Vectors in two and three dimensions — Question 268
Original QCE Vault practice · 5 marks
Q268 · Practice questionTechnology-freeComplex familiar5 marks
QUESTION 268 (5 marks)
A nondegenerate parallelogram ABCD in three-dimensional space has position vectors a,b,c,d in cyclic order. Let M and N be the midpoints of AB and DC. a)Prove that the diagonals AC and BD have the same midpoint.
b)Prove that MN is parallel to AD and has the same length.
WORKED SOLUTION
Practice marking scheme
5 marksANSWER(a) Both diagonal midpoints are (a+c)/2=(b+d)/2. (b) MN=AD. Worked solution
(a) Opposite sides have equal vectors: b−a=c−d. Rearranging gives a+c=b+d. Dividing by 2 proves that the two diagonal midpoints coincide. (b) m=(a+b)/2 and n=(d+c)/2. With c=b+d−a, n−m=d−a. Thus the vectors are equal, establishing both direction and length.
Equivalent justified methods accepted; respect any requested proof method.
Exact values unless specified. Sketches assessed by mathematical features, not artistic quality.
Part a: Write a correct vector equality for opposite sides.
Part a: Deduce a+c=b+d. Part a: Compare the diagonal midpoints explicitly.
Part b: Express both midpoints and subtract.
Part b: Obtain MN=d−a and interpret it. Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.
View the QCAA syllabusHow many marks did you earn?Compare your working with the guide above.
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