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Find the vector equation of the line passing through the point A(2,-1,1) and parallel to the line joining the points B (-1,4,1) and C(1,2,2) .Also find the Cartesian equations of the line . |
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Answer» Vector equation of the given line: the p.v of B `=(hat(i) +hat(j) +hat(k)) " and " p.v " of " C = (hat(i) +2hat(j) +2hat(k))` `:. , vec(BC) =(p.v " or " C) =(p .v. " of " B)` `=(hat(i) +2hat(j)+2hat(k)) -(-hat(i) +4hat(j) +hat(k)) =(2hat(i) -2hat(j) +hat(k))` The p.v. of A is `vec( r) =2hat(i) -hat(j) +hat(k)` `:. ` the vector equation of the given line is `vec( r) =vec(r )_(1) +lambda (vec(BC))` `hArr vec(r ) =(2hat(i) -hat(j) +hat(k)) +lambda(2hat(i) -2hat(j) +hat(k))` Cartesian equations of the given line: Taking `vec(r) =x hat(i) +yhat(j) +zhat(k)` equations (i) becomes `(xhat(i) +yhat(j) +zhat(k))=(2hat(i)-hat(j)+hat(k))+lambda(2hat(i) -2hat(j)+hat(k))` `hArr (xhat(i) +yhat(j)+zhat(k))=(2+2lambda)hat(i) +(-1-2lambda) hat(j) +(1+lambda) hat(k)` `hArr x =2 + 2lambda , y= 1- 2lambda " and " z=1+ lambda` `hArr (x-2)/(2)=(y+1)/(-2) =(z-1)/(1)=lambda` Hence `(x-2)/(2)=(y+1)/(-2) =(z-1)/(1)` are the required equation of the given line in the Cartesian form. |
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