Question

Consider the vectors a =3i +j −k, b =i +j +4k, c=i +3j +k, d =−i −3j +k, g =−3i −j +k. Which pairs (if any) of these vectors are perpendicular each others?

Answer 1

Answer:

a and b are perpendicular to each other, as are b and d, b and g

Step-by-step explanation:

To check whether two vectors are perpendicular to each other, we need the angle between these vectors to be 90 degrees.

We can find the angle between to vectors a and b from the following relation:

The cosine of the angle $\theta$ between two vectors is equal to the dot product of this vectors divided by the product of vector magnitude.

So

$cos(\theta) = \frac{a.b}{|a||b|}$

cos(90) = 0, so when the dot product between vectors a and b is 0, it means that these vectors are perpendicular to each other.

Now, for your exercise, let's compute the dot product between these vectors.

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a.b = (3,1,-1).(1,1,4) = 3+1-4 = 0

So a and b are perpendicular to each other.

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a.c = (3,1,-1).(1,3,1) = 3+3-1 = 5

a and c are not perpendicular to each other.

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a.d = (3,1,-1).(-1,-3,1) = -3-3-1 = -7

So not perpendicular

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a.g = (3,1,-1).(-3,-1,1) = -9-1-1 = -11

Not

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b.c = (1,1,4).(1,3,1) = 1+3+4 = 8

Not

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b.d = (1,1,4).(-1,-3,1) = -1 -3 +4 = 0

b.d = 0, so b and d are perpendicular to each other

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b.g = (1,1,4).(-3,-1,1) = -3-1+4 = 0

b.g = 0, perpendicular

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c.d = (1,3,1).(-1,-3,1) = -1-9+1 = -9

No

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c.g = (1,3,1).(-3,-1,1) = -3-3+1 = -5

No

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d.g = (-1,-3,1).(-3,-1,1) = 3+3+1 = 7

No