sqrt{3^2 + 3^2 + 0^2} \)
\( = \sqrt{9 + 9 + 0} \)
\( = \sqrt{18} \)
Since \( AB = BC \), two sides of the triangle are equal in length. Therefore, triangle ABC is an isosceles triangle.

(ii) Let \( A(0, 7, 10) \), \( B(-1, 6, 6) \), and \( C(-4, 9, 6) \) be the given points.
First, we calculate AB:
\( AB = \sqrt{(-1-0)^2 + (6-7)^2 + (6-10)^2} \)
\( = \sqrt{(-1)^2 + (-1)^2 + (-4)^2} \)
\( = \sqrt{1 + 1 + 16} \)
\( = \sqrt{18} = 3\sqrt{2} \)
Next, we calculate BC:
\( BC = \sqrt{(-4-(-1))^2 + (9-6)^2 + (6-6)^2} \)
\( = \sqrt{(-3)^2 + 3^2 + 0^2} \)
\( = \sqrt{9 + 9 + 0} \)
\( = \sqrt{18} = 3\sqrt{2} \)
Then, we calculate AC:
\( AC = \sqrt{(-4-0)^2 + (9-7)^2 + (6-10)^2} \)
\( = \sqrt{(-4)^2 + 2^2 + (-4)^2} \)
\( = \sqrt{16 + 4 + 16} \)
\( = \sqrt{36} = 6 \)
Now, we check if the sum of the squares of two sides equals the square of the third side:
\( AB^2 + BC^2 = (\sqrt{18})^2 + (\sqrt{18})^2 = 18 + 18 = 36 \).
Also, \( AC^2 = 6^2 = 36 \).
Since \( AB^2 + BC^2 = AC^2 \), the triangle ABC satisfies the Pythagorean theorem, which confirms it is a right-angled triangle with the right angle at B.

(iii) Let the given points be \( A(-1, 2, 1) \), \( B(1, -2, 5) \), \( C(4, -7, 8) \), and \( D(2, -3, 4) \). We use the distance formula to find the lengths of all sides and diagonals.
Length of AB:
\( AB = \sqrt{(1-(-1))^2 + (-2-2)^2 + (5-1)^2} \)
\( = \sqrt{2^2 + (-4)^2 + 4^2} \)
\( = \sqrt{4 + 16 + 16} = \sqrt{36} = 6 \)
Length of BC:
\( BC = \sqrt{(4-1)^2 + (-7-(-2))^2 + (8-5)^2} \)
\( = \sqrt{3^2 + (-5)^2 + 3^2} \)
\( = \sqrt{9 + 25 + 9} = \sqrt{43} \)
Length of CD:
\( CD = \sqrt{(2-4)^2 + (-3-(-7))^2 + (4-8)^2} \)
\( = \sqrt{(-2)^2 + 4^2 + (-4)^2} \)
\( = \sqrt{4 + 16 + 16} = \sqrt{36} = 6 \)
Length of DA:
\( DA = \sqrt{(-1-2)^2 + (2-(-3))^2 + (1-4)^2} \)
\( = \sqrt{(-3)^2 + 5^2 + (-3)^2} \)
\( = \sqrt{9 + 25 + 9} = \sqrt{43} \)
We notice that \( AB = CD = 6 \) and \( BC = DA = \sqrt{43} \). This indicates that the opposite sides are equal.
Now, we calculate the lengths of the diagonals.
Length of diagonal AC:
\( AC = \sqrt{(4-(-1))^2 + (-7-2)^2 + (8-1)^2} \)
\( = \sqrt{5^2 + (-9)^2 + 7^2} \)
\( = \sqrt{25 + 81 + 49} = \sqrt{155} \)
Length of diagonal BD:
\( BD = \sqrt{(2-1)^2 + (-3-(-2))^2 + (4-5)^2} \)
\( = \sqrt{1^2 + (-1)^2 + (-1)^2} \)
\( = \sqrt{1 + 1 + 1} = \sqrt{3} \)
Since the opposite sides are equal (\( AB = CD \) and \( BC = DA \)) and the diagonals are not equal (\( AC \neq BD \)), the quadrilateral ABCD is a parallelogram.
In simple words: To confirm triangle types: for isosceles, check if two sides are equal; for a right-angled triangle, see if the square of the longest side equals the sum of squares of the other two sides. For a parallelogram, verify if opposite sides are equal in length and diagonals are not equal.

Exam Tip: Always calculate all relevant side lengths or diagonal lengths using the 3D distance formula to verify geometric properties. Pay close attention to negative signs in coordinate differences.

