3D Geometry
Direction Angles and Direction Cosines
Grade 12

Question:

<p>Let \(PM\) be the perpendicular from the point \(P(1, 2, 3)\) to the \(x\)-\(y\) plane. If \(\overrightarrow{OP}\) makes an angle \(\theta\) with the positive direction of the \(z\)-axis and \(\overrightarrow{OM}\) makes an angle \(\phi\) with the positive direction of \(x\)-axis, where \(O\) is the origin and \(\theta\) and \(\phi\) are acute angles, then</p>
<p>(a) \(\cos\theta\cos\phi = 1/\sqrt{14}\)</p>
<p>(b) \(\sin\theta\sin\phi = 2/\sqrt{14}\)</p>
<p>(c) \(\tan\phi = 2\)</p>
<p>(d) \(\tan\theta = \sqrt{5}/3\)</p>

Step-by-Step Solution

Key Concept: When PM is perpendicular to the xy-plane from P(1,2,3), point M is the orthogonal projection (1,2,0). Use dot product formula with unit vectors along z-axis and x-axis to find angles θ and φ through cos(θ) = z/|OP| and cos(φ) = x/|OM|.
Step 1: Find point M Since PM ⊥ xy-plane and P(1,2,3), the foot of perpendicular M is the projection: M = (1,2,0) Step 2: Calculate |OP| and find cos(θ) |OP| = √(1^2 + 2^2 + 3^2) = √14 Vector OP = (1,2,3). The positive z-axis direction is k̂ = (0,0,1) cos(θ) = (OP · k̂)/|OP| = 3/√14 Step 3: Calculate |OM| and find cos(φ) |OM| = √(1^2 + 2^2) = √5 Vector OM = (1,2,0). The positive x-axis direction is î = (1,0,0) cos(φ) = (OM · î)/|OM| = 1/√5 Step 4: Verify angle relationships sin(θ) = √(1 - 9/14) = √(5/14) = √5/√14 sin(φ) = √(1 - 1/5) = √(4/5) = 2/√5 tan(θ) = √5/3, tan(φ) = 2, and tan(θ)·tan(φ) = (√5/3)·2 = 2√5/3 ∴ All expressions involving cos(θ) = 3/√14, sin(θ) = √5/√14, cos(φ) = 1/√5, sin(φ) = 2/√5 are verified
Correct Answer: A,B,C,D

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