Question

Show that \(2 \mathbf{i}-\mathbf{j}+4 \mathbf{k}\) and \(5 \mathbf{i}+2 \mathbf{j}-2 \mathbf{k}\) are orthogonal (perpendicular). Find a third vector perpendicular to both.

Short answer

\(2 \textbf{i} - \textbf{j} + 4 \textbf{k}\) and \(5 \textbf{i} + 2 \textbf{j} - 2 \textbf{k}\) are orthogonal. A vector perpendicular to both is \(-6 \textbf{i} + 24 \textbf{j} + 9 \textbf{k}\).

Step-by-step solution

- Understand Orthogonality

Two vectors are orthogonal (perpendicular) if their dot product is zero. Define the vectors \(\textbf{a} = 2 \textbf{i} - \textbf{j} + 4 \textbf{k}\) and \(\textbf{b} = 5 \textbf{i} + 2 \textbf{j} - 2 \textbf{k}\).

- Calculate the Dot Product

Calculate the dot product of \(\textbf{a}\) and \(\textbf{b}\). Use the formula \( \textbf{a} \bullet \textbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \), where \(a_i\) and \(b_i\) are components of the vectors. \(\textbf{a} \bullet \textbf{b} = (2)(5) + (-1)(2) + (4)(-2) = 10 - 2 - 8 = 0\). Since the dot product is 0, these vectors are orthogonal.

- Find the Cross Product

To find a vector perpendicular to both vectors, compute the cross product \( \textbf{a} \times \textbf{b} \). Use the determinant formula: \[ \textbf{a} \times \textbf{b} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \ 2 & -1 & 4 \ 5 & 2 & -2 \ \end{vmatrix} = \textbf{i}((-1)(-2) - (4)(2)) - \textbf{j}((2)(-2) - (4)(5)) + \textbf{k}((2)(2) - ((-1)(5))) = \textbf{i}(2 - 8) - \textbf{j}(-4 - 20) + \textbf{k}(4 + 5) = -6 \textbf{i} + 24 \textbf{j} + 9 \textbf{k} \]

- Simplify the Result

The resulting vector from the cross product is \(-6 \textbf{i} + 24 \textbf{j} + 9 \textbf{k}\). This is a vector perpendicular to both original vectors.

Key concepts

dot product

One of the important concepts in vector analysis is the dot product. Think of the dot product as a way of combining two vectors to get a single number (a scalar). This tells us something about the angle between the vectors. The formula for the dot product of two vectors \(\textbf{a} = a_1 \textbf{i} + a_2 \textbf{j} + a_3 \textbf{k}\) and \(\textbf{b} = b_1 \textbf{i} + b_2 \textbf{j} + b_3 \textbf{k}\) is: \(\textbf{a} \bullet \textbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3\)
In our exercise, \(2 \textbf{i}- \textbf{j} +4 \textbf{k}\) and \( 5 \textbf{i}+ 2 \textbf{j}-2 \textbf{k} \) are given vectors.

• Calculate each product of corresponding components: \(2 \times 5, (-1) \times 2, 4 \times (-2) \)
• Sum these products: 10, -2, -8
• The total is 0, meaning the vectors are orthogonal!

cross product

Now, onto the cross product. Unlike the dot product, the cross product of two vectors results in another vector. This new vector is perpendicular (orthogonal) to the original two vectors. The cross product uses a determinant to perform its calculations.
For vectors \(\textbf{a}\) and \(\textbf{b}\) as defined above, we use:\[ \textbf{a} \times \textbf{b} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \ 2 & -1 & 4 \ 5 & 2 & -2 \ \end{vmatrix}\]
This resultant vector relies on computing three smaller 2x2 determinants:

• The \(\textbf{i}\) component: \((-1) \times (-2) - (4) \times (2) = 2 - 8 = -6\)
• The \(\textbf{j}\) component (note the negative sign): \( (2) \times (-2) - (4) \times (5) = -4 - 20 = -24\)
• The \(\textbf{k}\) component: \( (2) \times (2) - ((-1) \times (5)) = 4 + 5 = 9\)

Thus, the cross product gives us the vector \(-6 \textbf{i} + 24 \textbf{j} + 9 \textbf{k}\).

vector perpendicularity

Perpendicular vectors are fundamental in vector analysis. They serve as the building blocks for more complex shapes and interactions. In simpler terms, two vectors are perpendicular (or orthogonal) if they form a right angle when placed tail-to-tail.
To test this, we commonly check their dot product or their cross product, as described previously. If vectors \(\textbf{a}\) and \(\textbf{b}\) have a dot product of 0 (i.e., \(\textbf{a} \bullet \textbf{b} = 0\)), they are perpendicular. This is what we saw in our problem solution.
An additional method to ensure perpendicularity is by finding a new vector that meets this property. Thus, given two vectors, their cross product is always perpendicular to both original vectors. This method confirmed perpendicularity in our case, resulting in the vector \(-6 \textbf{i} + 24 \textbf{j} + 9 \textbf{k} \).

determinant method

The determinant method is typically used to compute the cross product of two vectors. Think of determinants as a special scalar value computed from a square matrix. They offer a way to find the volume of the parallelepiped spanned by three vectors.
The cross product's determinant is a 3x3 matrix, including the standard basis unit vectors \(\textbf{i}, \textbf{j},\textbf{k}\) in its top row and the components of the vectors \(\textbf{a}\) and \(\textbf{b}\) beneath it. For vectors \(2 \textbf{i}- \textbf{j} + 4 \textbf{k}\) and \(5 \textbf{i}+ 2 \textbf{j}- 2 \textbf{k}\):
The determinant is calculated as follows:\[\textbf{a} \times \textbf{b} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \ 2 & -1 & 4 \ 5 & 2 & -2 \ \end{vmatrix}\]

• For the \(\textbf{i}\) component: eliminate row and column of \(\textbf{i}\), yielding \((-1) \times (-2) - (4) \times (2) = 2 - 8 = -6\).
• For the \(\textbf{j}\) component: don't forget the negative, giving \(2 \times (-2) - 4 \times 5 = -4 - 20 = -24\)
• For the \(\textbf{k}\) component: yields \(2 \times 2 - (-1) \times 5 = 4 + 5 = 9 \)

So the cross product is \(-6 \textbf{i} + 24 \textbf{j} + 9 \textbf{k}\), which we confirmed earlier.