ISC (Commerce) Class 12CISCE
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Find the Value of P, So that the Lines l_1:(1-x)/3=(7y-14)/p=(z-3)/2 and l_2=(7-7x)/3p=(y-5)/1=(6-z)/5 Are Perpendicular to Each Other. - ISC (Commerce) Class 12 - Mathematics

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Question

 

Find the value of p, so that the lines `l_1:(1-x)/3=(7y-14)/p=(z-3)/2 and l_2=(7-7x)/3p=(y-5)/1=(6-z)/5 ` are perpendicular to each other. Also find the equations of a line passing through a point (3, 2, – 4) and parallel to line l1.

 

Solution

The equation of line L1 :

`(1-x)/3=(7y-14)/p=(z-3)/2`

`=>(x-1)/(-3)=(y-2)/(p/7)=(z-3)/2.....(1)`

The equation of line L2 :

`(7-7x)/3p=(y-5)/1=(6-z)/5`

`=>(x-1)/((-3p)/7)=(y-5)/1=(z-6)/(-5).....(2)`

Since line L1 and L2 are perpendicular to each other, we have

`-3xx((-3p)/7)+p/7xx1+2xx(-5)=0`

`=>(9p)/7+p/7=10`

`=>10p=70`

`=>p=7`

Thus equations of lines L1 and L2 are:

`(x-1)/(-3)=(y-2)/1=(z-3)/2`

`(x-1)/(-3)=(y-5)/1=(z-6)/(-5)`

Thus the equation of the line passing through the point (3,2, -4) and parallel to the line L1 is:

`(x-3)/(-3)=(y-2)/1=(z+4)/2`

  Is there an error in this question or solution?
Solution Find the Value of P, So that the Lines l_1:(1-x)/3=(7y-14)/p=(z-3)/2 and l_2=(7-7x)/3p=(y-5)/1=(6-z)/5 Are Perpendicular to Each Other. Concept: Equation of a Line in Space.
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