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If the straight line π₯ minus 10 all over six is equal to π¦ plus six all over eight is equal to π§ plus two all over π is parallel to π₯ minus one all over negative 12 is equal to π¦ plus three all over π is equal to π§ plus one all over 14, find π plus π.
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In this question, weβre given the equation of two lines and weβre told that these two lines are parallel.
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We need to use these to determine the value of π plus π.
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To do this, we start by seeing that our lines are given in symmetric form.
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And we recall the symmetric form of a straight line is the form π₯ minus π₯ zero all over π is equal to π¦ minus π¦ zero all over π is equal to π§ minus π§ zero all divided by π.
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And this is a very useful form to write our straight line in because we can read off some information about our line.
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First, we see if we substitute π₯ is equal to π₯ zero, π¦ is equal to π¦ zero, and π§ is equal to π§ zero into this equation, then we get zero is equal to zero is equal to zero.
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Therefore, any straight line given in this form will pass through the point π₯ zero, π¦ zero, π§ zero.
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However, there is another useful piece of information we can get from the symmetric form of this line.
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The direction vector of this straight line is going to be the vector π, π, π.
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In this question, we need to use the fact that the two lines are parallel to determine the value of π plus π.
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And to do this, weβre going to need to recall exactly what we mean when we say two straight lines are parallel.
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We recall that we say that two straight lines are parallel if their direction vectors are nonzero scalar multiples of each other.
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So to determine whether these two straight lines are parallel or not, weβre going to need to determine whether their direction vectors are scalar multiples of each other.
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And because both of these lines are given in symmetric form, we can just read off their direction vectors.
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For our first straight line, the direction vector is going to be the vector six, eight, π.
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And we can do exactly the same for our second straight line.
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Its direction vector is going to be the vector negative 12, π, 14.
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Finally, because we know these two straight lines are parallel to each other, we know that these two vectors have to be nonzero scalar multiples of each other.
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In other words, the vector six, eight, π must be equal to π multiplied by the vector negative 12, π, 14, where π is some nonzero scalar because we know that these lines are parallel.
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So letβs solve these two vectors being equal to each other.
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We need to evaluate the scalar multiplication.
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Remember, we need to do this by multiplying every component in our vector by π.
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Multiplying through by π, we get the vector six, eight, π is equal to the vector negative 12π, ππ, 14π.
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And remember, for two vectors to be equal to each other, they must have the same dimension and their components must be equal.
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We can use this to find the values of π, π, and π.
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Letβs start by finding the value of π by setting the first component of these two vectors to be equal.
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Doing this, we get that six should be equal to negative 12π.
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And of course, we can solve for π by dividing through by negative 12.
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We get that π is equal to negative six divided by 12, which we can simplify to give us that π is equal to negative one-half.
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Now that we know the value of π, we can substitute this into our vector.
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Clearing some space and substituting in π is equal to negative one-half into our vector equation, we get that the vector six, eight, π is equal to the vector negative 12 times negative one-half, negative π over two, 14 times negative one-half.
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And of course, we can simplify this.
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In the first component on the vector on the right-hand side, negative 12 times negative one-half is equal to six.
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And in the third component of this vector, 14 multiplied by negative one-half is equal to negative seven.
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So weβve simplified the vector on the right-hand side to give us the vector six, negative π over two, negative seven.
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Now, since these two vectors are equal, all of their components must be equal.
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We can now equate the second components of these two vectors.
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This gives us that eight is equal to negative π divided by two.
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And of course, we can solve for our value of π by multiplying this equation through by negative two.
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We get that π is equal to negative two times eight, which is, of course, equal to negative 16.
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Of course, we can also equate the third components of these two vectors.
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And doing this, we see that our value of π has to be equal to negative seven.
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So weβve found the value of π and the value of π.
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And weβre asked in this question to find the value of π plus π.
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So we add these values together.
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π plus π is negative seven plus negative 16, which we can calculate is negative 23.
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Therefore, we were able to show if the straight line π₯ minus 10 all over six is equal to π¦ plus six all over eight is equal to π§ plus two all over π is parallel to the straight line π₯ minus one all over negative 12 is equal to π¦ plus three all over π is equal to π§ plus one all over 14, then π plus π must be equal to negative 23.