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Geometry and Trigonometry - 3D Geometry

Grade 11IB_AA

Review the key concepts, formulae, and examples before starting your quiz.

πŸ”‘Concepts

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The length of the space diagonal in a rectangular prism (cuboid) is the distance between two opposite vertices. For a cuboid with dimensions aa, bb, and cc, it is given by d=a2+b2+c2d = \sqrt{a^2 + b^2 + c^2}.

A 3D cuboid showing the space diagonal 'd' from one bottom corner to the opposite top corner.
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The angle between a line and a plane is the angle between the line and its orthogonal projection onto that plane. In a right pyramid, this often involves finding the angle between a slant edge and the diagonal of the base.

A right triangle formed by the vertical height, the projection on the base, and the slant edge of a pyramid.
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The distance between two points (x1,y1,z1)(x_1, y_1, z_1) and (x2,y2,z2)(x_2, y_2, z_2) in 3D Cartesian space is the length of the straight line segment connecting them, derived from the Pythagorean theorem.

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Surface area and volume of 3D shapes often require finding 'hidden' lengths, such as the slant height (ll) of a cone or pyramid, using l=h2+r2l = \sqrt{h^2 + r^2} where hh is the vertical height.

πŸ“Formulae

d=(x2βˆ’x1)2+(y2βˆ’y1)2+(z2βˆ’z1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}

M=(x1+x22,y1+y22,z1+z22)M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right)

Vpyramid=13Γ—baseΒ areaΓ—heightV_{pyramid} = \frac{1}{3} \times \text{base area} \times \text{height}

Vsphere=43Ο€r3V_{sphere} = \frac{4}{3}\pi r^3

Asphere=4Ο€r2A_{sphere} = 4\pi r^2

Vcone=13Ο€r2hV_{cone} = \frac{1}{3}\pi r^2 h

πŸ’‘Examples

Problem 1:

Calculate the distance between the points A(1,βˆ’2,4)A(1, -2, 4) and B(3,4,βˆ’1)B(3, 4, -1).

Solution:

d=(3βˆ’1)2+(4βˆ’(βˆ’2))2+(βˆ’1βˆ’4)2d = \sqrt{(3 - 1)^2 + (4 - (-2))^2 + (-1 - 4)^2} d=22+62+(βˆ’5)2d = \sqrt{2^2 + 6^2 + (-5)^2} d=4+36+25d = \sqrt{4 + 36 + 25} d=65β‰ˆ8.06d = \sqrt{65} \approx 8.06

Explanation:

Apply the 3D distance formula by substituting the coordinates of AA and BB into Ξ”x2+Ξ”y2+Ξ”z2\sqrt{\Delta x^2 + \Delta y^2 + \Delta z^2}.

Problem 2:

A square-based pyramid has a base side length of 1010 cm and a vertical height of 1212 cm. Find the angle that a slant edge makes with the base.

Solution:

  1. Find the distance from a corner of the base to the center of the base: The diagonal of the base is dbase=102+102=200=102d_{base} = \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2}. The distance from the corner to the center is r=1022=52r = \frac{10\sqrt{2}}{2} = 5\sqrt{2}.
  2. Use the tangent ratio in the right-angled triangle formed by the height (h=12h=12), the distance to center (r=52r=5\sqrt{2}), and the slant edge: tan⁑(ΞΈ)=1252\tan(\theta) = \frac{12}{5\sqrt{2}} ΞΈ=arctan⁑(1252)β‰ˆ59.5∘\theta = \arctan\left(\frac{12}{5\sqrt{2}}\right) \approx 59.5^\circ

Explanation:

First, find the horizontal distance from the base corner to the center of the base (half the diagonal). Then, use the vertical height and this horizontal distance as the opposite and adjacent sides of a right-angled triangle to find the angle.

Problem 3:

A cuboid has dimensions 44 cm by 33 cm by 1212 cm. Find the length of the space diagonal.

Solution:

d=42+32+122d = \sqrt{4^2 + 3^2 + 12^2} d=16+9+144d = \sqrt{16 + 9 + 144} d=169d = \sqrt{169} d=13Β cmd = 13 \text{ cm}

Explanation:

Use the 3D Pythagorean theorem d=l2+w2+h2d = \sqrt{l^2 + w^2 + h^2} to find the distance between two opposite corners of the cuboid.

Problem 4:

A right triangular prism has a length of 1515 cm. The triangular cross-section is a right-angled triangle with legs of 55 cm and 1212 cm. Find the distance between the two furthest vertices of the prism.

A right triangular prism with dimensions 5, 12, and 15, showing the diagonal between opposite vertices.

Solution:

  1. Identify the dimensions: a=5a = 5, b=12b = 12, and height (length of prism) L=15L = 15.
  2. The distance between the two furthest vertices is the space diagonal of the bounding box (or the hypotenuse of the triangle formed by the base hypotenuse and the prism length).
  3. First, find the hypotenuse of the triangular base: c=52+122=25+144=13Β cmc = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = 13\text{ cm}
  4. Now find the space diagonal (dd) using the base hypotenuse and the prism length: d=c2+L2=132+152d = \sqrt{c^2 + L^2} = \sqrt{13^2 + 15^2} d=169+225=394β‰ˆ19.8Β cmd = \sqrt{169 + 225} = \sqrt{394} \approx 19.8\text{ cm}

Explanation:

The furthest distance in a prism is between opposite corners. We first solve for the diagonal of one face (the hypotenuse of the triangle) and then use that result with the prism's length in a second Pythagorean calculation.

Problem 5:

A right cone has a base radius of r=6r = 6 cm and a slant height l=10l = 10 cm. Calculate the vertical height hh of the cone and the angle ΞΈ\theta that the slant edge makes with the horizontal base.

A cross-section of a cone showing the vertical height h, radius r, and slant height l forming a right-angled triangle.

Solution:

  1. Using the Pythagorean theorem in the right-angled triangle formed by the radius, vertical height, and slant height: h=l2βˆ’r2h = \sqrt{l^2 - r^2} h=102βˆ’62=100βˆ’36=64=8Β cmh = \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8\text{ cm}

  2. To find the angle ΞΈ\theta between the slant height and the base radius: cos⁑(ΞΈ)=adjhyp=rl\cos(\theta) = \frac{\text{adj}}{\text{hyp}} = \frac{r}{l} cos⁑(ΞΈ)=610=0.6\cos(\theta) = \frac{6}{10} = 0.6 ΞΈ=arccos⁑(0.6)β‰ˆ53.1∘\theta = \arccos(0.6) \approx 53.1^{\circ}

Explanation:

In 3D geometry, many problems involving cones or pyramids can be reduced to 2D right-angled triangle problems. Here, the vertical height, radius, and slant height form a right triangle. We apply the Pythagorean theorem for length and basic trigonometry for the angle of elevation.