In this section, we will explore how to apply Pythagoras' theorem and trigonometry in three-dimensional (3D) spaces. Understanding these concepts is crucial for solving problems related to finding angles between lines and planes, as well as calculating distances in 3D. These skills are essential for tackling real-world problems and are a key part of the CIE IGCSE syllabus.
Understanding 3D Geometry
Before diving into complex calculations, it's important to grasp the basics of 3D geometry. In a 3D space, points are represented by coordinates (x,y,z), where z represents the third dimension.
Key Concepts
Points in 3D: Coordinates (x,y,z) used to locate points in space.
Distance Formula: The formula to calculate the distance between two points in 3D.
Vectors: Representing direction and magnitude in 3D.
Pythagoras' Theorem in 3D
Pythagoras' theorem is not limited to 2D spaces. In 3D, it helps us calculate the distance between two points, which is crucial for many geometric and trigonometric applications.
Distance Between Two Points: The formula for calculating the distance between points A(x1,y1,z1) and B(x2,y2,z2) is (x2−x1)2+(y2−y1)2+(z2−z1)2.
Example 1: Calculating Distance in 3D
Calculate the distance between the points A(1,2,3) and B(4,5,6).
Solution:
Using the distance formula, the distance is:
d=(4−1)2+(5−2)2+(6−3)2$d = \sqrt{27}
<p></p>d = 3\sqrt{3} \text{ units}.<h2id="trigonometry−in−3−d"><strong>Trigonometryin3D</strong></h2><p>Trigonometryextendsinto3Dbyusinganglesbetweenvectorsandplanes.Thisisfundamentalforfindinganglesin3Dstructuresandforunderstandingthespatialrelationshipsbetweendifferentgeometricentities.</p><imgsrc="https://tutorchase−production.s3.eu−west−2.amazonaws.com/e2b421c2−09d7−4c6b−951a−f874694efe24−file.png"alt="Trigonometryin3D"style="width:500px;height:411px"width="500"height="411"><p></p><h3><strong>KeyTrigonometricRatios</strong></h3><ul><li><strong>Sine,Cosine,andTangent</strong>in3Dinvolveanglesbetweenlinesandplanesorbetweentwointersectinglines.</li></ul><h3><strong>Example2:FindingtheAnglebetweenaLineandaPlane</strong></h3><p>Imaginealineintersectingaplaneatapoint,andwewishtofindtheangleofintersection.</p><p></p><h4><strong>Solution</strong>:</h4><p>First,determinethedirectionvectorofthelineandthenormalvectoroftheplane.Theangle\thetabetweenthelineandtheplanecanbefoundusingthedotproductformulaandtrigonometricidentities.</p><p></p><h2id="solving−3−d−problems"><strong>Solving3DProblems</strong></h2><p>Whensolving3Dproblems,it′sessentialtovisualisethegeometricconfigurationandapplythecorrectformulaetofinddistancesorangles.</p><p></p><h3><strong>Example3:CalculatingDistancesin3DSpace</strong></h3><p>GivenpointsC(7, -2, 1)andD(-1, 4, 5),calculatethedistancebetweenthem.</p><p></p><h4><strong>Solution</strong>:</h4><p>Usingthedistanceformula,thedistanceis</p><p></p>d = \sqrt{(-1-7)^2 + (4+2)^2 + (5-1)^2}<p></p>d = \sqrt{64 + 36 + 16}<p></p><p></p>d = \sqrt{116}<p></p>d = 2\sqrt{29} \text { units}<p></p><h2id="practice−questions"><strong>PracticeQuestions</strong></h2><h3><strong>Problem1:</strong></h3><p>FindthedistancebetweenthepointsE(2, -3, 5)andF(-3, 2, -4).</p><p></p><h4><strong>Solution:</strong></h4><p>Tofindthedistancebetweentwopointsin3D,youusetheformula:</p><p></p>d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}<p></p><p>PlugginginthecoordinatesofEandFyields:</p><p></p>d = \sqrt{(-3 - 2)^2 + (2 + 3)^2 + (-4 - 5)^2}<p></p>d = \sqrt{25 + 25 + 81}<p></p>d = \sqrt{131}<p></p><h3><strong>Problem2:</strong></h3><p>DeterminetheanglebetweenthelinepassingthroughpointsG(1, 1, 1)andH(4, 4, 4)andtheplanedefinedbytheequationx + y + z = 12.</p><p></p><h4><strong>Solution:</strong></h4><p>Theangle\thetabetweenalineandaplanecanbefoundusingthedotproductoftheline′sdirectionvectorandtheplane′snormalvector.ForthelinethroughGandH,thedirectionvectordis(3, 3, 3)(thedifferencebetweenHandG).</p><p>ThenormalvectorNoftheplanex + y + z = 12is(1, 1, 1).Theangleisgivenby:</p><p></p>\cos(\theta) = \frac{d \cdot N}{|d||N|}<p></p>d \cdot N = (3)(1) + (3)(1) + (3)(1) = 9<p></p>|d| = \sqrt{3^2 + 3^2 + 3^2} = \sqrt{27}<p></p>|N| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}<p></p>\cos(\theta) = \frac{9}{\sqrt{27}\sqrt{3}}<p></p>\cos (\theta) = \frac{9}{\sqrt{81}} = \frac{9}{9} = 1<p></p><p>Theresultimpliesthat\theta = 0^\circ(orequivalently,thelineandtheplane′snormalvectorareparallel),whichalignswiththeconclusionthattherewasamisunderstandinginmypreviousexplanation.Theaccurateinterpretationisthatwhenthecosineoftheangleequals1,itmeanstheanglebetweenthelinedirectionvectorandtheplane′snormalvectoris0^\circ$, indicating they are parallel. This is a unique situation given the specific vectors involved and does not generally represent the angle between a line and a plane in 3D space.
Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.
Oxford University - PhD Mathematics
Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.
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