InterviewSolution
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InterviewSolution
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Find the vector and Cartesian equations of the line passing through the point (1,2-4) and perpendicular to each of the lines `(x-8)/(3)=(y+19)/(-16) =(z-10)/(7) " and " (x-15)/(3)=(y+29)/(8) =(z-5)/(-5)` |
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Answer» The given lines are `(x-8)/(3)=(y+19)/(-16) =(z-10)/(7)` `" and " (x-15)/(3)=(y+29)/(8) =(z-5)/(-5)` Let a,b,c be the direction rations of the required line . Then it being perpendicular to each of the lines (i)and (ii) we have 3a-16b+7c=0 and 3a+8b-5c =0 On solving these equations by cross multiplications we get `(a)/((80-56)) =(b)/((21+15))=(c)/((24+48))` `rArr (a)/(24)=(b)/(36) =(c )/(72) rArr (a)/(2)=(b)/(3)=(c)/(6)` Thus the desired line has direction rations 2,3,6 So we have to find the equations of a line passing through te point A(1,2-4) and having 2,3,6 as its directions rations So, the required equations in Cartesian form are `((x-1))/(2)=((y-2))/(3) =((z+4))/(6)` The position vector of point A is `vec(r ) =(hat(i) +2hat(j) -4hat(k))` Also the required line has direction rations 2,3,6 and so it is parallel to the vector `vec( m) =(hat(i) +3hat(j) +6hat(k))` So , its equation in vector form is `vec(r ) =vec(r )_(1) +lambda vec( m) rArr vec(r ) =(hat(i)+ 2hat(j) -4hat(k)) + lambda (2hat(i) +3hat(j) +6hat(k))` |
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