Linear algebra and analytic geometry. Vector algebra

Problems on vectors, angles and products.

Problems

Problem 1 (Dividing a segment into three equal parts). The segment with endpoints $A (2; - 5; 4)$ and $B (8; 1; - 2)$ is divided by the points $M$ and $P$ into three equal parts (with $P$ closer to $B$ than to $A$ ). Find the coordinates of $P$ .

Solution.

The division points are easiest to find through the vector $A B$ : if $M$ and $P$ divide the segment into three equal parts, then $A M = \frac{1}{3} A B$ , $A P = \frac{2}{3} A B$ .

Compute the vector itself:

A B = B - A = (8 - 2; 1 - (- 5); - 2 - 4) = (6; 6; - 6) .

The point $P$ is closer to $B$ , so it lies two thirds of the way from $A$ :

P = A + \frac{2}{3} A B = (2; - 5; 4) + \frac{2}{3} (6; 6; - 6) = (2 + 4; - 5 + 4; 4 - 4) .

Check: $P B = (2; 2; - 2) = \frac{1}{3} A B$ — the remaining third, as it should be.

Answer: $P (6; - 1; 0)$ .

Problem 2 (Finding a vertex of a triangle). In the triangle $A B C$ :

the point $M (4; 3; - 1)$ is the midpoint of the side $A B$ ,
the point $P (1; 0; 5)$ is the midpoint of the side $B C$ ,
the point $K (- 2; 6; - 3)$ is the midpoint of the side $A C$ .

Find the coordinates of the vertex $C$ .

Solution.

Denote the position vectors of the vertices by $A$ , $B$ , $C$ . The midpoint conditions give the system

\frac{A + B}{2} = M, \frac{B + C}{2} = P, \frac{A + C}{2} = K .

Add all three equations: the left-hand side becomes $A + B + C$ and the right-hand side $M + P + K$ , that is, $A + B + C = M + P + K$ .

The vertex $C$ is not on the side $A B$ , whose midpoint is $M$ . So subtract $A + B = 2 M$ : $C = (M + P + K) - 2 M = P + K - M$ .

Substitute the coordinates:

C = (1; 0; 5) + (- 2; 6; - 3) - (4; 3; - 1) = (1 - 2 - 4; 0 + 6 - 3; 5 - 3 + 1) .

Check: by the same trick $A = M + K - P = (1; 9; - 9)$ and $B = M + P - K = (7; - 3; 7)$ . Then $\frac{B + C}{2} = (1; 0; 5) = P$ and $\frac{A + C}{2} = (- 2; 6; - 3) = K$ — both midpoints agree.

Answer: $C (- 5; 3; 3)$ .

Problem 3 (A vector in terms of basis vectors). $A B C D E F$ is a regular hexagon, $O$ is its centre, and $M$ is the midpoint of the side $B C$ . Let $O C = a$ and $O E = b$ .

Express the vector $O M$ in terms of $a$ and $b$ .

Solution.

The key property of a regular hexagon: its opposite vertices are symmetric about the centre. The vertices $B$ and $E$ are three sides apart, so they are opposite, and therefore $O B = - O E = - b$ .

The point $M$ is the midpoint of $B C$ , and the midpoint of a segment is half the sum of the position vectors of its ends:

O M = \frac{O B + O C}{2} .

It remains to substitute:

O M = \frac{- b + a}{2} = \frac{1}{2} a - \frac{1}{2} b .

A check in coordinates: placing $O$ at the origin and the vertices on the unit circle $6 0^{\circ}$ apart gives $a = (- \frac{1}{2}; \frac{3}{2})$ , $b = (- \frac{1}{2}; - \frac{3}{2})$ , $O M = (0; \frac{3}{2})$ — exactly $\frac{1}{2} (a - b)$ .

Answer: $O M = \frac{a - b}{2}$ .

Problem 4 (The angle between vectors). The vectors $a$ and $b$ are given in a rectangular coordinate system. It is known that $∣ a ∣ = 3$ , $∣ b ∣ = 7$ , $∣2 a - b ∣ = 61$ .

Find the angle between the vectors $a$ and $b$ .

Solution.

The length is known not for the vectors themselves but for a combination of them, so expand the squared length through the dot product:

∣2 a - b ∣^{2} = (2 a - b) \cdot (2 a - b) = 4∣ a ∣^{2} - 4 (a \cdot b) + ∣ b ∣^{2} .

Substitute the data $∣ a ∣ = 3$ , $∣ b ∣ = 7$ , $∣2 a - b ∣^{2} = 61$ :

61 = 4 \cdot 9 - 4 (a \cdot b) + 49 = 85 - 4 (a \cdot b),

whence $a \cdot b = 6$ .

Now the cosine of the angle follows from the definition of the dot product:

cos φ = \frac{a \cdot b}{∣ a ∣ ∣ b ∣} = \frac{6}{3 \cdot 7} = \frac{2}{7} .

Answer: $φ = arccos \frac{2}{7} \approx 73. 4^{\circ}$ .

Problem 5 (The angle between vectors in a triangle). The vertices of a triangle are $A (4; - 2; - 5)$ , $B (- 2; 3; - 3)$ , $C (3; - 5; 4)$ .

Find the angle between the vectors $A B$ and $A C$ .

