Orthogonal vectors are defined and examples are presented along with their detailed solutions.

Two vectors \( \textbf x =
\begin{bmatrix}
x_1 \\
x_2 \\
. \\
. \\
. \\
x_n
\end{bmatrix}
\) and
\( \textbf y =
\begin{bmatrix}
y_1 \\
y_2 \\
. \\
. \\
. \\
y_n
\end{bmatrix}
\)
are orthogonal if and only if their inner product \( \textbf x \cdot \textbf y \) is equal to zero.

Hence

Vectors \( \textbf x \) and \( \textbf y \) are orthogonal \( \iff \) \( \textbf x \cdot \textbf y = x_1 y_1 + x_2 y_2 + ... + x_n y_n = 0 \)

Below are shown two 2-dimensional vetors \( \textbf x = \begin{bmatrix}
3 \\
2
\end{bmatrix} \) and \( \textbf y = \begin{bmatrix}
-2 \\
3
\end{bmatrix} \) whose inner product is given by \( \textbf x \cdot \textbf y = (3)(-2)+(-2)(3) = 6 - 6 = 0 \) and therefore are orthogonal. They make and angle of \( 90^{\circ} \) or \( \dfrac{\pi}{2} \)

Below are shown two 3-dimensional vetors \( \textbf u = \begin{bmatrix}
-2 \\
2 \\
3
\end{bmatrix} \) and \( \textbf v = \begin{bmatrix}
2\\
-1 \\
2
\end{bmatrix} \) whose inner product is given by \( \textbf x \cdot \textbf y = (-2)(2)+(2)(-1) +(3)(2) = -4 - 2 + 6 = 0 \) and therefore are orthogonal. They make and angle of \( 90^{\circ} \) or \( \dfrac{\pi}{2} \)

Example 1

a) Are the vectors \( \textbf x = \begin{bmatrix}
-1 \\
3 \\
1
\end{bmatrix} \) and
\( y = \begin{bmatrix}
-2 \\
3 \\
-11
\end{bmatrix} \) orthogonal?

b) Are the vectors \( \textbf u = \begin{bmatrix}
0 \\
-3 \\
1 \\
8
\end{bmatrix} \) and
\( \textbf v = \begin{bmatrix}
3 \\
5 \\
-2 \\
2
\end{bmatrix} \) orthogonal?

Solution to Example 1

a)

The inner product of vactors \( \textbf x \) and \( \textbf y \) is given by

\( \textbf x \cdot \textbf y = = (-1)(-2)+(3)(3)+(1)(-11) = 2 +9 -11 = 0 \)

The inner product is equal to zero and therefore the two vectors \( \textbf x \) and \( \textbf y \) are orthogonal.

b)

The inner product of vactors \( \textbf u \) and \( \textbf v \) is given by

\( \textbf u \cdot \textbf v = = (0)(3) + (-3)(5) + (1)(-2) + (8)(2) = 0 -15 -2 + 16 = -1 \)

The inner product is NOT equal to zero and therefore the two vectors \( \textbf x \) and \( \textbf y \) are NOT orthogonal.

Example 2

Let vectors \( \textbf u = \begin{bmatrix}
1\\
0 \\
-1
\end{bmatrix} \) ,
\( \textbf v = \begin{bmatrix}
-2 \\
1 \\
-2
\end{bmatrix} \) . Find a vector \( \textbf w \) that is orthogonal to vectors \( \textbf u \) and \( \textbf v \).

Solution to Example 2

Using determinants of matrices, the cross product of vectors
\( \textbf u = \begin{bmatrix}
u_x \\
u_y \\
u_z
\end{bmatrix}
\) and \( \textbf v = \begin{bmatrix}
v_x \\
v_y \\
v_z
\end{bmatrix} \) is a vector defined by:

that is orthogonal to both vectors \( \textbf u \) and \( \textbf v \).

Hence one way to find vector \( \textbf w \) orthogonal to vectors \( \textbf u \) and \( \textbf v \) is to calculate the cross product as defined above.

Note that all vectors parallel to vector \( \textbf w \) which are of the form \( k \textbf w \) , where \( k \) is an integer, are orthogonal to the vectors \( \textbf u \) and \( \textbf v \).

As an exercise, check that any vector given by \( k \textbf w \) is orthogonal to both vectors \( \textbf u \) and \( \textbf v \).

Example 3

Find the constants \( a \) and \( b \) such that the vectors \( \textbf u
\begin{bmatrix}
a\\
4 \\
-b
\end{bmatrix}
\) and
\( \textbf u
\begin{bmatrix}
a\\
1 \\
b
\end{bmatrix}
\) are orthogonal and \( a = b + 1\).

Solution to Example 3

Vectors \( \textbf u \) and \( \textbf v \) are orthogonal, hence their inner product is equal to zero

\( a^2 + 4 - b^2 = 0 \)

Given that \( a = b + 1\),substitute \( a \) by \( b + 1\) in the above equation

\( (b+1)^2 + 4 - b^2 = 0 \)

Expand the above equation and simplify

\( 2 b + 5 = 0 \)

Solve for \( b \)

\( b = -\dfrac{5}{2} \)

Find \( a \)

\( a = b + 1 = -\dfrac{5}{2} + 1 = -\dfrac{3}{2} \)

Hence vectors \( \textbf u \) and \( \textbf v \) defined above are orthogonal for \( a = -\dfrac{3}{2} \) and \( b = -\dfrac{5}{2} \) .

