Question

The unit vector which is perpendicular to \((2,-4,3)\) and \((5,0,1)\), is (A) \([\\{4 / \sqrt{(585)\\}},\\{13 / \sqrt{(585)\\},\\{20 / \sqrt{(585})\\}]}\) (B) \([\\{(-4) / \sqrt{(585)\\}},\\{13 / \sqrt{(585)\\}},\\{(-20) / \sqrt{(585)\\}}]\) (C) \([\\{(-4) / \sqrt{(585)\\}},\\{(-13) / \sqrt{(585)\\},\\{20 / \sqrt{(585})\\}]}\) (D) \([\\{(-4) / \sqrt{(585)\\},\\{13 / \sqrt{(585})\\},\\{20 / \sqrt{(585})\\}]}\)

Short answer

(D) \(\left[ \frac{-4}{\sqrt{585}}, \frac{13}{\sqrt{585}}, \frac{20}{\sqrt{585}} \right]\)

Step-by-step solution

Calculate the cross product of the given vectors

To find a vector perpendicular to the given vectors, we calculate the cross product. We know that cross product of \((a_1, a_2, a_3)\) and \((b_1, b_2, b_3)\) is given by \((a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1)\). Let vector \(A = (2,-4,3)\) and vector \(B = (5,0,1)\). Calculating the cross product of vector A and B, we get: \((-4\times1 - 3\times0, 3\times5 - 2\times1, 2\times0 - (-4)\times5) = (-4, 13, 20)\)

Calculate the magnitude of the cross product vector

Now we need to find the magnitude of the vector \((-4, 13, 20)\) which is given by \(\sqrt{(-4)^2 + 13^2 + 20^2}\). Calculation gives: \(\sqrt{(-4)^2 + 13^2 + 20^2} = \sqrt{16 + 169 + 400} = \sqrt{585}\)

Convert the cross product vector into a unit vector

To convert the cross product vector \((-4, 13, 20)\) into a unit vector, we need to divide each component by its magnitude obtained in Step 2. So, the unit vector is: \(\frac{(-4, 13, 20)}{\sqrt{585}} = \left( \frac{-4}{\sqrt{585}}, \frac{13}{\sqrt{585}}, \frac{20}{\sqrt{585}} \right)\)

Compare the calculated unit vector with the given options

The calculated unit vector is \(\left( \frac{-4}{\sqrt{585}}, \frac{13}{\sqrt{585}}, \frac{20}{\sqrt{585}} \right)\). Comparing this to the given options, we can see that it matches option (D). So, the correct answer is: (D) \(\left[\\{(-4) / \sqrt{(585)\\},\\{13 / \sqrt{(585})\\},\\{20 / \sqrt{(585})\\}]\right]\)

Key concepts

Unit Vector

In vector mathematics, a unit vector is a vector with a magnitude of one. It retains the direction of the original vector but scales its length to unity. This is significant because any vector can be broken down into its direction and magnitude, and converting to a unit vector isolates just the direction component. To obtain a unit vector from any non-zero vector, divide the vector by its magnitude.

For example, if we have a vector \(\vec{v} = (x, y, z)\), its magnitude is calculated using the formula \(\|\vec{v}\| = \sqrt{x^2 + y^2 + z^2}\). We get the unit vector \(\hat{v}\) by dividing \(\vec{v}\) by \(\|\vec{v}\|\), resulting in \(\hat{v} = (\frac{x}{\|\vec{v}\|}, \frac{y}{\|\vec{v}\|}, \frac{z}{\|\vec{v}\|})\). Thus, the magnitude of \(\hat{v}\) is 1, which means it only points in the direction of \(\vec{v}\) but does not have any length other than 1.

Vector Magnitude

The magnitude of a vector, often referred to as the vector's length, is a measure of how long the vector is. For a vector \(\vec{v}\) with components \(\vec{v} = (x, y, z)\), the magnitude is calculated using the Euclidean distance formula: \(\|\vec{v}\| = \sqrt{x^2 + y^2 + z^2}\).

In our example, the magnitude of the cross product vector \((−4, 13, 20)\) is \(\sqrt{(-4)^2 + 13^2 + 20^2} = \sqrt{585}\). This is a critical step in transforming any vector into a unit vector, as dividing by this magnitude scales the original vector down to a unit length while preserving its direction.

Vector Perpendicularity

Two vectors are perpendicular (or orthogonal) to each other if the angle between them is 90 degrees (π/2 radians). One way to ascertain whether two vectors are perpendicular is to compute their dot product; if the result is zero, the vectors are perpendicular. However, when we want to find a vector that is perpendicular to two given vectors in three-dimensional space, we use the cross product.

The cross product of vectors \(\vec{A}\) and \(\vec{B}\), denoted as \(\vec{A} \times \vec{B}\), results in a vector that is perpendicular to both \(\vec{A}\) and \(\vec{B}\). If the vectors are co-planar, the magnitude of the cross product will also reflect the area of the parallelogram spanned by the vectors. In the context of the exercise, the cross product was used to find a vector orthogonal to both \(\vec{A} = (2, -4, 3)\) and \(\vec{B} = (5, 0, 1)\), resulting in \(\vec{A} \times \vec{B} = (-4, 13, 20)\), which is then normalized to get the unit vector.