Question Video: Finding the Value of an Unknown in the Equation of a Line in Three Dimensions given the Equation of Another Line Perpendicular to It | Nagwa Question Video: Finding the Value of an Unknown in the Equation of a Line in Three Dimensions given the Equation of Another Line Perpendicular to It | Nagwa

Question Video: Finding the Value of an Unknown in the Equation of a Line in Three Dimensions given the Equation of Another Line Perpendicular to It Mathematics • Third Year of Secondary School

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Given that the lines (π₯ β 8)/3 = (π¦ + 4)/5 = (π§ + 6)/β2 and (π₯ β 10)/β5 = (π¦ + 7)/9 = (π§ β 3)/π are perpendicular, what is π?

02:30

Video Transcript

Given that the lines π₯ minus eight over three is equal to π¦ plus four over five is equal to π§ plus six over negative two and π₯ minus 10 over negative five is equal to π¦ plus seven over nine is equal to π§ minus three over π are perpendicular, what is π?

In this question, weβre given the equations of two lines written in Cartesian form. And we can see in one of the equations of these lines, we have an unknown value of π. We need to determine the value of π such that the two lines are perpendicular. So to answer this question, we should start by recalling how do we determine if two lines are perpendicular. And we know if two lines are perpendicular, then the direction vectors of each line must also be perpendicular. In particular, if we call these direction vectors π sub one and π sub two, then since these vectors are perpendicular, the dot product of the two vectors must be equal to zero.

This means we can determine if two lines are perpendicular by calculating the dot product of their direction vectors. And we can find the direction vectors of these two lines by noting theyβre written in Cartesian form. The components of the direction vector will be equal to the denominator of each part of the equation. We get the vector π sub one is equal to three, five, negative two. And we can do the exact same thing to find the direction vector of the second line. π sub two is the vector negative five, nine, π.

We can now calculate the dot product between these two vectors. And to do this, we recall to find the dot product of two vectors of the same dimension, we need to find the sum of the products of the corresponding components. This gives us three times negative five plus five times nine plus negative two multiplied by π. And remember, we want our lines to be perpendicular. So this dot product needs to be equal to zero. So we can set the left-hand side of this equation equal to zero.

We now want to solve this equation for π. Weβll do this by simplifying and rearranging. First, we can simplify the equation to get zero is equal to negative 15 plus 45 minus two π. We can subtract 30 from both sides of the equation to get negative 30 is equal to negative two π. Then we can solve for π by dividing both sides of the equation through by negative two. We get π is equal to 15, which is our final answer.

Therefore, we were able to show for the lines π₯ minus eight over three is equal to π¦ plus four over five is equal to π§ plus six over negative two and π₯ minus 10 over negative five is equal to π¦ plus seven over nine is equal to π§ minus three over π to be perpendicular, the value of π must be 15.

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