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Secondary school › Vectors

Vectors in coordinates

Position vectors, length, equal vectors, operations, unit vectors and components. 9 problems with step-by-step solutions.

Solve each problem in your notebook first, then open the solution.

Problem 1 Quadrants and position vectors

a) Name the quadrant or axis of each point: \(A(4, -1)\), \(B(-2, -5)\), \(C(0, 3)\), \(D(-6, 2)\), \(E(7, 0)\).

b) Find the position vector of \(\vec{PQ}\) for \(P(3, -2)\) and \(Q(-1, 4)\).

Show solution

a) \(A\): IV; \(B\): III; \(C\): on the \(y\)-axis; \(D\): II; \(E\): on the \(x\)-axis.

b) \(\vec{PQ} = (x_Q - x_P,\ y_Q - y_P) = (-4, 6)\).

Problem 2 Length of a vector

a) Find the length of \(\vec{u} = (5, -12)\), \(\vec{v} = (0, -7)\) and \(\vec{w} = (\sqrt3, 1)\).

b) Find the length of \(\vec{AB}\) for \(A(-2, 1)\) and \(B(4, 9)\).

c) \(A(a, 1)\), \(B(3, 5)\) and \(|AB| = 5\). Find \(a\).

Show solution

a) \(|\vec{u}| = \sqrt{25 + 144} = 13\); \(|\vec{v}| = 7\); \(|\vec{w}| = \sqrt{3 + 1} = 2\).

b) \(\vec{AB} = (6, 8)\), \(|\vec{AB}| = 10\).

c) \((3 - a)^2 + 16 = 25\), so \((3 - a)^2 = 9\) and \(a = 0\) or \(a = 6\).

Problem 3 Equal vectors

a) \(\vec{u} = (a - b, 4)\) and \(\vec{v} = (6, a + b)\) are equal. Find \(a\) and \(b\).

b) The vector \(\vec{v} = (5, -3)\) has initial point \((-1, 2)\). Find its terminal point.

c) \(\vec{MN} = (4, -1)\), \(M(1, k)\) and \(N(p, 3)\). Find \(p + k\).

Show solution

a) \(a - b = 6\) and \(a + b = 4\): \(a = 5\), \(b = -1\).

b) \((-1 + 5,\ 2 - 3) = (4, -1)\).

c) \(p - 1 = 4\), so \(p = 5\); \(3 - k = -1\), so \(k = 4\). \(p + k = 9\).

Problem 4 Parallelogram vertex

a) \(A(1, 2)\), \(B(5, 3)\), \(C(4, 7)\) are three vertices of the parallelogram \(ABCD\). Find \(D\).

b) \(A(2, -1)\), \(B(-3, 4)\), \(C(1, 0)\) and \(D(a, b)\) are given with \(\vec{AC} = \vec{BD}\). Find \(D\) and \(|\vec{AD}|\).

Show solution

a) \(\vec{AD} = \vec{BC} = (-1, 4)\), so \(D = (1 - 1,\ 2 + 4) = (0, 6)\).

b) \(\vec{AC} = (-1, 1)\) and \(\vec{BD} = (a + 3, b - 4)\). So \(a = -4\), \(b = 5\): \(D(-4, 5)\).

\(\vec{AD} = (-6, 6)\), \(|\vec{AD}| = 6\sqrt2\).

Problem 5 Operations

\(\vec{u} = (2, -3)\), \(\vec{v} = (-1, 4)\) and \(\vec{w} = (3, 0)\). Find:

a) \(\vec{u} + \vec{v}\)   b) \(\vec{u} - \vec{w}\)   c) \(3\vec{u} - 2\vec{v}\)   d) \(|2\vec{u} + \vec{v} - \vec{w}|\)   e) \(\vec{x}\) such that \(2\vec{x} + \vec{u} = \vec{w}\)

Show solution
  • a) \((1, 1)\)
  • b) \((-1, -3)\)
  • c) \((6, -9) - (-2, 8) = (8, -17)\)
  • d) \((4, -6) + (-1, 4) - (3, 0) = (0, -2)\), length \(2\)
  • e) \(2\vec{x} = \vec{w} - \vec{u} = (1, 3)\), so \(\vec{x} = \left(\frac12, \frac32\right)\)