Solution.

The angle at the vertex $A$ is the angle between the vectors of the sides leaving $A$ . First, the vectors themselves:

A B = B - A = (- 2 - 4; 3 + 2; - 3 + 5) = (- 6; 5; 2),

A C = C - A = (3 - 4; - 5 + 2; 4 + 5) = (- 1; - 3; 9) .

The dot product:

A B \cdot A C = (- 6) (- 1) + 5 \cdot (- 3) + 2 \cdot 9 = 6 - 15 + 18 = 9.

The lengths:

∣ A B ∣ = 36 + 25 + 4 = 65, ∣ A C ∣ = 1 + 9 + 81 = 91 .

Hence

cos φ = \frac{9}{65 \cdot 91} .

The radicand factorises: $65 = 5 \cdot 13$ , $91 = 7 \cdot 13$ , so $65 \cdot 91 = 1 3^{2} \cdot 35$ and $65 \cdot 91 = 1335$ . Then

cos φ = \frac{9}{13 35} = \frac{9 35}{455} \approx 0.117.

The typical mistake here is to take $B A$ instead of $A B$ , or to lose a sign in a coordinate difference: the angle then comes out obtuse instead of acute.

Answer: $φ = arccos \frac{9 35}{455} \approx 83. 3^{\circ}$ .

Problem 6 (The area of a parallelogram). In a rectangular coordinate system $∣ a ∣ = 3$ , $∣ b ∣ = 4$ , and the angle between $a$ and $b$ is $12 0^{\circ}$ .

Find the area of the parallelogram spanned by the vectors $3 a + b$ and $a + 2 b$ .

Solution.

The area of the parallelogram spanned by two vectors equals the length of their cross product. Expand the product by bilinearity:

[3 a + b, a + 2 b] = 3 [a, a] + 6 [a, b] + [b, a] + 2 [b, b] .

The cross product of a vector with itself is zero, and $[b, a] = - [a, b]$ , so what remains is $6 [a, b] - [a, b] = 5 [a, b]$ .

The length of the original product:

[a, b] = ∣ a ∣ ∣ b ∣ sin 12 0^{\circ} = 3 \cdot 4 \cdot \frac{3}{2} = 6.

So $S = 5 \cdot 6 = 30$ .

It is easy to go wrong twice here: forgetting that $[b, a]$ enters with a minus sign (the coefficient then comes out as $7$ and the answer as $42$ ), and using the cosine instead of the sine.

Answer: $S = 30$ .

Problem 7 (The cross product). In a right-handed orthonormal basis $i, j, k$ the vectors $a = (1; - 2; 2)$ and $b = (- 3; 4; 1)$ are given.

Find the cross product $[2 a - b, a + b]$ .

Solution.

First simplify the expression itself, without touching the coordinates:

[2 a - b, a + b] = 2 [a, a] + 2 [a, b] - [b, a] - [b, b] = 3 [a, b] .

One cross product is left instead of expanding two long sums. Compute it as a determinant:

[a, b] = i 1 - 3 j - 2 4 k 21 = i ((- 2) \cdot 1 - 2 \cdot 4) - j (1 \cdot 1 - 2 \cdot (- 3)) + k (1 \cdot 4 - (- 2) (- 3)) .

Component by component: $- 2 - 8 = - 10$ ; $- (1 + 6) = - 7$ ; $4 - 6 = - 2$ . That is, $[a, b] = (- 10; - 7; - 2)$ .

Multiply by 3: $3 [a, b] = (- 30; - 21; - 6)$ .

Check: the result must be orthogonal to both original vectors. $(- 30; - 21; - 6) \cdot (1; - 2; 2) = - 30 + 42 - 12 = 0$ and $(- 30; - 21; - 6) \cdot (- 3; 4; 1) = 90 - 84 - 6 = 0$ — it agrees. This check catches both a lost minus in the second component (it is taken with a minus sign) and swapped rows of the determinant.

Answer: $(- 30; - 21; - 6)$ .

Problem 8 (The scalar triple product). The vectors

a = - i + j - 3 k, b = 2 i + j + 2 k, c = 3 i - 2 j - k

are given.

Find the scalar triple product $⟨ a, b, c ⟩$ .

Solution.

The scalar triple product equals the determinant of the matrix whose rows are the coordinates of the vectors:

⟨ a, b, c ⟩ = - 1 23 11 - 2 - 3 2 - 1 .

Expand along the first row:

- 1 \cdot 1 - 2 2 - 1 - 1 \cdot 23 2 - 1 + (- 3) \cdot 23 1 - 2 .

The minors: $1 \cdot (- 1) - 2 \cdot (- 2) = 3$ ; $2 \cdot (- 1) - 2 \cdot 3 = - 8$ ; $2 \cdot (- 2) - 1 \cdot 3 = - 7$ .

Add up:

- 1 \cdot 3 - 1 \cdot (- 8) + (- 3) \cdot (- 7) = - 3 + 8 + 21 = 26.

The sign is positive, so the triple $a, b, c$ is right-handed, and the absolute value $26$ is the volume of the parallelepiped they span.

The minus sign in front of the second minor is what gets lost most often: expansion along the first row alternates $+, -, +$ .

Answer: $26$ .

Problems

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