MATH 120/125Linear Algebra M.Solomonovich
Assignment # 2 SOLUTIONSVector algebra: dot product; cross product; triple product.
- It is known that Find all the values of t, for which vectors 3a + tb and 3a – tb are orthogonal.
Solution.
Let us write the orthogonality condition for the vectors 3a + tb and 3a – tb:
- It is given:. Find the norm of the vector m + 3n.
Solution.
For any vector v, . Then we can find the norm of m + 3n:
- a, b, and c are three unit vectors satisfying the condition . Evaluate .
Let us multiply the given condition by a; we shall obtain . Similarly, by multiplying the given condition by b and c we can obtain: . Let us add the obtained three equalities: , which, taking into account that (since a, b, and c are unit vectors) and the commutative property of dot product, can be rewritten as .
It is interesting to observe the geometric meaning of this result.
First, let us notice that for a unit vector u, its projection onto a vector v is equal to the cosine of the angle between v and u: .
Also, a projection of a vector onto a unit vector is equal to their dot product (why?).
Vectors a, b, and c form a regular triangle with the sides of length 1(see the figure below). Then, =.
- Vectors a and b are given by their components in the standard basis: a = (1, -2, 2); b = (1, 0, 1). Find: (i) the orthogonal projection of a onto a – 2b; (ii) an angle between a and a + b.
Solution.
; Then .
(It is not specified whether the scalar or vector projection to be found; the vector projection is ).
We can determine the cosine of the angle: .
- Evaluate the area of a parallelogram formed by the vectors 3a + b and a – b, if .
Solution.
- Vectors a and b are given by their components in the standard basis :
a = (2, -1, 1); b = (1, 0, -1). Find the unit vector perpendicular to both vectors a and b and forming an obtuse angle with the positive Y-semi-axis.
Solution.
A vector v perpendicular to a and b will be a scalar multiple of their cross product.
. Then a unit vector collinear with v will be .
Since the required vector forms an obtuse angle with the Y-axis, its second component must be negative (the cosine of an obtuse angle is negative); hence the required vector is .
- Four points are given by their coordinates: A(1, 0, 2); B(0, 1, 3); C(1, 1, -1); and D(1, 1, 4) .
(i)Evaluate the area of triangle BCD;
(ii)Determine, if these four points are lying in one plane;
(iii)Evaluate the volume of the pyramid ABCD.
Solution.
(i)The area is a half of the area of a parallelogram formed by any two vectors that represent sides of the triangle: .
(ii)Four points are coplanar iff any four vectors joining all of them are coplanar. The latter are coplanar iff their triple product is zero. ; hence the points are not coplanar.
(iii)