Example 4

\( \textbf u \), \( \textbf v \) and \( \textbf w \) are vectors in \( R^n \) such that \[ \textbf w = \textbf v - 4 \alpha \textbf u \] where \( \alpha \) is a real number.

Find \( \alpha \) if \( \textbf u \) and \( \textbf w \) are orthogonal, the norm of \( \textbf u \) is equal to 5 and \( \textbf u^T \textbf v = 3 \).

Solution to Example 4

Since \( \textbf u \) and \( \textbf w \) are orthogonal, their inner product is equal to zero; hence

\( \textbf u \cdot \textbf w = 0 \)

Given that \( \textbf w = \textbf v - 4 \alpha \textbf u \), substitute \( \textbf w \) by \( \textbf v - 4 \alpha \textbf u \) in the above equation

\( \textbf u \cdot (\textbf v - 4 \alpha \textbf u) = 0 \)

Use distributivity and associativity of inner product to expand the left side of the above equation

\( \textbf u \cdot \textbf v - 4 \alpha \textbf u \cdot \textbf u = 0 \) (I)

The expression \( \textbf u^T \textbf v \) is another form in writing the inner product of \( \textbf u \) and \( \textbf v \) ; hence

\( \textbf u \cdot \textbf v = \textbf u^T \textbf v = 3 \)

The norm of a vector is defined as

\( ||\textbf u|| = \sqrt {\textbf u \cdot \textbf u} \)

Given that the norm of \( \textbf u \) written as \( ||\textbf u|| \) is given and equal to 5, we have

\( \textbf u \cdot \textbf u = ||\textbf u||^2 = 5^2 \)

Substitute \( \textbf u \cdot \textbf v \) and \( \textbf u \cdot \textbf u \) by their values in equation (I) above to obtain

\( 3 - 4 \alpha (25) = 0 \)

Solve the above equation for \( \alpha \) to obtain

\( \alpha = \dfrac{3}{100} \)

Example 5

Find all values of \( \theta \) such that the vectors \( \textbf u
\begin{bmatrix}
\sin(\theta) \\
\cos(\theta)
\end{bmatrix}
\) and
\( \textbf v
\begin{bmatrix}
-1 \\
1
\end{bmatrix}
\) are orthognal.

Solution to Example 5

For the vectors \( \textbf u \) and \( \textbf v \) to be orthogonal, the inner product must be equal to 0. Hence

\( \sin(\theta) (-1) + \cos(\theta) (1) = 0 \)

The above equations may be written as

\( \sin(\theta) = \cos(\theta) \)

Divide both sides of the above equation by \( \cos(\theta) \)

\( \dfrac{\sin(\theta)}{\cos(\theta)} = \dfrac{\cos(\theta)}{\cos(\theta)} \)

Simplify and rewrite as

\( \tan (\theta ) = 1 \)

Solve for \( \theta \)

\( \theta = \dfrac{\pi}{4} + n \pi \)

Example 6

Let \( \textbf u =
\begin{bmatrix}
\sin(\theta) \cos (\phi) \\
\sin(\theta) \sin (\phi) \\
\cos (\theta)
\end{bmatrix}
\) ,
\( \textbf v =
\begin{bmatrix}
\cos(\theta) \cos (\phi) \\
\cos(\theta) \sin (\phi) \\
-\sin (\theta)
\end{bmatrix}
\) and
\( \textbf w =
\begin{bmatrix}
-\sin (\phi) \\
\cos (\phi) \\
0
\end{bmatrix}
\). Proove that each of these vectors is orthogonal to the other two vectors.

Solution to Example 7

We calculate the inner product of each pair of vectors

\( \textbf u \cdot \textbf v = [\sin(\theta) \cos (\phi)] \; [\cos(\theta) \cos (\phi)] + [\sin(\theta) \sin (\phi)] \; [\cos(\theta) \sin (\phi)] + [\cos (\theta)] \; [-\sin (\theta)] \)

Factor \( \sin(\theta) \cos(\theta) \) out of the first and second terms

\( = [\sin(\theta) \cos(\theta) ] \; [\cos^2 (\phi) + \sin^2 (\phi)] - \cos (\theta) \sin (\theta) \)

Use the identity \( \cos^2 (\phi) + \sin^2 (\phi) = 1 \) to simplify

\( = \sin(\theta) \cos(\theta) - \cos (\theta) \sin (\theta) = 0 \)

\( \textbf u \cdot \textbf w = [\sin(\theta) \cos (\phi)] \; [-\sin (\phi)] + [\sin(\theta) \sin (\phi)] \; [\cos (\phi)] +[\cos (\theta)][0] \)

Factor \( \sin(\theta) \cos(\phi) \) out of the first and second terms and simplify

\( = [ \sin(\theta) \cos(\phi) ] \; [ -\sin (\phi) + \sin (\phi)] + 0 = 0 \)

\( \textbf v \cdot \textbf w = [\cos(\theta) \cos (\phi)] \; [-\sin (\phi)] + [\cos(\theta) \sin (\phi)] \; [\cos (\phi)] +[-\sin (\theta)] \; [0] \)

Factor \( \cos(\theta) \cos(\phi) \) out of the first and second terms and simplify

\( = [ \cos(\theta) \cos(\phi) ] \; [ -\sin (\phi) + \sin (\phi)] + 0 = 0 \)

The inner product of each pair of vectors is equal to zero and therefore each vector is orthogonal to the other two vectors.

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