Question 4. Find the equation of the set of points P, which are equidistant from (1, 2, 3) and (3, 2, – 1).
Answer: Let \( A(1, 2, 3) \) and \( B(3, 2, -1) \) be the two given points. Let \( P(x, y, z) \) be a point that is equidistant from A and B.
According to the condition, the distance from P to A (PA) must be equal to the distance from P to B (PB).
So, \( PA = PB \).
Squaring both sides gives us \( PA^2 = PB^2 \).
Using the distance formula, we write the equation:
\( (x-1)^2 + (y-2)^2 + (z-3)^2 = (x-3)^2 + (y-2)^2 + (z-(-1))^2 \)
We can cancel the \( (y-2)^2 \) term from both sides, as it is identical.
\( (x-1)^2 + (z-3)^2 = (x-3)^2 + (z+1)^2 \)
Now, expand the squared terms:
\( x^2 - 2x + 1 + z^2 - 6z + 9 = x^2 - 6x + 9 + z^2 + 2z + 1 \)
Simplify the equation by canceling \( x^2 \) and \( z^2 \) from both sides and collecting similar terms:
\( -2x - 6z + 10 = -6x + 2z + 10 \)
Subtract 10 from both sides:
\( -2x - 6z = -6x + 2z \)
Move all terms involving x and z to one side:
\( -2x + 6x - 6z - 2z = 0 \)
\( 4x - 8z = 0 \)
Divide the entire equation by 4:
\( x - 2z = 0 \)
This is the required equation for the set of points P.
In simple words: A point that is equally far from two other points will always be on a specific plane. To find this plane's equation, set the squared distances from the moving point to each fixed point equal to each other. Simplify the equation to get your final answer.

Exam Tip: When finding a locus of points equidistant from two fixed points, always square the distances to remove square roots and simplify the algebra. Remember that this results in the equation of a plane perpendicular to the line segment connecting the two fixed points.

Question 5. Find the equation of set of points P, the sum of whose distances from A(4, 0, 0) and B(- 4, 0, 0) is equal to 10.
Answer: Let \( A(4, 0, 0) \) and \( B(-4, 0, 0) \) be the two fixed points. Let \( P(x, y, z) \) be a point such that the sum of its distances from A and B is 10.
So, \( PA + PB = 10 \).
Using the distance formula, we can write:
\( \sqrt{(x-4)^2 + y^2 + z^2} + \sqrt{(x-(-4))^2 + y^2 + z^2} = 10 \)
\( \sqrt{(x-4)^2 + y^2 + z^2} = 10 - \sqrt{(x+4)^2 + y^2 + z^2} \)
Square both sides of the equation:
\( (x-4)^2 + y^2 + z^2 = \left(10 - \sqrt{(x+4)^2 + y^2 + z^2}\right)^2 \)
\( (x-4)^2 + y^2 + z^2 = 100 - 20\sqrt{(x+4)^2 + y^2 + z^2} + (x+4)^2 + y^2 + z^2 \)
Expand \( (x-4)^2 \) and \( (x+4)^2 \):
\( x^2 - 8x + 16 + y^2 + z^2 = 100 - 20\sqrt{(x+4)^2 + y^2 + z^2} + x^2 + 8x + 16 + y^2 + z^2 \)
Cancel \( x^2, y^2, z^2 \) and 16 from both sides:
\( -8x = 100 - 20\sqrt{(x+4)^2 + y^2 + z^2} + 8x \)
Move all terms without the square root to the left side:
\( -8x - 8x - 100 = -20\sqrt{(x+4)^2 + y^2 + z^2} \)
\( -16x - 100 = -20\sqrt{(x+4)^2 + y^2 + z^2} \)
Divide the entire equation by -4 to simplify:
\( 4x + 25 = 5\sqrt{(x+4)^2 + y^2 + z^2} \)
Square both sides again to remove the remaining square root:
\( (4x+25)^2 = \left(5\sqrt{(x+4)^2 + y^2 + z^2}\right)^2 \)
\( 16x^2 + 200x + 625 = 25((x+4)^2 + y^2 + z^2) \)
\( 16x^2 + 200x + 625 = 25(x^2 + 8x + 16 + y^2 + z^2) \)
\( 16x^2 + 200x + 625 = 25x^2 + 200x + 400 + 25y^2 + 25z^2 \)
Subtract \( 16x^2 + 200x \) from both sides:
\( 625 = 9x^2 + 400 + 25y^2 + 25z^2 \)
Rearrange the terms to get the final equation:
\( 9x^2 + 25y^2 + 25z^2 = 625 - 400 \)
\( 9x^2 + 25y^2 + 25z^2 = 225 \)
This is the equation of the set of points P, which describes an ellipsoid.
In simple words: When the total distance from a point to two fixed points is constant, the set of all such points forms a shape called an ellipsoid. To find its equation, you set up the distance sum, move one square root to the other side, and square both sides twice to get rid of all square roots, then simplify.

Exam Tip: Questions involving the sum or difference of distances from two fixed points usually lead to conic sections (ellipses, hyperbolas) in 2D, and ellipsoids or hyperboloids in 3D. The key is careful algebraic manipulation, especially squaring terms with square roots.