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Question Video: Identifying the Parametric Equation of a Line Given Its Vector Equation Mathematics

The vector equation of a straight line is given by 𝐫 = βŒ©βˆ’1, 3βŒͺ + π‘˜ 〈5, 2βŒͺ. Which of the following pairs of parametric equations represents this straight line? [A] π‘₯ = 2 + 5π‘˜, 𝑦 = 3 βˆ’ π‘˜ [B] π‘₯ = 2 + 3π‘˜, 𝑦 = 5 βˆ’ π‘˜ [C] π‘₯ = 5 βˆ’ π‘˜, 𝑦 = 2 + 3π‘˜ [D] π‘₯ = 3 + 2π‘˜, 𝑦 = βˆ’1 + 5π‘˜ [E] π‘₯ = βˆ’1 + 5π‘˜, 𝑦 = 3 + 2π‘˜

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Video Transcript

The vector equation of a straight line is given by 𝐫 equals negative one, three plus π‘˜ times five, two. Which of the following pairs of parametric equations represents this straight line? Is it option (A) π‘₯ equals two plus five π‘˜, and 𝑦 equals three minus π‘˜? Option (B) π‘₯ equals two plus three π‘˜, and 𝑦 equals five minus π‘˜. Option (C) π‘₯ equals five minus π‘˜, and 𝑦 equals two plus three π‘˜. Option (D) π‘₯ equals three plus two π‘˜, and 𝑦 equals negative one plus five π‘˜. Or is it option (E) π‘₯ equals negative one plus five π‘˜, and 𝑦 equals three plus two π‘˜?

In this question, we are given a vector equation for a straight line and asked to use this to identify which of five given pairs of parametric equations also represents the line. To do this, let’s start by recalling what is meant by a vector equation of a line. It’s an equation of the form 𝐫 equals 𝐫 sub zero plus π‘˜ times 𝐝. 𝐫 sub zero is the position vector of any point on the line, and 𝐝 is a nonzero vector parallel to the line, called a direction vector.

For our vector equation of a line, we see that 𝐫 sub zero is the vector negative one, three and 𝐝 is the vector five, two. This gives us information about the line. We can note that the line must pass through the point negative one, three and the line must be parallel to the vector five, two.

There are many ways we can now go about answering this question. One way is to use this vector equation of the line to construct a pair of parametric equations for the line. We can do this by evaluating the vector expression on the right-hand side of the equation. We evaluate the scalar product by multiplying every component by the scalar π‘˜. We obtain negative one, three plus five π‘˜, two π‘˜. Now, we add the vectors together by adding their corresponding components. We get the vector negative one plus five π‘˜, three plus two π‘˜.

For any real value of the parameter π‘˜, this gives us the position vector of a point on the line. We can call this vector π‘₯, 𝑦. For these vectors to be equal, all of their corresponding components must be equal. Equating the components then gives us that π‘₯ equals negative one plus five π‘˜ and 𝑦 equals three plus two π‘˜, that is, a pair of parametric equations representing the line. We can see that this matches option (E).

It is worth noting that we have only shown that this is one possible pair of parametric equations for the line. In fact, just like vector equations, there are infinitely many choices for the pair of parametric equations of a line. In general, the pair of parametric equations will be in the form π‘₯ equals π‘₯ sub zero plus π‘Žπ‘˜ and 𝑦 equals 𝑦 sub zero plus π‘π‘˜, where π‘₯ sub zero, 𝑦 sub zero is any point on the line and the vector π‘Ž, 𝑏 is any direction vector of the line.

We can use this to find more pairs of parametric equations of the line or show that a pair of parametric equations does not represent the line. For example, we can check the direction vectors of the other four options. We can see that none of these vectors are parallel to the vector five, two. This means that none of these options represent the same line as the vector equation of the line. Hence, the answer is option (E). A pair of parametric equations for the line is π‘₯ equals negative one plus five π‘˜ and 𝑦 equals three plus two π‘˜.

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