Question Video: Finding the Unknown Component of a Vector Parallel to Another Mathematics

Given that 𝐀 = βŸ¨βˆ’6, βˆ’15⟩, 𝐁 = βŸ¨π‘˜, βˆ’10⟩, and 𝐀 βˆ₯ 𝐁, find the value of π‘˜.

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Video Transcript

Given that vector 𝐀 is equal to negative six, negative 15 and vector 𝐁 is equal to π‘˜, negative 10 and the vector 𝐀 is parallel to the vector 𝐁, find the value of π‘˜.

In this question, we’re given two vectors in terms of their components: the vector 𝐀 and the vector 𝐁. And we’re also told that these two vectors are parallel. We need to use this to find the value of π‘˜, which is one of the components of vector 𝐁. To answer this question, let’s start by recalling what it means to say vector 𝐀 is parallel to vector 𝐁. We say that two vectors are parallel if they’re nonscalar multiples of each other. In other words, because vector 𝐀 is parallel to vector 𝐁, there must exist some scalar π‘š such that 𝐀 is equal to π‘š times 𝐁.

We can substitute the expressions we’re given for vector 𝐀 and vector 𝐁 into this equation. This gives us that the vector negative six, negative 15 will be equal to π‘š multiplied by the vector π‘˜, negative 10. We can then simplify the right-hand side of this equation by evaluating the scalar multiplication. Remember, to multiply a vector by a scalar, we just multiply each of the components by the scalar. This gives us that the vector negative six, negative 15 will be equal to the vector π‘˜π‘š, negative 10π‘š. Since these two vectors must be equal, the first component of each vector must be equal and the second component of each vector must be equal.

This gives us two equations. Equating the first component of each vector gives us that negative six must be equal to π‘˜ times π‘š. And equating the second component of each vector gives us that negative 15 must be equal to negative 10 times π‘š. We can solve the second equation for the value of π‘š. We just need to divide both sides of our equation through by negative 10. This gives us that π‘š is equal to negative 15 divided by negative 10, which if we divide both our numerator and our denominator by negative five, we see that π‘š is equal to three over two. We can then substitute this value of π‘š into our first equation.

Doing this gives us that negative six must be equal to π‘˜ multiplied by three over two. And we can then solve this equation for π‘˜ by dividing both sides of our equation by three over two, which is of course the same as multiplying by two over three. This gives us that π‘˜ is equal to negative six multiplied by two-thirds, which we can then evaluate. Negative six divided by three is equal to negative two, which gives us that π‘˜ is equal to negative two multiplied by two, which is negative four. Therefore, we were able to show if 𝐀 is the vector negative six, negative 15 and 𝐁 is the vector π‘˜, negative 10 and the vector 𝐀 is parallel to the vector 𝐁, then the value of π‘˜ must be equal to negative four.

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