The cartesian equations of a line are 3x -3 =2y +1=5-6z (a) Write these equations in standard form and find the direction ratios of the given line . (b) write the equations for the given line in vector form.
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(a) The given equations may be written as
`3(x-1) =2(y+(1)/(2)) =-6 (z-(5)/(6))`
`rArr ((x-1))/(((1)/(3)))= ((y+(1)/(2)))/(((1)/(2)))= ((z-(5)/(6)))/(((-1)/(6)))`
`rArr ((x-1))/(2)= ((y+(1)/(2)))/(3)=((z+(5)/(6)))/(-1)`
This is the standard form of the given equations in cartesian form. Clearly its direction rations are 2,3,-1
(b) The given line passes through the point `A(1,(-1)/(2),(5)/(6))` and it is parallel to the vector `vec( m) =(2hat(i) +3hat(j) -hat(k))`
The position vector of the point A is `vec(r )_(1) =( hat(i) -(1)/(2) hat(j)+ (5)/(6) hat(k))`
So the vector equation of the given line is
`vec(r ) =vec( r) _(1) +lambdavec(m ) rArr vec( r) =(hat(i) -(1)/(2) hat(j) +(5)/(6) hat(k)) + lambda (2hat(i) +3hat(j) -hat(k))`