Lesson Video: Dot Product in 3D | Nagwa Lesson Video: Dot Product in 3D | Nagwa

# Lesson Video: Dot Product in 3D Mathematics

In this video, we will learn how to find the dot product of two vectors in 3D.

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### Video Transcript

In this video, we will learn how to find a dot product of two vectors in three dimensions. We will begin by looking at what of a vector in three dimensions looks like and some of its key properties.

A three-dimensional vector is an ordered triple such that vector π has components π one, π two, and π three. This can also be written as π one π plus π two π plus π three π, where π, π, and π at the standard unit vectors perpendicular to each other in three dimensions. If vector π has components π one, π two, and π three and vector π has components π one, π two, and π three, then the dot product of vector π and vector π is equal to π one π one plus π two π two plus π three π three. We multiply the corresponding components in each vector and then find the sum of these values. This will give us a scalar quantity and not a vector.

The dot product of vector π and vector π is also equal to the magnitude of vector π multiplied by the magnitude of vector π multiplied by the cos of angle π, where π is the angle between the vectors. This value of π must lie between zero and π radians or zero and 180 degrees.

We recall that the magnitude of vector π is equal to the square root of π one squared plus π two squared plus π three squared. We square each of the components, find the sum of these three values, and then square root our answer. Once again, this will give us a scalar quantity. And the magnitude of any nonzero vector will be greater than zero. We will now look at a question where we need to calculate the dot products of two vectors.

Given that vector π¨ is equal to negative six, negative three, five and vector π© is equal to seven, negative four, negative one, determine the dot product of vector π¨ and vector π©.

The three-dimensional vector π¨ has components π one, π two, and π three, whereas vector π© has components π one, π two, and π three. The dot product of these two vectors is equal to π one multiplied by π one plus π two multiplied by π two plus π three multiplied by π three. We find the product of the corresponding components and then find the sum of these three values.

In this question, we need to multiply negative six by seven, negative three by negative four, and five by negative one. Multiplying a negative number by a positive number gives a negative answer. Therefore, negative six multiplied by seven is equal to negative 42. Multiplying two negative numbers together gives a positive answer. Therefore, negative three multiplied by negative four is equal to 12. Finally, five multiplied by negative one is negative five.

The dot product of vector π¨ and vector π© is equal to negative 42 plus 12 plus negative five. Negative 42 plus 12 is equal to negative 30. And subtracting five from this gives us negative 35. The dot product of the vectors negative six, negative three, five and seven, negative four, negative one is negative 35.

In our next question, we will consider the properties of two perpendicular vectors.

For what value of π are vectors π¨ seven, negative seven π, negative six and π© seven, negative three, π perpendicular?

We recall that the dot product of any two vectors π¨ and π© is equal to the magnitude of vector π¨ multiplied by the magnitude of vector π© multiplied by the cos of π, where π is the angle between the two vectors.

If two vectors are perpendicular, as in this question, the angle between them will be equal to 90 degrees. We know that the cos of 90 degrees is equal to zero. This leads us to the fact that if two vectors are perpendicular, their dot product is equal to zero.

To find the dot products of two vectors in three dimensions, we find the product of their corresponding components and then the sum of these three values. Seven multiplied by seven is equal to 49. Multiplying negative seven π by negative three gives us positive 21π, as multiplying two negatives gives a positive answer. Negative six multiplied by π is equal to negative six π. And adding this is the same as subtracting six π.

This gives us the equation zero is equal to 49 plus 21π minus six π. Subtracting 49 from both sides and collecting like terms gives us negative 49 is equal to 15π. Finally, dividing both sides of this equation by 15 gives us π is equal to negative 49 over 15. This is the value of π such that vectors π¨ and π© are perpendicular.

In our next question, we need to decide which statement is true about the two vectors.

Which of the following is true of the vectors π¨ equal to negative three, seven, negative eight and π© negative six, negative one, negative one. Is it (A) they are parallel, (B) they are perpendicular, or (C) they are neither parallel nor perpendicular?

In order to answer this question, we need to recall the properties of two vectors when theyβre parallel or perpendicular. Two vectors are parallel if vector π¨ is equal to π multiplied by vector π©, where π is a scalar quantity not equal to zero. This means that each of their components must be multiplied by the same scalar.

