ISC (Commerce) Class 12CISCE
Share

# 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

ConceptEquation of a Line in Space

#### 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?

#### Video TutorialsVIEW ALL 

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.
S