Introduction To Three- Dimensional Geometry — Flashcards
Telangana Open School (TOSS) · Class 12 · Mathematics
24 flashcards for Introduction To Three- Dimensional Geometry (Telangana Open School (TOSS) Class 12 Mathematics) to test yourself on key terms and facts.
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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: $$ = \sqrt{(8 - 2)^2 + (2 - 5)^2 + (-6 + 4)^2} = \sqrt{6^2 + (-3)^2 + (-2)^…
Show that points A(-2, 4, -3), B(4, -3, -2), and C(-3, -2, 4) form an equilateral triangle.
Answer
Step 1: Find AB: $$ 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…
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: $$ BC = \sqrt{(5 - 1)^2 + (4 - 4)^2 + (0 - 5)^2} = \sqrt{16 + 0 + 25} = \sqr…
Show that points (2, -3, 3), (1, 2, 4), and (3, -8, 2) are collinear.
Answer
Step 1: Find PQ: $$ 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}…
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: $$ AC^2 = (3 - 1)^…
Find the coordinates of the point dividing the line segment joining (2, -4, 3) and (-4, 5, -6) in the ratio 2:1 internally.
Answer
Use section formula (internal): $$ 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(…
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: $$ x = \frac{lx_2 - mx_1}{l - m} = \frac{2(1) - 3(-1)}{2 - 3} = \frac{2 + 3}{-1} = -5 $$ $$ y = \frac{2(-1) - 3(-3)}{-1} = \frac{-2 + 9}{-1} = -7 $$ $$ z = \frac{2(2) - 3…
Find the ratio in which the XY-plane divides the line segment joining (2, -3, 5) and (7, 1, 3).
Answer
Let ratio be l:m. The z-coordinate of the dividing point is: $$ 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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