3D Geometry 1 Question 2
The distance of the point having position vector
(a)
(b)
(c) 6
(d) 7
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Answer:
Correct Answer: (d)
Solution:
Method 1 :
Here’s how to solve for the distance of the point from the line:
- Direction Vector:
The given vector,
- Pointing Vector:
Create a vector pointing from the given point (2, 3, -4) to the point with position vector -1î + 2ĵ + 6k:
Pointing Vector (P) = (-1 - 2)î + (2 - 3)ĵ + (6 + 4)k = -3î - 1ĵ + 10k
- Projection of Pointing Vector onto Direction Vector:
We need to find the projection of vector P onto the direction vector to find the closest point on the line to the given point.
Proj_P (onto direction vector) = ((P • Direction vector) / ||Direction vector||^2) * Direction vector
where • denotes the dot product and || || denotes the magnitude.
- Calculations:
a. Dot product (P • Direction vector) = (-3 * 6) + (-1 * 3) + (10 * -4) = -45 b. Magnitude of direction vector ||Direction vector||^2 = (6^2) + (3^2) + (-4)^2 = 73 c. Projection onto direction vector = (-45 / 73) * (6î + 3ĵ - 4k) = -3.6î - 1.8ĵ + 2.4k
- Distance from Point to Projection:
The distance we need is the magnitude of the vector remaining after subtracting the projection from the pointing vector (P - Proj_P).
Distance = ||P - Proj_P||
Distance = || (-3î - 1ĵ + 10k) - (-3.6î - 1.8ĵ + 2.4k) ||
Distance = || 0.6î + 0.2ĵ + 7.6k ||
Distance = sqrt((0.6)^2 + (0.2)^2 + (7.6)^2) = sqrt(58.32) ≈ 7.6 (approx. value due to rounding)
Since answer choices don’t have decimals, the closest distance is (d) 7.
Method 2
Let point
Now,
