Introduction to 3D Geometry Miscellaneous Exercise: Class 11 NCERT Solutions

Consolidating Your 3D Geometry Toolkit

The miscellaneous exercise assumes you are comfortable with the core tools of 3D coordinate geometry. The challenge lies in identifying which tool to use and how to apply it to solve a given geometric problem. Before you tackle the problems, let's quickly recap the essential formulas that you'll be using frequently.

1. Distance Formula: To find the distance between two points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂), we use:

d = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²]
This is fundamental for proving properties of shapes based on side lengths (e.g., equilateral triangle, rhombus, parallelogram).

1. Section Formula: To find the coordinates of a point R that divides the line segment joining P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) in the ratio m:n.

• Internal Division: R = ( (mx₂ + nx₁)/(m+n), (my₂ + ny₁)/(m+n), (mz₂ + nz₁)/(m+n) )
• External Division: R = ( (mx₂ - nx₁)/(m-n), (my₂ - ny₁)/(m-n), (mz₂ - nz₁)/(m-n) )

A special case is the Mid-point Formula (when m=n=1): ( (x₁+x₂)/2, (y₁+y₂)/2, (z₁+z₂)/2 ).

1. Centroid of a Triangle: For a triangle with vertices (x₁, y₁, z₁), (x₂, y₂, z₂), and (x₃, y₃, z₃), the centroid is:

G = ( (x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3, (z₁+z₃+z₃)/3 )

The miscellaneous questions will require you to use these formulas strategically to prove geometric properties or find unknown coordinates or loci.

Worked Examples: From Problem to Solution

• Problem 1: Show that the points A(0, 7, 10), B(-1, 6, 6), and C(-4, 9, 6) form an isosceles right-angled triangle. Solution: Step 1: Use the distance formula to find the lengths of all three sides. AB² = (-1-0)² + (6-7)² + (6-10)² = (-1)² + (-1)² + (-4)² = 1 + 1 + 16 = 18 BC² = (-4 - (-1))² + (9-6)² + (6-6)² = (-3)² + 3² + 0² = 9 + 9 + 0 = 18 AC² = (-4-0)² + (9-7)² + (6-10)² = (-4)² + 2² + (-4)² = 16 + 4 + 16 = 36 Step 2: Analyze the side lengths to check for isosceles property. Since AB² = 18 and BC² = 18, we have AB = BC = √18 = 3√2. As two sides are equal, the triangle ABC is isosceles. Step 3: Use the Pythagorean theorem to check for the right-angle property. We observe that AB² + BC² = 18 + 18 = 36. And AC² = 36. Since AB² + BC² = AC², the triangle satisfies the Pythagorean theorem, with the right angle at vertex B. Conclusion: Therefore, the points A, B, and C form an isosceles right-angled triangle.
• Problem 2: Find the coordinates of the point which divides the line segment joining the points P(-2, 3, 5) and Q(1, -4, 6) in the ratio (i) 2:3 internally, and (ii) 2:3 externally. Solution: Let the given points be P(x₁, y₁, z₁) = (-2, 3, 5) and Q(x₂, y₂, z₂) = (1, -4, 6). (i) Internal Division (m=2, n=3) Step 1: Apply the internal section formula. The coordinates (x, y, z) are given by: x = (mx₂ + nx₁)/(m+n) = (21 + 3(-2))/(2+3) = (2 - 6)/5 = -4/5 y = (my₂ + ny₁)/(m+n) = (2(-4) + 33)/(2+3) = (-8 + 9)/5 = 1/5 z = (mz₂ + nz₁)/(m+n) = (26 + 35)/(2+3) = (12 + 15)/5 = 27/5 Answer (i): The point is (-4/5, 1/5, 27/5). (ii) External Division (m=2, n=3) Step 1: Apply the external section formula. The coordinates (x, y, z) are given by: x = (mx₂ - nx₁)/(m-n) = (21 - 3(-2))/(2-3) = (2 + 6)/(-1) = -8 y = (my₂ - ny₁)/(m-n) = (2(-4) - 33)/(2-3) = (-8 - 9)/(-1) = 17 z = (mz₂ - nz₁)/(m-n) = (26 - 35)/(2-3) = (12 - 15)/(-1) = 3 Answer (ii): The point is (-8, 17, 3).

