Question

Find a unit vector in the same direction as the vector \(A=4 i-2 j+4 k\), and another unit vector in the same direction as \(\mathbf{B}=-4 \mathrm{i}+3 \mathbf{k}\). Show that the vector sum of these unit vectors bisects the angle between \(\mathbf{A}\) and \(\mathbf{B}\). Hint : Sketch the rhombus having the two unit vectors as adjacent sides.

Short answer

The unit vector in the same direction as \(\textbf{A}\) is \(\textbf{u}_{A} = \frac{2}{3}\textbf{i} - \frac{1}{3}\textbf{j} + \frac{2}{3}\textbf{k}\), and for \(\textbf{B}\) it is \(\textbf{u}_{B} = -\frac{4}{5}\textbf{i} + \frac{3}{5}\textbf{k}\). Their sum bisects the angle between \(\textbf{A}\) and \(\textbf{B}\).

Step-by-step solution

Find the magnitude of vector A

First, compute the magnitude of vector \(\textbf{A} = 4\textbf{i} - 2\textbf{j} + 4\textbf{k} \). The magnitude \(\textbf{A} \) is given by \(\textbf{A}= \sqrt{(4)^2 + (-2)^2 + (4)^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6 \).

Compute the unit vector of A

Divide each component of \(\textbf{A}\) by its magnitude to get the unit vector: \(\textbf{u}_{A} = \frac{4}{6}\textbf{i} - \frac{2}{6}\textbf{j} + \frac{4}{6}\textbf{k} = \frac{2}{3}\textbf{i} - \frac{1}{3}\textbf{j} + \frac{2}{3}\textbf{k}\).

Find the magnitude of vector B

Compute the magnitude of vector \(\textbf{B} = -4\textbf{i} + 3\textbf{k}\). The magnitude \(\textbf{B}\) is given by \(\textbf{B}= \sqrt{(-4)^2 + (3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5\).

Compute the unit vector of B

Divide each component of \(\textbf{B}\) by its magnitude to get the unit vector: \(\textbf{u}_{B} = \frac{-4}{5}\textbf{i} + \frac{3}{5}\textbf{k} = -\frac{4}{5}\textbf{i} + \frac{3}{5}\textbf{k}\).

Find the vector sum of the unit vectors

Add the unit vectors \(\textbf{u}_{A} \) and \(\textbf{u}_{B}\): \(\textbf{u}_{A} + \textbf{u}_{B} = \frac{2}{3}\textbf{i} - \frac{1}{3}\textbf{j} + \frac{2}{3}\textbf{k} + -\frac{4}{5}\textbf{i} + \frac{3}{5}\textbf{k} \).

Simplify the vector sum

Combine the terms with the same unit vectors: \((\frac{2}{3} - \frac{4}{5})\textbf{i} + (-\frac{1}{3})\textbf{j} + (\frac{2}{3} + \frac{3}{5})\textbf{k} = (\frac{10}{15} - \frac{12}{15})\textbf{i} - \frac{1}{3}\textbf{j} + (\frac{10}{15} + \frac{9}{15})\textbf{k} = -\frac{2}{15}\textbf{i} - \frac{1}{3}\textbf{j} + \frac{19}{15}\textbf{k}\).

Show that the vector sum bisects the angle

Since the vector sum \(\textbf{u}_{A} + \textbf{u}_{B}\) is proportional to the direction of \(\textbf{A}\+\textbf{B}\), and if the two vectors are unit vectors, their sum will have equal contributions from both original directions. This implies that the sum vector bisects the angle between \(\textbf{A}\) and \(\textbf{B}\).

Key concepts

Unit Vector

A unit vector is a vector with a magnitude of exactly 1, representing direction only. To compute a unit vector, you need to divide each component of a vector by its magnitude.
The formula for finding the unit vector \(\textbf{u} \) of a vector \(\textbf{V} \) is given by:
\(\textbf{u} = \frac{\textbf{V}}{|\textbf{V}|}\).
In this problem, we computed the magnitudes of vectors \(\textbf{A} \) and \(\textbf{B} \), and then divided each component by their respective magnitudes to find the corresponding unit vectors. This process ensures that we maintain the direction of the original vectors while standardizing their length to 1.

Vector Magnitude

The magnitude of a vector represents its length. It’s computed as the square root of the sum of the squares of its components.
If you have a vector \(\textbf{V} = a\textbf{i} + b\textbf{j} + c\textbf{k}\), its magnitude \(|\textbf{V}|\) is given by:
\[|\textbf{V}| = \sqrt{a^2 + b^2 + c^2}\].
In our example, the magnitudes of \(\textbf{A}\) and \(\textbf{B}\) were calculated as follows:
\(|\textbf{A}| = 6\) and \(|\textbf{B}| = 5\).
This calculation is essential for normalizing the vector to obtain the unit vector.

Vector Sum

The vector sum is the addition of two vectors by adding their corresponding components. This concept is useful in understanding linear combinations of vectors.
For two vectors \(\textbf{u}_{A}\) and \(\textbf{u}_{B}\), the vector sum can be written as:
\(\textbf{u}_{A} + \textbf{u}_{B} = \frac{2}{3}\textbf{i} - \frac{1}{3}\textbf{j} + \frac{2}{3}\textbf{k} + -\frac{4}{5}\textbf{i} + \frac{3}{5}\textbf{k}\).
This step-by-step addition of the components resulted in:
\(\textbf{u}_{A} + \textbf{u}_{B} = -\frac{2}{15}\textbf{i} - \frac{1}{3}\textbf{j} + \frac{19}{15}\textbf{k}\).
This vector represents the combined effect of both unit vectors.

Angle Bisector

The bisector of the angle between two vectors divides the angle into two equal parts. When the sum of two unit vectors is taken, it results in a vector that is aligned along the angle bisector of the original vectors.
In this problem, the vector sum \(\textbf{u}_{A} + \textbf{u}_{B}\) represents a direction that evenly splits the angle between vectors \(\textbf{A}\) and \(\textbf{B}\).
This result is demonstrated by the symmetry of the unit vectors and proves that their sum bisects the angle. The direction vector reflects equal contribution from both unit vectors, confirming that it lies on the angle bisector.