Problem 6 Equations with vectors

a) \(M(2, 1)\), \(N(3, m)\), \(K(m, 4)\) and \(|\vec{MN} + \vec{NK}| = 5\). Find \(m\).

b) \(3\vec{OA} + 2\vec{OB} = (7, 4)\) and \(\vec{OA} - \vec{OB} = (-1, 3)\). Find \(\vec{OA}\) and \(\vec{OB}\).

c) \(\vec{AB} = (m - 1, m)\) and \(\vec{CD} = (m + 1, 2)\), and \(|\vec{AB} + \vec{CD}| = 10\). Find \(m\).

Show solution

a) \(\vec{MN} + \vec{NK} = \vec{MK} = (m - 2, 3)\). \((m - 2)^2 + 9 = 25\), so \(m - 2 = \pm 4\): \(m = 6\) or \(m = -2\).

b) Let \(\vec{OA} = \vec{x}\), \(\vec{OB} = \vec{y}\). From the second, \(\vec{x} = \vec{y} + (-1, 3)\). Then \(5\vec{y} + (-3, 9) = (7, 4)\), so \(\vec{y} = (2, -1)\) and \(\vec{x} = (1, 2)\).

c) \(\vec{AB} + \vec{CD} = (2m, m + 2)\). \(4m^2 + (m + 2)^2 = 100 \Rightarrow 5m^2 + 4m - 96 = 0 \Rightarrow m = 4\) or \(m = -\frac{24}{5}\).

Problem 7 Unit vectors

a) Find the unit vector in the direction of \(\vec{a} = (3, -4)\).

b) \(\vec{OA} = 2\vec{i} + 3\vec{j}\) and \(\vec{OB} = 5\vec{i} - \vec{j}\). Find the unit vector in the direction of \(\vec{AB}\).

c) \(\vec{OA} = 6\vec{i} + 8\vec{j}\) and \(C\) is on \(OA\) with \(|OC| = 4\). Find \(\vec{OC}\).

Show solution

a) \(|\vec{a}| = 5\): \(\left(\frac35, -\frac45\right)\).

b) \(\vec{AB} = 3\vec{i} - 4\vec{j}\), length 5: \(\frac35\vec{i} - \frac45\vec{j}\).

c) The unit vector of \(\vec{OA}\) is \(\frac{1}{10}(6, 8) = (0.6, 0.8)\). \(\vec{OC} = 4\cdot(0.6, 0.8) = 2.4\vec{i} + 3.2\vec{j}\).

Problem 8 Components and direction angle

a) Write \(\vec{u} = (-7, 2)\) with \(\vec{i}\) and \(\vec{j}\).

b) \(|\vec{v}| = 6\) and \(\vec{v}\) makes the angle \(150^\circ\) with the positive \(x\)-axis. Find its components.

c) Find the angle between the positive \(x\)-axis and \(\vec{w} = 2\vec{i} - 2\sqrt3\,\vec{j}\).

Show solution

a) \(\vec{u} = -7\vec{i} + 2\vec{j}\).

b) \(\vec{v} = (6\cos 150^\circ,\ 6\sin 150^\circ) = (-3\sqrt3,\ 3)\).

c) \(|\vec{w}| = 4\), \(\cos\theta = \frac12\), \(\sin\theta = -\frac{\sqrt3}{2}\): \(\theta = 300^\circ\) (or \(-60^\circ\)).

Problem 9 Right triangle in coordinates

\(A(1, 1)\), \(B(4, 5)\) and \(C(8, 2)\) are the vertices of a triangle. Find \(|AB|\), \(|BC|\), \(|AC|\) and show that the triangle is a right isosceles triangle.

Show solution

\(\vec{AB} = (3, 4)\), \(|AB| = 5\); \(\vec{BC} = (4, -3)\), \(|BC| = 5\); \(\vec{AC} = (7, 1)\), \(|AC| = \sqrt{50}\).

\(|AB|^2 + |BC|^2 = 25 + 25 = 50 = |AC|^2\), so the angle at \(B\) is right, and \(|AB| = |BC|\): right isosceles.

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