Two 4-Mark Vector Questions

This lesson works through two exam-style vector questions. The first finds the values of a parameter for which the angle between two vectors is obtuse, and the second proves the magnitude of a sum of three vectors that are each perpendicular to the sum of the other two.

Question 1: values of $\lambda$ for an obtuse angle

We are given the two vectors

$\vec a = 2\lambda^2\,\hat i + 4\lambda\,\hat j + \hat k, \qquad \vec b = 7\,\hat i - 2\,\hat j + \lambda\,\hat k.$

We want the values of $\lambda$ for which the angle $\theta$ between $\vec a$ and $\vec b$ is obtuse, that is $\theta > 90^\circ$.

The angle formula

The angle between two vectors satisfies

$\cos\theta = \frac{\vec a \cdot \vec b}{|\vec a|\,|\vec b|}.$

The dot product

The dot product is the sum of the products of corresponding components:

$\vec a \cdot \vec b = (2\lambda^2)(7) + (4\lambda)(-2) + (1)(\lambda) = 14\lambda^2 - 8\lambda + \lambda = 14\lambda^2 - 7\lambda.$

Obtuse means the cosine is negative

If $\theta$ is obtuse, then $\theta > 90^\circ$, so $\cos\theta < 0$. Therefore

$\frac{\vec a \cdot \vec b}{|\vec a|\,|\vec b|} < 0.$

The magnitudes $|\vec a|$ and $|\vec b|$ are always positive, so the denominator is positive. For the fraction to be negative, the numerator must be negative:

$\vec a \cdot \vec b < 0.$

Solving the inequality

$14\lambda^2 - 7\lambda < 0 \;\Rightarrow\; 7\lambda(2\lambda - 1) < 0.$

The critical points are $\lambda = 0$ and $\lambda = \tfrac{1}{2}$. The product $7\lambda(2\lambda - 1)$ is negative between these roots, so

$\lambda \in \left(0, \tfrac{1}{2}\right).$

The angle is obtuse precisely when $\lambda$ lies in the open interval $\left(0, \tfrac{1}{2}\right)$.

Question 2: magnitude of $\vec a + \vec b + \vec c$

Let $\vec a$, $\vec b$, $\vec c$ be three vectors such that each one is perpendicular to the sum of the other two, with $|\vec a| = 3$, $|\vec b| = 4$, and $|\vec c| = 5$. We prove that $|\vec a + \vec b + \vec c| = 5\sqrt{2}$.

The perpendicularity conditions

Since each vector is perpendicular to the sum of the other two, their dot products vanish:

$\vec a \cdot (\vec b + \vec c) = 0, \qquad \vec b \cdot (\vec a + \vec c) = 0, \qquad \vec c \cdot (\vec a + \vec b) = 0.$

Expanding the square of the magnitude

$|\vec a + \vec b + \vec c|^2 = (\vec a + \vec b + \vec c) \cdot (\vec a + \vec b + \vec c).$

Grouping the terms by the leading vector:

$= \vec a \cdot \vec a + \vec a \cdot (\vec b + \vec c) + \vec b \cdot \vec b + \vec b \cdot (\vec a + \vec c) + \vec c \cdot \vec c + \vec c \cdot (\vec a + \vec b).$

Each of the mixed terms is zero by the perpendicularity conditions, and $\vec a \cdot \vec a = |\vec a|^2$, so

$|\vec a + \vec b + \vec c|^2 = |\vec a|^2 + |\vec b|^2 + |\vec c|^2.$

Substituting the magnitudes

$|\vec a + \vec b + \vec c|^2 = 3^2 + 4^2 + 5^2 = 9 + 16 + 25 = 50.$

Therefore

$|\vec a + \vec b + \vec c| = \sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}.$

Hence proved.

Key takeaways

• An angle is obtuse exactly when the dot product of the two vectors is negative, since the magnitudes in the denominator are always positive.
• A parameter range for an obtuse angle comes from solving the quadratic inequality $\vec a \cdot \vec b < 0$.
• When each of three vectors is perpendicular to the sum of the other two, $|\vec a + \vec b + \vec c|^2 = |\vec a|^2 + |\vec b|^2 + |\vec c|^2$.