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# (a) Let $P$ be a point not on the line $L$ that passes through the points $Q$ and $R$. Show that the distance $d$ from the point $P$ to the line $L$ is$$d = \frac{\mid a \times b \mid}{\mid a \mid}$$where $a = \vec{QR}$ and $b = \vec{QP}$ (b) Use the formula in part (a) to find the distance from the point $P (1, 1, 1)$ to the line through $Q (0, 6, 8)$ and $R (-1, 4, 7)$

## b. $\sqrt{\frac{97}{3}}$

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##### Lily A.

Johns Hopkins University

##### Catherine R.

Missouri State University

##### Heather Z.

Oregon State University

##### Michael J.

Idaho State University

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### Video Transcript

All right. So in this problem we have points Q. And R. And then this line L passes through both Q. And R. And then P. Is some point over here that's not on the line now. Um And they said the vector A. Goes from cuba are and the vector B. Goes from Q. Two P. And then ignore these great numbers for now. But that's those will be the values um or the coordinates of P. Q. And R. In part beat. But first let's do part A. Which is to show that the distance from P. To the line L. So that distance there which is perpendicular is the cross product is a cross B over the magnitude of a crusty over the magnitude of that. So for this we need to remember um the formula for the magnitude of a cross B. And we can write that as the magnitude of A times the magnitude of B. Times the sine of the angle between them. The angle between them is that angle theta. So what is this um formula evaluate to? Well that's magnitude of eight times magnitude of be over sine theta divided by magnitude of A. And now we can cancel out magnitude of a. From the top and bottom. So we're just left with a magnitude of be signed theater. Let's think about that. Um Meg if we think back to geometry in this right angle triangle then B. Is the hypotenuse magnitude of B. Is the hypotenuse. And so then um that's the opposite side. So that the length of that is in fact hypotenuse time signed data, magnitude of the signed data. Um Okay, so that proves that that formula always works. Why do we want that formula? Um when we could just calculate it as the magnitude of the time signed data? Um It's because we don't always know tha tha exactly. In fact here now we have these three sets of coordinates and we're going to use the formula to calculate it. Because it would be a little bit of a pain to find data. You would have to use the dot product or across product anyways, so we'll just use the formula and calculate magnitude of a Crosby over the magnitude of A. So let's think about that. Um So let's first calculate the vectors A. And B. So the vector A A. Goes from cuba are so we're going to take the coordinates of ar minus the coordinates of Q. So negative one, negative 24 minus six. And then seven minus eight. Negative one. And then the coordinates of so be it will be the coordinates of p minus the coordinates of Q. So positive 11 minus six is negative 51 minus seven is one minus eight is negative seven. Okay, that's a lot of negative signs. I'm just gonna use let's see the fact that mag uh a cross b magnitude of a Cross B is equal to the magnitude of negative Crosby. So that way I can do the cross product of this vector with positive numbers. Cross products have a lot of negative numbers in them. So I'm gonna just try to reduce the number of negative numbers here. Okay. And then um there's lots of ways to calculate cross product. I like to write it as the determinant of this. Three by three. Matrix with I. J. K. In the top and then um A. And the second row or negative A. In this case in the second row, negative B. In the third row. Okay, so then we get Um Let's see two times. So we're going to take I right? Um and then times the two by two determinant. So two times negative seven is negative 14 minus negative five, negative nine. And then jay we're going to do that determine it. So one times negative seven is negative, seven minus one is negative eight. I failed in reducing the number of negative sense. Oh no, I didn't. Because there's we've got alternate signs up here. Okay, I didn't fully fail. So that's actually plus eight, apologize for that. Do not forget that negative sign there. And then uh for K we take this two by two determinant, it's a positive again, that's good. Um So negative five minus two, negative seven. Okay, it's still a lot of negative sense. So that's the magnitude of a cross B Is the magnitude of that vector. So the root of and the negative signs don't matter when they take magnitude nine squared plus eight squared plus seven squared Whatever that is. So the route of, let's see 81 plus 64 Plus 49. That's 100 Route of 194. Almost 14. Um Okay. And then the magnitude of A. Is the same as the magnitude of negative A. But it doesn't matter we're finding magnitude of that. So the root of one plus four plus one. That's route six. Yeah. That part wasn't hard. Um Okay so then the answer to the question is the distance from P. Two L. Is the root of 194. Over the route of six. and if it doesn't ask you around um you should probably leave it an exact value.

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##### Lily A.

Johns Hopkins University

##### Catherine R.

Missouri State University

##### Heather Z.

Oregon State University

##### Michael J.

Idaho State University

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