Common Pitfalls and Exam Strategy

Miscellaneous exercises are designed to catch common errors. Be mindful of these traps:

• Incomplete Proofs: When asked to prove a shape is a parallelogram, showing opposite sides are equal (AB=CD and BC=DA) is one way. A faster way is often to show that the midpoint of diagonal AC is the same as the midpoint of diagonal BD. If you need to prove it's a rhombus, you must show all four sides are equal. For a rectangle, prove it's a parallelogram and that its diagonals are equal.
• Sign Errors: The distance and section formulas involve many subtractions and squares. Be extremely careful with negative signs. A small error like (-4 - (-1)) = -5 instead of -3 will lead to a completely wrong answer.
• Confusing Formulas: Double-check if you're using the correct formula. Don't mix up the internal and external division formulas (the only difference is the sign in the numerator and denominator).
• Locus Problems: For locus problems, start with the given geometric condition (e.g., PA = PB or PA² + PB² = k²). Substitute the coordinates (x, y, z) for the moving point P. The final answer should be an equation in terms of x, y, and z that represents the path or surface.

Practice Questions with Solutions

• Q: Find the ratio in which the YZ-plane divides the line segment formed by joining the points (-2, 4, 7) and (3, -5, 8). A: Step 1: Let the YZ-plane divide the line segment joining P(-2, 4, 7) and Q(3, -5, 8) in the ratio k:1 at point R(x, y, z). Step 2: Any point on the YZ-plane has its x-coordinate equal to 0. We use the section formula for the x-coordinate. x = (k3 + 1(-2)) / (k+1) Step 3: Set the x-coordinate to 0 and solve for k. 0 = (3k - 2) / (k+1) 3k - 2 = 0 3k = 2 k = 2/3 Final answer: The required ratio is 2:3.
• Q: 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 D. A: Step 1: Let the coordinates of the fourth vertex D be (x, y, z). In a parallelogram, the diagonals bisect each other. Therefore, the midpoint of diagonal AC is the same as the midpoint of diagonal BD. Step 2: Find the midpoint of AC. Midpoint of AC = ((3 + (-1))/2, (-1 + 1)/2, (2 + 2)/2) = (2/2, 0/2, 4/2) = (1, 0, 2). Step 3: Find the midpoint of BD and equate it to the midpoint of AC. Midpoint of BD = ((1 + x)/2, (2 + y)/2, (-4 + z)/2). Equating the coordinates: (1+x)/2 = 1 => 1+x = 2 => x=1. (2+y)/2 = 0 => 2+y = 0 => y=-2. (-4+z)/2 = 2 => -4+z = 4 => z=8. Final answer: The coordinates of the fourth vertex D are (1, -2, 8).
• Q: Show that the points P(–2, 3, 5), Q(1, 2, 3) and R(7, 0, –1) are collinear. A: Step 1: To prove collinearity, we can show that the sum of the distances of any two segments equals the third (e.g., PQ + QR = PR). We use the distance formula. Step 2: Calculate the distances PQ, QR, and PR. PQ = √((1 - (-2))² + (2 - 3)² + (3 - 5)²) = √(3² + (-1)² + (-2)²) = √(9 + 1 + 4) = √14. QR = √((7 - 1)² + (0 - 2)² + (-1 - 3)²) = √(6² + (-2)² + (-4)²) = √(36 + 4 + 16) = √56 = √(4 14) = 2√14. PR = √((7 - (-2))² + (0 - 3)² + (-1 - 5)²) = √(9² + (-3)² + (-6)²) = √(81 + 9 + 36) = √126 = √(9 14) = 3√14. Step 3: Check if the sum of two smaller distances equals the largest distance. PQ + QR = √14 + 2√14 = 3√14. This is equal to the distance PR. Final answer: Since PQ + QR = PR, the points P, Q, and R are collinear.
• Q: Find the locus of the point P for which the sum of the squares of its distances from the points A(3, 4, 5) and B(–1, 3, –7) is equal to 2k². A: Step 1: Let the coordinates of the point P be (x, y, z). The given condition is PA² + PB² = 2k². Step 2: Use the distance formula to express PA² and PB². PA² = (x - 3)² + (y - 4)² + (z - 5)² PB² = (x - (-1))² + (y - 3)² + (z - (-7))² = (x + 1)² + (y - 3)² + (z + 7)² Step 3: Substitute these into the given condition and simplify. [(x - 3)² + (y - 4)² + (z - 5)²] + [(x + 1)² + (y - 3)² + (z + 7)²] = 2k² (x² - 6x + 9 + y² - 8y + 16 + z² - 10z + 25) + (x² + 2x + 1 + y² - 6y + 9 + z² + 14z + 49) = 2k² 2x² + 2y² + 2z² - 4x - 14y + 4z + 109 = 2k² Step 4: Write the final equation of the locus. 2x² + 2y² + 2z² - 4x - 14y + 4z + 109 - 2k² = 0. Dividing by 2 (optional, but cleaner): x² + y² + z² - 2x - 7y + 2z + (109/2) - k² = 0. Final answer: The locus is the equation 2x² + 2y² + 2z² - 4x - 14y + 4z + 109 - 2k² = 0, which represents a sphere.

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