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# Show that the lines with symmetric equations $x = y = z$ and $x + 1 = \frac{y}{2} = \frac{z}{3}$ are skew, and find the distance between these lines.

## $\frac{1}{\sqrt{6}}$

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the question they are asking to show that the lines with symmetric equations. Executive I go to said and explains when he called to my bed to go to the adversary R Schiewe lines and also to find out distance between these lines. So in questions either asking too show that whether these two lines are skew lines so shaking for lines L one and L two, R skew lines. Therefore the first equation is given as X equal to equal to zero, let it be equal to T. Therefore the value of X equal to T by equal to T and Z is equal to T. Therefore from here we can find the normal to display to this line that is In one vector is equal to 111, the coefficients of the term t. And the coordinates that is the points passing through this line is equal to 000. And the equation of the second line is given as X plus one is equal to Why by two equal to zero by three equal to asset. So this is equal to X Equal to as -1, Why equal to 2? S and zero equal to three. Is therefore we find normal to this line that is equal to the coefficients of the variables. S This is equal to 12 and three. And here the Coinates that passes to this line is equal to -1. The constant terms with these variable that is minus 10 and zero. Therefore, after finding these two for checking whether these two lines are skew lines, so it has to be neither people nor perpendicular to satisfy this condition. So, since in one and in two are not multiples or equal vectors. So L one and L two are not parallel lines. And the second criteria to check is whether these two lines are perpendicular to. So in order to take that we have to find the yeah whether these two lines are perpendicular or not, if they're not perpendicular, then only they are skew lines. So in order to find this of condition, we have to evaluate the equation from this formula that is As shown in this figure. Uh there are two lines that are not parallel like this, maybe this is line one and this is line to, they are uh there is a line that passes through the these lines, There are two points, that is, This is a point P1 and this is a point to and uh and there is a normal and that is perpendicular to these two lines. So therefore in order to find the, to check whether these two lines are perpendicular or not, we have to find the um yeah Value of P one and P two and put it in the equation that is P one, P 2, victor dot In one Victor Cross into Victor. And if this is equal to zero, not equal to zero then it's not perpendicular. L one, L 2, not perpendicular to each other. Therefore, in order to find the value of P one, P 2. As we have already found the points of the lines. L one, L 2. So P one is equal to okay 000 and B two is equal to -100. So therefore okay therefore the Putting the finding the distance P one p 2 vector we get this to be equal to -100. So after finding the value of P one P two we find across product of the normal victors that are Given in one and 2. Yeah that is in one cross and to victor is equal to I. Cobb Jacob. Okay cap and the values of N one R 111 and the value of into our 123. Therefore calculating this value, we get it to be equal to. It will do if we can clear the value we get icap 3 -2 -1 2 -1. So this is equal to icap minus two Jacob let's kick up. So therefore the value of this equation P one, P two vector dart in one cap crossing to cap is equal to the value of you won't be too was -100. So -100. the value of the In one cross into vector that is 1 -2 and one. So This value is equal to -1. Okay. Uh huh So this is not equal to zero. Therefore Lying L one and L two. R not perpendicular to each other. Um By the scalar triple. Yeah rule. Mhm. So this is a scalar product skele triple production and therefore the lines are not parallel and not perpendicular. Therefore L. One and L. Two lines are skew lines. So the question for a is drug question A that to show that the lines are skew lines are approved Now the second question they are asking that that to find the distance between these lines. So in order to find the distance between the lines the formula is this is equal to let the decision between the lines B. D. So this is equal to the scalar value of P one P two victor, not the normal victims too. Body lines and divided by the scalar value of the normal vector of these lines. So this is equal to since p one P two is equal to minus 100. So the dot product with the normal victor which is the cross product value of both. The normal victories of the lines. That is equal to one -2 and one. Mhm. Yeah the scalar value of this divided by route. Under the value of the yeah Skill er value of the normal victor. That is one square place to square plus one square. So this is equal to yeah the scarier value of -1, divided by route under one squared plus two squared plus one squared is equal to six. so this is equal to one by route six. Therefore the shortest distance between these two lines is equal to one by route six, and this is the answer the given question yeah.

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