Lesson Video: Parallel and Perpendicular Vectors in Space | Nagwa Lesson Video: Parallel and Perpendicular Vectors in Space | Nagwa

Lesson Video: Parallel and Perpendicular Vectors in Space Mathematics

In this video, we will learn how to recognize parallel and perpendicular vectors in space.

16:25

Video Transcript

In this video, we will learn how to recognize parallel and perpendicular vectors in space. We will begin by looking at the conditions that must be true for two vectors to be parallel or perpendicular.

Two vectors 𝐀 and 𝐁 are parallel if and only if they are scalar multiples of each other. Vector 𝐀 must be equal to π‘˜ multiplied by vector 𝐁, where π‘˜ is a constant not equal to zero. If vector 𝐀 has components π‘Ž sub π‘₯, π‘Ž sub 𝑦, and π‘Ž sub 𝑧 and vector 𝐁 has components 𝑏 sub π‘₯, 𝑏 sub 𝑦, 𝑏 sub 𝑧, then the vectors are parallel if the ratio of their components are equal. π‘Ž sub π‘₯ over 𝑏 sub π‘₯ must be equal to π‘Ž sub 𝑦 over 𝑏 sub 𝑦, which is also equal to π‘Ž sub 𝑧 over 𝑏 sub 𝑧. To prove that two vectors are parallel, we must show that these conditions are true.

Let’s now consider the conditions for two vectors to be perpendicular. Two vectors 𝐀 and 𝐁 are perpendicular if and only if their scalar product, sometimes known as the dot product, is equal to zero. If vector 𝐀 once again has components π‘Ž sub π‘₯, π‘Ž sub 𝑦, π‘Ž sub 𝑧 and vector 𝐁 has components 𝑏 sub π‘₯, 𝑏 sub 𝑦, 𝑏 sub 𝑧, then the vectors are perpendicular if π‘Ž sub π‘₯, 𝑏 sub π‘₯ plus π‘Ž sub 𝑦, 𝑏 sub 𝑦 plus π‘Ž sub 𝑧, 𝑏 sub 𝑧 is equal to zero. We find the sum of the product of the π‘₯-, 𝑦-, and 𝑧-components, and if this is equal to zero, the vectors are perpendicular.

Once again, to prove that two vectors are perpendicular, we must show that the above conditions are true. We will now look at some questions where we need to determine whether vectors are perpendicular or parallel.

Determine whether the following is true or false: If the component of a vector in the direction of another vector is zero, then the two are parallel.

Let’s begin by considering two vectors 𝐀 and 𝐁 as shown. The component of vector 𝐀 on vector 𝐁 is shown in the diagram. We can see that this creates a right angle triangle. And we can let the angle at the origin be πœƒ. In right angle trigonometry, the cosine ratio is equal to the adjacent over the hypotenuse. We know that the hypotenuse is equal to the magnitude of vector 𝐀. And if we let the component of vector 𝐀 on vector 𝐁 be equal to π‘₯, then cos πœƒ is equal to π‘₯ over the magnitude of vector 𝐀. Rearranging this equation, we get that π‘₯ is equal to the magnitude of vector 𝐀 multiplied by cos πœƒ.

We’re told in the question that π‘₯ is equal to zero. Therefore, zero is equal to the magnitude of vector 𝐀 multiplied by cos πœƒ. The magnitude of any of vector must be positive; it could not equal zero. This means that cos πœƒ must be equal to zero. πœƒ is therefore equal to 90 degrees. This means that the two vectors are actually perpendicular. We can therefore conclude that the statement is false. The vectors are not parallel but are perpendicular.

In our next question, we will find the missing value using a pair of parallel vectors.

Find the values of π‘š and 𝑛 so that vector two 𝐒 plus seven 𝐣 plus π‘šπ€ is parallel to vector six 𝐒 plus 𝑛𝐣 minus 21𝐀.

