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Introduction To Three- Dimensional Geometry

Telangana Open School (TOSS) · Class 12 · Mathematics

Flashcards for Introduction To Three- Dimensional Geometry — Telangana Open School (TOSS) Class 12 Mathematics. Quick Q&A cards covering key concepts, definitions, and formulas.

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24 Flashcards
Card 1Distance Between Two Points

Find the distance between points (2, 5, -4) and (8, 2, -6).

Answer

Step 1: Use the distance formula: $$ |PQ| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} $$ Step 2: Plug in values: $$ = \sqrt{(8 - 2)^2 + (2 - 5)^2 + (-6 + 4)^2} = \sqrt{6^2 + (-3)^2 + (-2)^

Card 2Distance Between Two Points

Show that points A(-2, 4, -3), B(4, -3, -2), and C(-3, -2, 4) form an equilateral triangle.

Answer

Step 1: Find AB: $$ AB = \sqrt{(4 + 2)^2 + (-3 - 4)^2 + (-2 + 3)^2} = \sqrt{36 + 49 + 1} = \sqrt{86} $$ Step 2: Find BC: $$ BC = \sqrt{(-3 - 4)^2 + (-2 + 3)^2 + (4 + 2)^2} = \sqrt{49 + 1 + 36} = \sqrt

Card 3Distance Between Two Points

Verify if points A(-1, 2, 3), B(1, 4, 5), and C(5, 4, 0) form a triangle.

Answer

Step 1: Find AB: $$ AB = \sqrt{(1 + 1)^2 + (4 - 2)^2 + (5 - 3)^2} = \sqrt{4 + 4 + 4} = \sqrt{12} ≈ 3.46 $$ Step 2: Find BC: $$ BC = \sqrt{(5 - 1)^2 + (4 - 4)^2 + (0 - 5)^2} = \sqrt{16 + 0 + 25} = \sqr

Card 4Distance Between Two Points

Show that points (2, -3, 3), (1, 2, 4), and (3, -8, 2) are collinear.

Answer

Step 1: Find PQ: $$ PQ = \sqrt{(1 - 2)^2 + (2 + 3)^2 + (4 - 3)^2} = \sqrt{1 + 25 + 1} = \sqrt{27} = 3\sqrt{3} $$ Step 2: Find QR: $$ QR = \sqrt{(3 - 1)^2 + (-8 - 2)^2 + (2 - 4)^2} = \sqrt{4 + 100 + 4}

Card 5Distance Between Two Points

Prove that points A(1,2,-2), B(2,3,-4), C(3,4,-3) form a right-angled triangle.

Answer

Step 1: Find AB²: $$ AB^2 = (2 - 1)^2 + (3 - 2)^2 + (-4 + 2)^2 = 1 + 1 + 4 = 6 $$ Step 2: Find BC²: $$ BC^2 = (3 - 2)^2 + (4 - 3)^2 + (-3 + 4)^2 = 1 + 1 + 1 = 3 $$ Step 3: Find AC²: $$ AC^2 = (3 - 1)^

Card 6Section Formula in 3D

Find the coordinates of the point dividing the line segment joining (2, -4, 3) and (-4, 5, -6) in the ratio 2:1 internally.

Answer

Use section formula (internal): $$ x = \frac{lx_2 + mx_1}{l + m} = \frac{2(-4) + 1(2)}{2 + 1} = \frac{-8 + 2}{3} = -2 $$ $$ y = \frac{2(5) + 1(-4)}{3} = \frac{10 - 4}{3} = 2 $$ $$ z = \frac{2(-6) + 1(

Card 7Section Formula in 3D

Find the point dividing the join of (-1, -3, 2) and (1, -1, 2) externally in the ratio 2:3.

Answer

Use external section formula: $$ x = \frac{lx_2 - mx_1}{l - m} = \frac{2(1) - 3(-1)}{2 - 3} = \frac{2 + 3}{-1} = -5 $$ $$ y = \frac{2(-1) - 3(-3)}{-1} = \frac{-2 + 9}{-1} = -7 $$ $$ z = \frac{2(2) - 3

Card 8Section Formula in 3D

Find the ratio in which the XY-plane divides the line segment joining (2, -3, 5) and (7, 1, 3).

Answer

Let ratio be l:m. The z-coordinate of the dividing point is: $$ z = \frac{l(3) + m(5)}{l + m} $$ Since it lies on XY-plane, z = 0: $$ \frac{3l + 5m}{l + m} = 0 \Rightarrow 3l + 5m = 0 \Rightarrow \fra

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What are the important topics in Introduction To Three- Dimensional Geometry for Telangana Open School (TOSS) Class 12 Mathematics?
Introduction To Three- Dimensional Geometry covers several key topics that are frequently asked in Telangana Open School (TOSS) Class 12 board exams. Focus on the core concepts listed on this page and practise related questions to build confidence.
How to score full marks in Introduction To Three- Dimensional Geometry — Telangana Open School (TOSS) Class 12 Mathematics?
Understand the core concepts first, then work through the 57 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.
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There are 24 flashcards for Introduction To Three- Dimensional Geometry covering key definitions, formulas, and concepts. Use them daily for 10–15 minutes for best results.

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