### Video Transcript

If π¨ is the vector π, three and π© is the vector six, π minus six and vector π¨ is perpendicular to vector π©, then π is equal to blank. Is it option (A) one, option (B) negative one, option (C) two, or is it option (D) negative two?

In this question, weβre given two vectors in terms of their components, the vector π¨ and the vector π©. And weβre also told that these two vectors are perpendicular. We need to use this to determine the value of the scalar π. To answer this question, letβs start by recalling what it means for two vectors to be perpendicular.

We say that two vectors are perpendicular if their dot product is equal to zero. In other words, the dot product between vector π¨ and vector π© must be equal to zero. To evaluate the dot product between these two vectors, weβll start by writing them in component form. We want to find the dot product between the vector π, three and the vector six, π minus six. Remember, to find the dot product between two vectors, we need to find the sum of the product of the corresponding components of the two vectors.

So we need to start by multiplying the first two components of the vectors together. Thatβs π multiplied by six. And then we need to add on the product of the second components of both vectors. Thatβs three multiplied by π minus six. And because we know our two vectors are perpendicular, this sum must be equal to zero. We can then simplify the left-hand side of this equation. π times six is equal to six π, and we can distribute three over our parentheses to get three π minus 18.

Now, we just need to solve this equation for π. First, six π plus three π is equal to nine π. This gives us nine π minus 18 is equal to zero. We can simplify further by adding 18 to both sides of this equation, giving us that nine π must be equal to 18.

Finally, we can solve for π by dividing both sides of our equation through by nine. This gives us that π is equal to two. Therefore, we were able to show if π¨ is the vector π, three and π© is the vector six, π minus six and π¨ is perpendicular to vector π©, then the value of π must be equal to two, which is option (C).