We know that two vectors are parallel if they are scalar multiples of each other. Vector 𝐀 must be equal to π‘˜ multiplied by vector 𝐁, where π‘˜ is a constant not equal to zero. In this question, two 𝐒 plus seven 𝐣 plus π‘šπ€ must be equal to the constant π‘˜ multiplied by six 𝐒 plus 𝑛 𝐣 minus 21𝐀. By distributing the parentheses or expanding the brackets on the right-hand side and then equating components, we see that two is equal to six π‘˜. Equating the 𝐣-components, we have seven is equal to π‘›π‘˜. Equating the 𝐀-components, we have π‘š is equal to negative 21 multiplied by the constant π‘˜.

We can solve the first equation by dividing both sides by six. Therefore, π‘˜ is equal to two over six or two-sixths. This fraction can be simplified so that π‘˜ is equal to one-third. We can then substitute this value into our other two equations. Seven is equal to 𝑛 multiplied by one-third. Multiplying both sides of this equation by three gives us 𝑛 is equal to 21. In our third equation, π‘š is equal to negative 21 multiplied by one-third. One-third of 21 is equal to seven. Therefore, one-third of negative 21 is equal to negative seven. If our two vectors are parallel, then π‘š is equal to negative seven and 𝑛 is equal to 21.

An alternative method here would be to write each of the components as a ratio. We could do this either way round. By putting the larger 𝐒-component on top, we have six over two is equal to 𝑛 over seven which is equal to negative 21 over π‘š. Six divided by two is equal to three. So we can therefore solve three is equal to 𝑛 over seven and three is equal to negative 21 over π‘š to calculate the values of π‘š and 𝑛. This would once again give us answers of 𝑛 equals 21 and π‘š equals negative seven.

In our next question, we will identify the vector that is not perpendicular to a given vector.

Which of the following vectors is not perpendicular to the line whose direction vector 𝐫 is six, negative five? Is it (A) 𝐫 is equal to negative five, negative six? (B) 𝐫 is equal to five, six. (C) 𝐫 is equal to 10, 12. Or (D) 𝐫 is equal to 12, negative 10.

We recall that two vectors are perpendicular if their scalar or dot product is equal to zero. To calculate the scalar product, we find the sum of the product of the individual components. For option (A), we need to multiply six by negative five and then add negative five multiplied by negative six. Multiplying a positive number by a negative number gives a negative answer. And multiplying two negative numbers together gives a positive answer. This leaves us with negative 30 plus 30. As this is equal to zero, the two vectors are perpendicular. Option (A) is therefore not the correct answer.

Repeating this process for option (B), we have six multiplied by five plus negative five multiplied by six. This simplifies to 30 plus negative 30. Once again, this is equal to zero. So option (B) is not correct. In option (C), we have six multiplied by 10 plus negative five multiplied by 12. Six multiplied by 10 is equal to 60, and negative five multiplied by 12 is equal to negative 60. Option (C) is not the correct answer as the scalar product equals zero.

In option (D), our calculation is six multiplied by 12 plus negative five multiplied by negative 10. Six multiplied by 12 is 72, and negative five multiplied by negative 10 is positive 50. This means that the scalar product is equal to 122. This is not equal to zero. Option (D) is therefore the correct answer. The vector 12, negative 10 is not perpendicular to the line whose direction vector is six, negative five. This is because their scalar product is not equal to zero.

In our next question, we need to identify whether two vectors are parallel, perpendicular, or neither.

Given the two vectors 𝐀 is equal to eight 𝐒 minus seven 𝐣 plus 𝐀 and 𝐁 is equal to 64𝐒 minus 56𝐣 plus eight 𝐀, determine whether these two vectors are parallel, perpendicular, or otherwise.

We know that two vectors are parallel if they are scalar multiples of each other. Vector 𝐀 must be equal to π‘˜ multiplied by vector 𝐁. Two vectors 𝐀 and 𝐁 are perpendicular on the other hand, if their scalar or dot product is equal to zero. Let’s firstly consider whether our two vectors 𝐀 and 𝐁 are parallel. If one vector is a scalar multiple of another vector, then the ratio of their individual components must be equal. In this case, 64 over eight must be equal to negative 56 over negative seven, which must be equal to eight over one. 64 divided by eight is equal to eight, and eight divided by one is equal to eight.

