Two vectors a and b are given. (a) Find a vector...

Two vectors a and b are given. (a) Find a vector perpendicular to both a and b. (b) Find a unit vector perpendicular to both a and b.
$$
\mathbf{a}=3 \mathbf{j}+5 \mathbf{k}, \quad \mathbf{b}=-\mathbf{i}+2 \mathbf{k}
$$

Key Concepts

Vector Cross Product

The cross product is an operation on two vectors in three-dimensional space that results in a new vector which is perpendicular to both original vectors. This concept is crucial for finding a normal vector to a plane defined by the two vectors and relies on the magnitude of the vectors as well as the sine of the angle between them.

Unit Vector

A unit vector is a vector that has a magnitude of one and indicates direction. It is obtained by dividing any non-zero vector by its magnitude, often used to normalize a vector, which is essential when the direction is required without scaling by any length.

Normalization

Normalization is the process of converting a vector into a unit vector. This is done by dividing the vector by its magnitude, ensuring that the new vector maintains its original direction while standardizing its length to one.

Transcript

00:01 This question gives us two vectors and asks us to find a vector that's perpendicular to both of them.

00:05 We have a special tool for this, and that's going to be the cross product.

00:09 We know that when we have two vectors a cross b, that's always going to be perpendicular to a and perpendicular to b.

00:19 This was shown in your book by calculating a cross b dot product a.

00:26 The dot product, when two vectors are perpendicular, is always going to be equal to zero.

00:33 And we proved that no matter what a and b are, a crossb .a will equal zero.

00:39 Same is true for b, so a cross b is perpendicular both to a and to b.

00:45 So let's jump into it.

00:47 The vectors that we're given are a, which is equal to 3j plus 5k.

00:54 Now this is a linear combination of our unit vectors, which we can pretty easily turn into just one vector.

01:03 It's going to be 0, 3, 5.

01:08 The second vector that we're given is b, which is equal to negative i plus 2k.

01:16 Again, that's going to be negative 1, 0, and 2.

01:21 So let's start doing this.

01:23 We know that to take the cross product, we are going to be finding the determinant of a 3x3.

01:30 Matrix whose first row are our unit vectors i jk.

01:34 Our third and four or second and third row will be our vectors, so 035 and negative 1 02.

01:44 So let's start.

01:46 We know that taking the determinant of a 3x3 matrix we're going to do co -factor decomposition and i'm going to use the first row because it's going to be far easier than anything else.

01:56 So what we'll have is the determinant of minor 1 -1 which is just the vector or the matrix without row 1.

02:02 In column 1, 305, 2, multiply that by i.

02:08 Then we're going to subtract, remember we're subtracting because of where j is in the matrix because it's element 1 ,2, and 1 and 2 add up to an odd number.

02:18 We're going to have to subtract it.

02:20 So we're subtracting minor 1, 2, which is 0 negative 1, 5, 2 times j, and then we're adding, again, we're adding minor 1 3, which is 0, 1, 3 ,0 times k, our unit vector k.

02:38 So now we can do this...

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