Angles and perpendicular vectors

What you will be able to do

• find the angle between two vectors, and between two lines;
• test whether two vectors are perpendicular, and find the value that makes them so;
• use the dot product in triangles, parallelograms and cubes.

Is the corner square?

A builder checks that a corner is a right angle by measuring. A drawing on a computer cannot be measured with a tape. It needs a test that uses numbers only. The dot product is that test.

The angle between two vectors

Rearrange the definition of the dot product:

The inverse cosine gives exactly one angle in this range, so there is no choice to make. If the value is $-1$ the angle is $\pi$, which means the vectors point in opposite directions.

Take a cube of edge $1$ with one corner at the origin. Its edges along the axes are $\hat i$, $\hat j$, $\hat k$. The diagonal of the front face is $\hat i + \hat j$, and the diagonal of the right face is $\hat j + \hat k$.

1. Two face diagonals. $\cos\theta = \dfrac{(1)(0) + (1)(1) + (0)(1)}{\sqrt{2}\sqrt{2}} = \dfrac{1}{2}$, so $\theta = 60^\circ$.
2. A body diagonal and an edge. The body diagonal is $\hat i + \hat j + \hat k$. With the edge $\hat i$, $\cos\theta = \dfrac{1}{\sqrt{3}}$, so $\theta \approx 54.74^\circ$.
3. Two body diagonals. The diagonals are $\hat i + \hat j + \hat k$ and $\hat i + \hat j - \hat k$. Then $\cos\phi = \dfrac{1 + 1 - 1}{3} = \dfrac{1}{3}$ and $\phi \approx 70.53^\circ$. The other angle at the centre is $180^\circ - 70.53^\circ = 109.47^\circ$.

Perpendicular vectors

If $\vec a$ and $\vec b$ are non-zero, then

The vectors $2\hat i - 3\hat j + \hat k$ and $4\hat i + \hat j - 5\hat k$ have dot product $8 - 3 - 5 = 0$. So they are perpendicular. To find a missing value, put the dot product equal to zero. For $\vec a = \lambda\hat i + 2\hat j - \hat k$ and $\vec b = 3\hat i - \hat j + 4\hat k$ the condition is $3\lambda - 2 - 4 = 0$, so $\lambda = 2$.

In the plane, a vector at right angles to $x\hat i + y\hat j$ is $-y\hat i + x\hat j$, because $x(-y) + y(x) = 0$. A unit vector perpendicular to $3\hat i + 4\hat j$ is $\pm\dfrac{-4\hat i + 3\hat j}{5}$.

Useful facts

• Diagonals of a parallelogram. For a parallelogram with sides $\vec a$ and $\vec b$, the diagonals are $\vec a + \vec b$ and $\vec a - \vec b$. Their dot product is $(\vec a + \vec b)\cdot(\vec a - \vec b) = \lvert\vec a\rvert^2 - \lvert\vec b\rvert^2$. So the diagonals are perpendicular exactly when the sides are equal, which is a rhombus.
• Equal diagonals. $\lvert\vec a + \vec b\rvert = \lvert\vec a - \vec b\rvert$ exactly when $\vec a\cdot\vec b = 0$. The parallelogram is then a rectangle.
• A right angle in a triangle. The angle at a vertex is a right angle when the dot product of the two vectors along its sides is zero.
• Direction cosines. Two lines with direction cosines $l_1, m_1, n_1$ and $l_2, m_2, n_2$ have $\cos\theta = l_1l_2 + m_1m_2 + n_1n_2$, because the direction cosines are the components of unit vectors.

Lines and vectors. Two lines meet in two angles that add up to $180^\circ$. For the angle between two lines, the books take the acute one, so use $\cos\theta = \dfrac{\lvert\vec a\cdot\vec b\rvert}{\lvert\vec a\rvert\lvert\vec b\rvert}$. For the angle between two vectors, the angle may be obtuse, so keep the sign.

A worked example

Show that the triangle $PQR$ with $P(3, 1, -1)$, $Q(5, 3, 0)$ and $R(7, -1, -5)$ has a right angle at $P$.

1. The sides from $P$. $\overrightarrow{PQ} = (2, 2, 1)$ and $\overrightarrow{PR} = (4, -2, -4)$.
2. The dot product. $2(4) + 2(-2) + 1(-4) = 8 - 4 - 4 = 0$. So $\overrightarrow{PQ} \perp \overrightarrow{PR}$.
3. A check with Pythagoras. $\lvert\overrightarrow{PQ}\rvert^2 = 9$ and $\lvert\overrightarrow{PR}\rvert^2 = 36$. $\overrightarrow{QR} = (2, -4, -5)$ has $\lvert\overrightarrow{QR}\rvert^2 = 45 = 9 + 36$.

Practise with new numbers

Try it in Python

The program finds the angles of the cube in the lesson and tests a right angle.

Common mistakes

• Using the angle between the lines as obtuse. For lines, take the acute angle with $\lvert\vec a\cdot\vec b\rvert$.
• Forgetting to divide by both magnitudes.
• Calling a vector perpendicular to itself. Only the zero vector has $\vec a\cdot\vec a = 0$.
• Using the wrong vectors at a corner. For the angle at $P$ use $\overrightarrow{PQ}$ and $\overrightarrow{PR}$, both starting at $P$.
• Taking $\cos^{-1}$ of a number outside $[-1, 1]$. That means an arithmetic mistake.

Remember these

• $\cos\theta = \dfrac{\vec a\cdot\vec b}{\lvert\vec a\rvert\lvert\vec b\rvert}$, and $0 \le \theta \le \pi$.
• $\vec a \perp \vec b \iff \vec a\cdot\vec b = 0$, for non-zero vectors.
• $(\vec a + \vec b)\cdot(\vec a - \vec b) = \lvert\vec a\rvert^2 - \lvert\vec b\rvert^2$.
• The angle between lines uses $\lvert\vec a\cdot\vec b\rvert$, so it is acute or right.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the angle between $\vec a = \hat i - \hat j$ and $\vec b = \hat j - \hat k$.

2. For what value of $\lambda$ are $\vec a = \lambda\hat i + 3\hat j - 2\hat k$ and $\vec b = 3\hat i - \hat j + 3\hat k$ perpendicular?

3. If $\lvert\vec a\rvert = \lvert\vec b\rvert$, show that $\vec a + \vec b$ and $\vec a - \vec b$ are perpendicular.

4. Find a unit vector in the plane that is perpendicular to $5\hat i + 12\hat j$.

5. The dot product of two direction vectors of two lines is $-6$, and their magnitudes are $3$ and $4$. What is the angle between the two lines?

References

• NCERT, Mathematics Part II for Class XII, Chapter 10, section on the scalar product and its applications.
• OpenStax, Calculus Volume 3, Chapter 2, section on the dot product.