Video Transcript
Fill in the blank: If 𝐀 and 𝐁 are two perpendicular vectors and vector 𝐀 is equal to two, two, negative six, then vector 𝐁 may be equal to what. Is it (A) four, four, negative 12; (B) three, negative three, two; (C) negative three, negative three, two; (D) negative three, two, three; or (E) three, three, two?
We are told in this question that the two vectors are perpendicular. And we recall that if two vectors 𝐮 and 𝐯 are perpendicular, their dot or scalar product is equal to zero. In this question, we’ll need to find the dot product of vector 𝐀 and the five options to see which one equals zero.
Let’s begin with option (A). We calculate the dot product of any two vectors by multiplying the corresponding components and then finding the sum of these values. For option (A), we have two multiplied by four plus two multiplied by four plus negative six multiplied by negative 12. This simplifies to eight plus eight plus 72, which is equal to 88. As this is not equal to zero, option (A) is not correct.
Let’s now consider option (B), the vector three, negative three, two. This time, the dot product is equal to two multiplied by three plus two multiplied by negative three plus negative six multiplied by two. This is equal to six plus negative six plus negative 12, which simplifies to negative 12 and is once again not equal to zero. Option (B) is not perpendicular to vector 𝐀.
We repeat this process for option (C), giving us two multiplied by negative three plus two multiplied by negative three plus negative six multiplied by two. This is equal to negative six plus negative six plus negative 12, which is equal to negative 24 and once again is not equal to zero. We can therefore rule out option (C).
Option (D) is the vector negative three, two, three. This time, the dot product is equal to two multiplied by negative three plus two multiplied by two plus negative six multiplied by three. This is equal to negative six plus four plus negative 18. This simplifies to negative 20, which once again is not equal to zero.
Our final option (E) is the vector three, three, two. This time, the dot product is equal to two multiplied by three plus two multiplied by three plus negative six multiplied by two. This is equal to six plus six plus negative 12. Six plus six equals 12, and adding negative 12 to this gives us zero. This means that the dot product of vectors two, two, negative six and three, three, two is equal to zero.
We can therefore conclude that out of the five options given, the vector that is perpendicular to vector 𝐀 is three, three, two.
