Three-Dimensional Geometry — Practice

Question 1

Find the vector and Cartesian equations of the line through (1, 2, −4) with direction ratios (2, −1, 3).

Accepted Answer

VECTOR FORM: r⃗ = a⃗ + λb⃗ where a⃗ = î + 2ĵ − 4k̂ and b⃗ = 2î − ĵ + 3k̂. r⃗ = (î + 2ĵ − 4k̂) + λ(2î − ĵ + 3k̂). CARTESIAN FORM: (x−1)/2 = (y−2)/(−1) = (z+4)/3. (Using point (1,2,−4) and DRs (2,−1,3).)

Question 2

Find the distance of the point (2, 3, −5) from the plane x + 2y − 2z = 9.

Accepted Answer

Plane equation: x + 2y − 2z − 9 = 0. Here a=1, b=2, c=−2, d=−9. Point: (x₁, y₁, z₁) = (2, 3, −5). DISTANCE = |ax₁ + by₁ + cz₁ + d| / √(a²+b²+c²) = |1·2 + 2·3 + (−2)·(−5) + (−9)| / √(1+4+4) = |2 + 6 + 10 − 9| / √9 = |9| / 3 = 9/3 = 3 units.

Question 3

Find the shortest distance between the lines r⃗ = (î + 2ĵ + k̂) + λ(î − ĵ + k̂) and r⃗ = (2î − ĵ − k̂) + μ(2î + ĵ + 2k̂).

Accepted Answer

Here a₁⃗ = î+2ĵ+k̂, b₁⃗ = î−ĵ+k̂, a₂⃗ = 2î−ĵ−k̂, b₂⃗ = 2î+ĵ+2k̂. a₂⃗ − a₁⃗ = (2−1)î + (−1−2)ĵ + (−1−1)k̂ = î − 3ĵ − 2k̂. b₁⃗ × b₂⃗ = |î  ĵ  k̂; 1 −1 1; 2  1  2| = î(−2−1) − ĵ(2−2) + k̂(1+2) = −3î − 0ĵ + 3k̂ = −3î + 3k̂. |b₁⃗ × b₂⃗| = √(9+0+9) = √18 = 3√2. (a₂⃗−a₁⃗)·(b₁⃗×b₂⃗) = (1)(−3) + (−3)(0) + (−2)(3) = −3 + 0 − 6 = −9. SHORTEST DISTANCE = |−9| / 3√2 = 9/(3√2) = 3/√2 = 3√2/2 units.

Practice — Three-Dimensional Geometry

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