Question

Find the coordinates of the point where the line through the points (3.-4.-5) and (2.-3.1) crosses the plane determined by the points (1,2,3),(4,2,-3) and (0,4,3)​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

∴The coordinate of the point is$(-\frac{29}{3} ,\frac{26}{3} ,33)$

Step-by-step explanation:

Given points of plane are (1,2,3) ,(4,2,-3) and (0,4,3).

Therefore the equation of plane $\left|\begin{array}{ccc}(x-1)&(y-2)&(z-3)\\4-1&2-2&-3-3\\0-4&4-2&3+3\end{array}\right|=0$

⇒$\left|\begin{array}{ccc}(x-1)&(y-2)&(z-3)\\3&0&-6\\-4&2&6\end{array}\right|=0$

⇒12(x-1)-(y-2)(18+24)+(z-3)(6-0)=0

⇒12x -12 - 42y+84 +6z-18=0

⇒12x - 42y +6z  +54= 0

⇒2x-7y +z +9 =0

The equation of straight line is

$\frac{x-3}{3-2}=\frac{y+4}{-4+3} =\frac{z+5}{-5-1}$

⇒$\frac{x-3}{1}=\frac{y+4}{-1} =\frac{z+5}{-6}= r (say)$

Any point on the straight line be P (r+3,-r-4,-6r-5).

Let p be lie on the plane. So p will be satisfy the equation of plane.

∴2(r+3)-7(-r-4)+(-6r-5)+9=0

⇒3r=-38

⇒r=$-\frac{38}{3}$

∴ The coordinate of p = $(-\frac{29}{3} ,\frac{26}{3} ,33)$