The dot product

Dot product, angle between vectors, perpendicular vectors, area, lines, projection and work. 11 problems with step-by-step solutions.

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

Problem 1 Dot product

a) Find \(\vec{u}\cdot\vec{v}\) for \(\vec{u} = (3, -2)\), \(\vec{v} = (4, 5)\).

b) Find \((\vec{u} + 2\vec{v})\cdot(\vec{u} - \vec{v})\) for \(\vec{u} = (1, 2)\), \(\vec{v} = (-2, 1)\).

c) Find the length of \(\vec{w} = (-3, 3)\) using the dot product.

Show solution

a) \(12 - 10 = 2\).

b) \(\vec{u} + 2\vec{v} = (-3, 4)\), \(\vec{u} - \vec{v} = (3, 1)\): \(-9 + 4 = -5\).

c) \(|\vec{w}| = \sqrt{\vec{w}\cdot\vec{w}} = \sqrt{18} = 3\sqrt2\).

Problem 2 Dot product with points

a) \(A(1, -2)\), \(B(3, 1)\), \(C(-1, 4)\). Find \(\vec{AB}\cdot\vec{CB}\).

b) \(A(-1, 2)\), \(B(2, 3)\), \(C(m, 1)\), \(D(0, -1)\) and \(\vec{AB}\cdot\vec{DC} = 10\). Find \(m\).

c) \(\vec{u} = (-5, m - 2)\), \(\vec{v} = (4, m + 2)\) and \(\vec{u}\cdot\vec{v} = 0\). Find \(m\).

Show solution

a) \(\vec{AB} = (2, 3)\), \(\vec{CB} = (4, -3)\): \(8 - 9 = -1\).

b) \(\vec{AB} = (3, 1)\), \(\vec{DC} = (m, 2)\): \(3m + 2 = 10\), \(m = \frac83\).

c) \(-20 + m^2 - 4 = 0\), \(m = \pm 2\sqrt6\).

Problem 3 Angle between vectors

a) Find the cosine of the angle between \(\vec{u} = (1, 3)\) and \(\vec{v} = (3, 1)\).

b) Find the angle between \(\vec{u} = (1, \sqrt3)\) and \(\vec{v} = (\sqrt3, 1)\).

c) Find the angle between \(\vec{p} = (2, -2)\) and \(\vec{q} = (-3, 0)\).

Show solution

\(\cos\theta = \dfrac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}|}\).

a) \(\frac{6}{\sqrt{10}\sqrt{10}} = \frac35\).

b) \(\frac{2\sqrt3}{2\cdot 2} = \frac{\sqrt3}{2}\), \(\theta = 30^\circ\).

c) \(\frac{-6}{2\sqrt2\cdot 3} = -\frac{\sqrt2}{2}\), \(\theta = 135^\circ\).

Problem 4 Angle in a triangle

a) \(A(1, 1)\), \(B(4, 2)\), \(C(2, 5)\). Find \(\cos A\).

b) The angle between \(\vec{u} = (a, 1)\) and \(\vec{v} = (1, a)\) is \(60^\circ\). Find \(a\).

Show solution

a) \(\vec{AB} = (3, 1)\), \(\vec{AC} = (1, 4)\): \(\cos A = \frac{3 + 4}{\sqrt{10}\sqrt{17}} = \frac{7}{\sqrt{170}} \approx 0.537\).

b) \(\frac{2a}{a^2 + 1} = \frac12 \Rightarrow a^2 - 4a + 1 = 0 \Rightarrow a = 2 \pm \sqrt3\).

Problem 5 Perpendicular vectors

a) Are \(\vec{u} = (4, -6)\) and \(\vec{v} = (3, 2)\) perpendicular? Are \((2, 5)\) and \((5, 2)\)?

b) \(\vec{u} = (a, 2)\) is perpendicular to \(\vec{v} = (5, -3)\). Find \(a\).

c) Write two vectors perpendicular to \(\vec{w} = (3, -1)\).

d) \(\vec{A} = (k, 3)\), \(\vec{B} = (-2, k - 1)\) and \(\vec{A}\perp\vec{B}\). Find \(|\vec{A} + \vec{B}|\).

Show solution

a) \(12 - 12 = 0\): yes. \(10 + 10 = 20 \ne 0\): no.

b) \(5a - 6 = 0\), \(a = \frac65\).

c) For example \((1, 3)\) and \((-2, -6)\) (any \(t(1, 3)\), \(t \ne 0\)).

d) \(-2k + 3k - 3 = 0\), \(k = 3\). \(\vec{A} + \vec{B} = (1, 5)\), length \(\sqrt{26}\).

Problem 6 Length from the dot product

a) \(|\vec{u}| = 3\), \(|\vec{v}| = 2\) and the angle between them is \(60^\circ\). Find \(|2\vec{u} - 3\vec{v}|\).

b) \(|\vec{a}| = 5\), \(|\vec{b}| = 3\), \(|\vec{a} + \vec{b}| = 7\). Find \(\vec{a}\cdot\vec{b}\) and the angle between \(\vec{a}\) and \(\vec{b}\).

