Question Video: Parallel and Perpendicular Lines | Nagwa Question Video: Parallel and Perpendicular Lines | Nagwa

Question Video: Parallel and Perpendicular Lines Mathematics • Third Year of Secondary School

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Consider the two lines π₯ = 4 + 2π‘, π¦ = 6 + π‘, π§ = 2 β 2π‘ and π« = β¨6, 7, 0β© + π‘β¨5, 4, 7β©. Determine whether they are parallel or perpendicular.

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Video Transcript

Consider the two lines π₯ equals four plus two π‘, π¦ equals six plus π‘, π§ equals two minus two π‘, and π« equals six, seven, zero plus π‘ times five, four, seven. Determine whether they are parallel or perpendicular.

Now we can determine whether two lines are parallel or perpendicular by considering their direction vectors. And so what weβre going to do is begin by rewriting the equation of our first line in vector form so we can then extract its direction vector. And of course, thatβs of the form π₯ nought, π¦ nought, π§ nought plus π‘ times π, π, π, where π, π, π is the direction vector and π₯ nought, π¦ nought, π§ nought is the point which the line passes through. To achieve this, we begin by writing each expression as the component of our vector. Then we separate the components as shown. So we get four, six, two plus two π‘, π‘, negative two π‘. And then we take out that constant factor of π‘. So the vector form of our first line is four, six, two plus π‘ times two, one, negative two.

So letβs write down then the direction vector of each of our lines. Letβs call our first line πΏ sub one. Its direction vector is two, one, negative two. Then the direction vector of our second line, weβll call that πΏ sub two, is five, four, seven. And then we recall that if two vectors are parallel, specifically direction vectors, then one is a scalar multiple of the other. So for our lines to be parallel, we can say that the direction vector two, one, negative two can be written as some scalar multiple π of the vector five, four, seven.

Now, if we look at the second component one and four, we might deduce that for this to be true, π would have to be equal to one-quarter, since one-quarter of four is equal to one. But one-quarter of five is not equal to two, and a quarter of seven is not equal to negative two. And so this is not true. These vectors cannot be written as scalar multiples of one another, and so the lines cannot be parallel. We might deduce then that theyβre perpendicular. But weβre going to check. For two vectors to be perpendicular, of course, their dot product or scalar products have to be equal to zero. They also have to intersect at a point, but weβll deal with that in a moment.

Letβs just double-check that their dot product is indeed equal to zero. Well, their dot product or their scalar product is two times five plus one times four plus negative two times seven, which is indeed equal to zero. So this element so far is good. We do, of course, need to double-check that they intersect at a point. Now if the two lines intersect at a point, there must be some parameter values π‘ sub one in our first equation and π‘ sub two in our second that make the π₯-coordinates, π¦-coordinates, and π§ coordinates, respectively, equal.

For our π₯-coordinates, letβs say thatβs four plus two π‘ sub one is equal to six plus five π‘ sub two. For our π¦-coordinate, itβs six plus π‘ sub one equals seven plus four π‘ sub two. And then we have a corresponding equation for our π§-coordinate. Solving any two of these equations simultaneously, we obtain π‘ sub one equals one and π‘ sub two equals zero. Of course, we can check these by then substituting into the third equation, the third one that we hadnβt used. Substituting π‘ sub one equals one into the equation of our first line and we find that this point of intersection is four plus two, six plus one, two minus two, which is six, seven, zero.

Letβs double-check. We get the same values when we substitute into πΏ sub two. As expected, we do indeed get the value of six, seven, zero. And of course, if we had substituted π‘ sub one equals one and π‘ sub two equal zero into that third equation that we didnβt use, we would have expected this result. So since the direction vectors of the two lines are perpendicular and they intersect at a point, we can say that the lines themselves must also be perpendicular.

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