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Find the vector equation of a line which is parallel to the vector `2 hat i- hat j+3 hat k`and which passes through the point (5, -2,4). Also reduce it toCartesian form. |
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Answer» Vector equation of the given line: The given line passes through the point `A(5,-2,4) " and " `it is parallel to the vector `vec(m) =(2hat(i) -hat(j) +3hat(k))` the position vector of A is given by `vec(r )_(1) =(5hat(i) -2hat(j) +4hat(k))` Hence the vector equations of the given line is `vec(r ) =vec(r )_(1) +lambdavec(m ) rArr vec( r) =(5hat(i) -2hat(j) +4hat(k)) +lambda (2hat(i) -hat(j) +3hat(k)) " ".....(i)` Cartesisan equation of the given line : Taking `vec(r ) =(xhat(i) +yhat(j) +zhat(k))` equation (i) becomes : `(xhat(i) +yhat(j) +zhat(k)) =(5hat(i) -2hat(j)+4hat(k)) +lambda(2hat(i)-hat(j)+3hat(k))` `rArr (xhat(i) +yhat(j) +zhat(k)) =(5+2lambda) hat(i) -(2+lambda ) hat(j) +(4 +3lambda) hat(k)` `rArr x=5 +2lambda ,y=-(2+lambda) " and " z=4 +3lambda` `rArr (x-5)/(2)=(y+2)/(-1) =(z-4)/(3)= lambda` Hence `(x-5)/(2)=(y+2)/(-1) =(z-4)/(3)` are the required equations of the given line in Cartesian form. |
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