Show solution

a) \(\vec{u}\cdot\vec{v} = 3\cdot 2\cdot\frac12 = 3\). \(|2\vec{u} - 3\vec{v}|^2 = 4\cdot 9 - 12\cdot 3 + 9\cdot 4 = 36\), so the length is \(6\).

b) \(49 = 25 + 2\vec{a}\cdot\vec{b} + 9\), so \(\vec{a}\cdot\vec{b} = \frac{15}{2}\). \(\cos\theta = \frac{7.5}{15} = \frac12\), \(\theta = 60^\circ\).

Problem 7 Geometry with the dot product

a) In an equilateral triangle \(ABC\) with side 6, \(M\) is the midpoint of \(BC\). Find \(\vec{AM}\cdot\vec{BC}\) and \(\vec{AB}\cdot\vec{AC}\).

b) In a square \(ABCD\) with side 6, \(E\) is the midpoint of \(CD\). Find \(\vec{AE}\cdot\vec{AB}\) and \(\vec{AE}\cdot\vec{BD}\).

Show solution

a) The median is also the altitude, so \(\vec{AM}\perp\vec{BC}\) and \(\vec{AM}\cdot\vec{BC} = 0\). \(\vec{AB}\cdot\vec{AC} = 6\cdot 6\cdot\cos 60^\circ = 18\).

b) Coordinates: \(A(0, 0)\), \(B(6, 0)\), \(C(6, 6)\), \(D(0, 6)\), \(E(3, 6)\). \(\vec{AE} = (3, 6)\), \(\vec{AB} = (6, 0)\): \(18\). \(\vec{BD} = (-6, 6)\): \(-18 + 36 = 18\).

Problem 8 Area of a triangle

a) Find the area of the triangle with vertices \(A(1, 2)\), \(B(5, 3)\), \(C(2, 6)\).

b) In triangle \(ABC\), \(|BA| = 5\), \(|BC| = 8\) and \(\vec{BA}\cdot\vec{BC} = 20\). Find the area.

Show solution

a) \(\vec{AB} = (4, 1)\), \(\vec{AC} = (1, 4)\). Area \(= \frac12|4\cdot 4 - 1\cdot 1| = \frac{15}{2}\).

b) \(\cos B = \frac{20}{40} = \frac12\), \(\sin B = \frac{\sqrt3}{2}\). Area \(= \frac12\cdot 5\cdot 8\cdot\frac{\sqrt3}{2} = 10\sqrt3\).

Problem 9 Line through a point

a) Find the equation of the line through \(A(2, -1)\) perpendicular to \(\vec{n} = (4, 3)\).

b) Find the equation of the line through \(B(-3, 2)\) parallel to \(\vec{d} = (2, 5)\).

Show solution

a) A point \(P(x, y)\) is on the line when \(\vec{AP}\cdot\vec{n} = 0\): \(4(x - 2) + 3(y + 1) = 0\), i.e. \(4x + 3y - 5 = 0\).

b) A normal vector is \((5, -2)\): \(5(x + 3) - 2(y - 2) = 0\), i.e. \(5x - 2y + 19 = 0\).

Problem 10 Scalar component (projection)

a) Find the scalar component of \(\vec{u} = (3, 4)\) along \(\vec{v} = (1, -2)\).

b) Find the scalar component of \(\vec{u} = (2, 5)\) along \(\vec{v} = (-3, 4)\) and the vector projection of \(\vec{u}\) on \(\vec{v}\).

Show solution

Scalar component: \(\dfrac{\vec{u}\cdot\vec{v}}{|\vec{v}|}\).

a) \(\frac{3 - 8}{\sqrt5} = -\sqrt5\).

b) \(\frac{-6 + 20}{5} = \frac{14}{5}\). Vector projection: \(\frac{14}{5}\cdot\frac{(-3, 4)}{5} = \left(-\frac{42}{25}, \frac{56}{25}\right)\).

Problem 11 Forces and work

a) A car weighing 12 000 N stands on a road inclined \(20^\circ\). Find the force needed to stop it rolling down and the force pressing on the road.

b) A force \(\vec{F} = (3, 5)\) N moves an object from \(A(1, 1)\) to \(B(5, 3)\) (in metres). Find the work done.

c) A sledge is pulled 50 m with a force of 30 N along a rope that makes \(40^\circ\) with the ground. Find the work.

Show solution

a) Along the road: \(12000\sin 20^\circ \approx 4104\) N. Against the road: \(12000\cos 20^\circ \approx 11276\) N.

b) \(\vec{AB} = (4, 2)\): \(W = \vec{F}\cdot\vec{AB} = 12 + 10 = 22\) J.

c) \(W = 30\cdot 50\cos 40^\circ \approx 1149\) J.