Video: Using Relations between Vectors to Solve Problems

Find the values of π‘š and 𝑛 so that vector 2𝑖 + 7𝑗 + π‘šπ‘˜ is parallel to vector 6𝑖 + 𝑛𝑗 βˆ’ 21π‘˜.

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Video Transcript

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

Let’s begin by recalling what it means for two vectors to be parallel to one another. Two vectors are parallel if they’re scalar multiples of one another. This means that we can multiply one vector by a single scalar to obtain the other. Or we can multiply the second vector by the reciprocal of this scalar to obtain the first.

We require then that our first vector two 𝑖 plus seven 𝑗 plus π‘šπ‘˜ is equal to some scalar 𝑐 multiplied by our second vector six 𝑖 plus 𝑛𝑗 minus 21π‘˜. We could also choose to express these vectors as column vectors. So alternatively, we could say that the column vector two, seven, π‘š must be equal to the scalar 𝑐 multiplied by the column vector six, 𝑛, negative 21.

Now to multiply a column vector by a scalar, we just multiply each component of our vector by that scalar. So on the right-hand side of our equation, 𝑐 multiplied by the column vector six, 𝑛, negative 21 is equal to the column vector six 𝑐, 𝑛𝑐, negative 21𝑐.

Next, we can equate the components of our vectors. Equating the 𝑖-components or the first components, we have the equation two equals six 𝑐. Or we can write this as six 𝑐 equals two. We can solve this equation by dividing by six to give 𝑐 is equal to two-sixths, which simplifies to one-third. So in fact, we’ve found the value of the scalar that makes these two vectors parallel.

Equating the 𝑗-components of our two vectors gives the equation 𝑛 multiplied by 𝑐 is equal to seven. We’ve just found that the value of 𝑐 is one-third. So substituting gives 𝑛 multiplied by one-third is equal to seven. To solve this equation, we multiply both sides by three to give 𝑛 is equal to 21. Finally, we can equate the π‘˜-components of our two factors to give π‘š is equal to negative 21𝑐.

Remember, we know that 𝑐 is equal to one-third. So substituting gives π‘š is equal to negative 21 multiplied by one-third, which is negative seven. By recalling then that if two vectors are parallel, they must be scalar multiples of one another and then equating components. We’ve found the values of π‘š and 𝑛 so that the two given vectors are parallel. π‘š is equal to negative seven, and 𝑛 is equal to 21.

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