Vector Algebra
Geometry using Vectors
Grade 12
Question:
<p>Given \(\overrightarrow{OA} = 7\hat{i} - 4\hat{j} + 7\hat{k}\), \(\overrightarrow{OB} = \hat{i} - 6\hat{j} + 10\hat{k}\), \(\overrightarrow{OC} = -\hat{i} - 3\hat{j} + 4\hat{k}\), \(\overrightarrow{OD} = 5\hat{i} - \hat{j} + \hat{k}\). Then \(ABCD\) is</p>
<p>A square</p>
<p>A rhombus but not a square</p>
<p>A rectangle but not a square</p>
<p>A parallelogram but not a rhombus</p>
Step-by-Step Solution
Key Concept: Four points form a parallelogram if and only if the diagonals bisect each other, which means the midpoint of AC equals the midpoint of BD. Calculate midpoints using the formula: midpoint = (position vector 1 + position vector 2)/2.
Step 1: Find midpoint of diagonal AC Midpoint of AC = (OA + OC)/2 = [(7-1)î + (-4-3)ĵ + (7+4)k̂]/2 = (6î - 7ĵ + 11k̂)/2 = 3î - 3.5ĵ + 5.5k̂ Step 2: Find midpoint of diagonal BD Midpoint of BD = (OB + OD)/2 = [(1+5)î + (-6-1)ĵ + (10+1)k̂]/2 = (6î - 7ĵ + 11k̂)/2 = 3î - 3.5ĵ + 5.5k̂ Step 3: Compare midpoints Since midpoint of AC = midpoint of BD, the diagonals bisect each other. This is the defining property of a parallelogram. ∴ Answer: B (ABCD is a parallelogram)
Correct Answer: B