Dividing a negative number by a negative number gives a positive answer. Therefore, negative 56 divided by negative seven is also equal to eight. We can therefore conclude that vector 𝐁 is equal to eight multiplied by vector 𝐀 or vector 𝐀 is equal to one-eighth of vector 𝐁. The two vectors 𝐀 and 𝐁 are therefore parallel. Whilst they cannot be parallel and perpendicular, let’s just check that the scalar product is not equal to zero. The 𝐒-components of our vector are eight and 64. The 𝐣-components are negative seven and negative 56. The 𝐀-components are one and eight. Eight multiplied by 64 is 512. Negative seven multiplied by negative 56 is 392. One multiplied by eight is equal to eight. The scalar product of vectors 𝐀 and 𝐁 is therefore equal to 912. As this is not equal to zero, the vectors 𝐀 and 𝐁 are not perpendicular.

In our final question, we will solve a problem involving a pair of perpendicular straight lines.

If the straight line π‘₯ plus eight over negative 10 is equal to 𝑦 plus eight over π‘š which is equal to 𝑧 plus 10 over negative eight is perpendicular to π‘₯ plus five over negative four which is equal to 𝑦 plus eight over 10 and 𝑧 equals eight, find π‘š.

We notice that both our straight lines in this question are given in symmetric form. π‘₯ minus π‘₯ sub zero over π‘Ž is equal to 𝑦 minus 𝑦 sub zero over 𝑏 which is equal to 𝑧 minus 𝑧 sub zero over 𝑐. Any straight line in this form can be rewritten in vector form π‘₯ sub zero, 𝑦 sub zero, 𝑧 sub zero plus 𝑑 multiplied by π‘Ž, 𝑏, 𝑐. π‘₯ sub zero, 𝑦 sub zero, 𝑧 sub zero is the initial position and π‘Ž, 𝑏, 𝑐 is the direction of the straight line.

The vector equation of our first line is therefore equal to negative eight, negative eight, negative 10 plus 𝑑 multiplied by negative 10, π‘š, negative eight. The second equation doesn’t appear to have a 𝑧-component; however, we are told that 𝑧 equals eight. Therefore, the initial position is negative five, negative eight, eight. As the 𝑧-component is remaining constant, the direction of the straight line is negative four, 10, zero. We know that when two straight lines are perpendicular, the scalar product of the directions equals zero. To calculate the scalar products, we find the sum of the products of the individual components.

This gives us negative 10 multiplied by negative four plus π‘š multiplied by 10 plus negative eight multiplied by zero. This must sum to zero. Negative 10 multiplied by negative four is equal to 40, π‘š multiplied by 10 is 10π‘š, and negative eight multiplied by zero is zero. Subtracting 40 from both sides of this equation gives us 10π‘š is equal to negative 40. We can then divide both sides of this equation by 10, giving us π‘š is equal to negative four. If the two straight lines are perpendicular, the value of π‘š is equal to negative four.

We will now summarize the key points from this video. Two vectors are parallel if vector 𝐀 is equal to π‘˜ multiplied by vector 𝐁, where π‘˜ is a constant not equal to zero. Any two vectors are only parallel if they are scalar multiples of each other. This also means that the ratio of their components are equal. Two vectors are perpendicular if the scalar product or dot product is equal to zero. To calculate the scalar product, we find the sum of the products of the individual components. By considering these two properties, we can identify whether two vectors are parallel, perpendicular, or otherwise.

When dealing with straight lines in vector form, we consider the direction of the vector to identify whether two straight lines are parallel or perpendicular. If a straight line has vector equation 𝐫 equal to π‘₯ sub zero, 𝑦 sub zero, 𝑧 sub zero plus 𝑑 multiplied by π‘Ž, 𝑏, 𝑐, it is the direction part π‘Ž, 𝑏, 𝑐 that is key.

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