Find the coordinates of a point P where the line through A(3,-4,-5) an

1.	Find the coordinates of a point P where the line through A(3,-4,-5) and B(2,-3,1) crosses the plane, passing through the point L(2,2,1),M(3,0,1) and N(4,-1,0).Also find the ratio in which P divides the line segment AB.No spam
Answer» $\large\underline{\sf{Solution-}}$ Given that, ➢ The line through A(3,-4,-5) and B(2,-3,1) crosses the plane, passing through the point L(2,2,1),M(3,0,1) and N(4,-1,0). We know, ➢ Equation of line passing through POINTS (a, b, c) and (d, E, f) is given by $\rm :\longmapsto\:\boxed{ \rm \:\dfrac{x - a}{d - a} = \dfrac{y - b}{e - b} = \dfrac{z - c}{f - c}}$ ➢ Hence, the equation of line passes through (3, - 4, - 5) and (2, - 3, 1) is $\rm :\longmapsto\:\rm \:\dfrac{x - 3}{2 - 3} = \dfrac{y + 4}{ - 3 + 4} = \dfrac{z + 5}{1 + 5} = k$ $\rm :\longmapsto\:\rm \:\dfrac{x - 3}{- 1} = \dfrac{y + 4}{1} = \dfrac{z + 5}{6} = k$ ➢ So, the coordinates of any point P on the line AB be $\bf :\longmapsto\:Coordinates \: of \: P = ( - k + 3, \: k - 4, \: 6k - 5)$ Now, Equation of plane passes through the point L (2,2,1), M (3,0,1) and N (4,-1,0) is $\rm \: \: \: \: \begin{gathered}\sf \left \| \begin{array}{ccc}x - 2&y - 2&z - 1\\3 - 2&0 - 2&1 - 1\\4 - 2& - 1 - 2&0 - 1\end{array}\right \| \end{gathered} = 0$ $\rm \: \: \: \: \begin{gathered}\sf \left \| \begin{array}{ccc}x - 2&y - 2&z - 1\\1&- 2&0\\2& -3&- 1\end{array}\right \| \end{gathered} = 0$ $\rm :\longmapsto\:(x - 2)(2 - 0) - (y - 2)( - 1 - 0) + (z - 1)( - 3 + 4) = 0$ $\rm :\longmapsto\:2x - 4 + y - 2 + z - 1 = 0$ $\rm :\longmapsto\:2x + y + z - 7 = 0$ Now, P lies in this plane, So, P must satisfy the equation of plane. $\rm :\longmapsto\:2( - k + 3) + k - 4 + 6k - 5 - 7 = 0$ $\rm :\longmapsto\:- 2k +6 +7 k - 16 = 0$ $\rm :\longmapsto\:5k - 10 = 0$ $\bf\implies \:k = 2$ Hence, $\rm :\longmapsto\:Coordinates \: of \: P = ( - 2 + 3, \: 2 - 4, \: 12 - 5)$ $\bf :\longmapsto\:Coordinates \: of \: P = ( 1, \: - 2, \:7)$ Let assume that P ( 1, - 2, 7 ) divides the line passes through (3, - 4, - 5) and (2, - 3, 1) in the ratio n : 1. So, By Section Formula, $\rm :\longmapsto\:(1, - 2,7) = \bigg(\dfrac{2n + 3}{n + 1}, \: \dfrac{ - 3n - 4}{n + 1}, \: \dfrac{n - 5}{n + 1} \bigg)$ So, on comparing x - coordinates on both sides, we get $\rm :\longmapsto\:\dfrac{2n + 3}{n + 1} = 1$ $\rm :\longmapsto\:2n + 3 = n + 1$ $\rm :\longmapsto\:2n - n = 1 - 3$ $\rm :\longmapsto\:n = - 2$ This implies, P 1, - 2, 7 ) divides the line passes through (3, - 4, - 5) and (2, - 3, 1) in the ratio 2 : 1 externally.

Find the coordinates of a point P where the line through A(3,-4,-5) and B(2,-3,1) crosses the plane, passing through the point L(2,2,1),M(3,0,1) and N(4,-1,0).Also find the ratio in which P divides the line segment AB.No spam

Answer»

$\large\underline{\sf{Solution-}}$

Given that,

➢ The line through A(3,-4,-5) and B(2,-3,1) crosses the plane, passing through the point L(2,2,1),M(3,0,1) and N(4,-1,0).

We know,

➢ Equation of line passing through POINTS (a, b, c) and (d, E, f) is given by

$\rm :\longmapsto\:\boxed{ \rm \:\dfrac{x - a}{d - a} = \dfrac{y - b}{e - b} = \dfrac{z - c}{f - c}}$

➢ Hence, the equation of line passes through (3, - 4, - 5) and (2, - 3, 1) is

$\rm :\longmapsto\:\rm \:\dfrac{x - 3}{2 - 3} = \dfrac{y + 4}{ - 3 + 4} = \dfrac{z + 5}{1 + 5} = k$

$\rm :\longmapsto\:\rm \:\dfrac{x - 3}{- 1} = \dfrac{y + 4}{1} = \dfrac{z + 5}{6} = k$

➢ So, the coordinates of any point P on the line AB be

$\bf :\longmapsto\:Coordinates \: of \: P = ( - k + 3, \: k - 4, \: 6k - 5)$

Now,

Equation of plane passes through the point L (2,2,1), M (3,0,1) and N (4,-1,0) is

$\rm \: \: \: \: \begin{gathered}\sf \left | \begin{array}{ccc}x - 2&y - 2&z - 1\\3 - 2&0 - 2&1 - 1\\4 - 2& - 1 - 2&0 - 1\end{array}\right | \end{gathered} = 0$