In this question, the value of π would have to satisfy the three equations. Negative three is equal to negative six π, seven is equal to negative one π, and negative eight is equal to negative one π. It is clear from the second and third equation that this will not be correct, as negative one π cannot be equal to seven and negative eight. Dividing both sides of the bottom equation by negative one gives us π is equal to eight, whereas dividing both sides of the second equation by negative one gives us π is equal to negative seven. If we divide both sides of the top equation by negative six, we get π is equal to one-half.

As the value of π is not the same for all three equations, we can conclude that vector π¨ is not equal to a scalar quantity π multiplied by vector π©. This means that the two vectors are not parallel and option (A) is incorrect.

We know that two vectors are perpendicular if their dot product is equal to zero. We calculate the dot product of two vectors by firstly multiplying their corresponding components. We then find the sum of these three values. Negative three multiplied by negative six is equal to 18. Seven multiplied by negative one is negative seven. And adding this is the same as subtracting seven. Finally, negative eight multiplied by negative one is equal to eight. The dot product of vector π¨ and vector π© is equal to 18 minus seven plus eight. This is equal to 19.

As this is not equal to zero, the dot product of vector π¨ and vector π© is not equal to zero. We can, therefore, conclude that the two vectors are not perpendicular, so option (B) is also incorrect. The correct answer is, therefore, option (C). Vector π¨ and vector π© are neither parallel nor perpendicular.

In our final question, we will calculate the dot product of two vectors given their magnitudes and the angle between them.

Suppose vector π¨ is equal to negative one, two, sevens, the magnitude of vector π© is equal to 13, and the angle between the two vectors is 135 degrees. Find the dot product of vector π¨ and vector π© to the nearest hundredth.

We can calculate the dot product of vector π¨ and vector π© by finding their magnitudes, multiplying these together, and then multiplying this by the cos of angle π, where π is the angle between the two vectors. In this question, we are told that the magnitude of vector π© is equal to 13. And the angle between the two vectors is 135 degrees. In order to calculate the dot product of vector π¨ and vector π©, we firstly need to calculate the magnitude of vector π¨.

Vector π¨ is written in terms of its three components, which we will call π sub one, π sub two, and π sub three. The magnitude of any vector is a scalar quantity. It is equal to the square root of π sub one squared plus π sub two squared plus π sub three squared. In this question, we begin by squaring negative one, two, and seven. These are equal to one four and 49, respectively. One plus four plus 49 is equal to 54. Therefore, the magnitude of vector π¨ is equal to the square root of 54.

Using the laws of radicals or surds, this can be rewritten as the square root of nine multiplied by the square root of six, as nine multiplied by six is 54. The square root of nine is equal to three. Therefore, the square root of 54 can be rewritten as three root six. This is the magnitude of vector π¨.

We can now calculate the dot product of vector π¨ and vector π©. It is equal to three root six multiplied by 13 multiplied by the cos of 135 degrees. The cos of 135 degrees is equal to negative root two over two. This means that the dot product is equal to negative 39 root 12 over two. Root 12 can be rewritten as root four multiplied by root three. As root four is equal to two, this simplifies to negative 39 root three.

This isnβt the end of the question, however, as we are asked to give our answer to the nearest hundredth. Negative 39 root three is equal to negative 67.5499 and so on. Rounding this to two decimal places or the nearest hundredth is negative 67.55. The dot product of vector π¨ and vector π© to the nearest hundredth is negative 67.55.

We will now summarize the key points from this video. We saw in this video that if vector π¨ has components π one, π two, and π three and vector π© has components π one, π two, π three, then the dot product of vectors π¨ and π© is equal to π one multiplied by π one plus π two multiplied by π two plus π three multiplied by π three. We find the product of the corresponding components and then the sum of these three values.

We also saw that we can calculate the dot product by multiplying the magnitude of vector π¨ by the magnitude of vector π© by cos of π, where π is the angle between the two vectors. The dot product will always give us a scalar quantity and not a vector quantity. The same is true when calculating the magnitude. Finally, we saw that if two vectors are perpendicular, then their dot product is equal to zero. This is because two perpendicular vectors will meet at right angles and the cos of 90 degrees is equal to zero. This in turn means that the dot product will be equal to zero.