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Otherwise the vectors pointing from the point (0, 1) to these points, that is(5, 8


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MA 1111: Linear Algebra I

Selected answers/solutions to the assignment due October 8, 2015

1.(a) No. Otherwise the vectors pointing from the point (0, 1) to these points, that is(5, 8) = (5, 9) − (0, 1)and(8, 13) = (8, 14) − (0, 1), would have been proportional, which is not the case. (But they are very close to being proportional, so drawing this on a grid paper might lead to a wrong answer).

(b) Yes. Computing the vectors along the sides of that triangle, we get the vectors (3, 4) = (1, 1) − (−2,−3), (8,−6) = (9,−5) − (1, 1), and (11,−2) = (9,−5) − (−2,−3). The scalar product of the first two is (3, 4)·(8,−6) =24−24 =0, so the cosine of the angle between them is 0, and these vectors form a right angle.

2. (−2, 0), (0,−2), or (2, 4). In general, if a, b, and c are given points, then the fourth point is one of a+b−c, b+c−a, and c+a−b. One of possible ideas is to use the parallelogram rule carefully. Another idea:

the midpoint of the segment connecting (a1, a2) to (b1, b2) has coordinates (a1+b2 1,a2+b2 2); use the fact that the center of a parallelogram is the midpoint of each of its diagonals.

3.(a) No. It is not linear inv, since

(u·(v1+v2))(v1+v2) − (u·v1)v1− (u·v2)v2=

(u·v1)v2+ (u·v2)v1.

That latter expression is nonzero for some choices ofu,v1,v2. For example, if u=v1=v2=i, the result is 2i.

(b) As we established in class, the cross product of two vectors is perpen- dicular to both of them, so(u×v)·v=0always. As such, it is a multilinear expression: 0+0=0.

4.We have |b| =√


5,|c| =√

1+32 =√

10, andb·c= 7, so we havecosϕ= 7


50. We also have|u|=√


38|, andu·v=18, socosϕ= 18


38 = 6

38. 5. (a) This area, as we know, is equal to the length of the vector product of these vectors. We have u ×v = (4,−1,−1), so the area is



2. (b) This area is the absolute value ofw·(u×v), that is|12−1|=11.

6.Note that the vectorsv−u= (1, 1, 3) andw−u= (2,−2,−1)are in this plane, so their cross product is perpendicular to this plane. We have

(v−u)×(w−u) = (5, 7,−4).

This vector is perpendicular to the plane (as is any scalar multiple of it).


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