Question

Find the equation of a plane passing through the points A(2,1,2) and B(4,-2,1) and perpendicular to the plane r(i-2k)=5. Also, find the coordinates of the point where the line passing through the points (3,4,1) and (5,1,6) crosses the plane thus obtained.

Answer 1

Answer:

Heree is ur ans.........

Answer 2

The equation of plane is 3x+y+z=9

The coordinate of the point of intersection of line and plane is $(\dfrac{7}{4},\dfrac{47}{8},-\dfrac{17}{8})$

Step-by-step explanation:

A plane passing through the points A(2,1,2) and B(4,-2,1).

The vector, AB = 2i - 3j - k

The required plane perpendicular to the plane, r(i-2k)=5

Normal vector of this plane, n' = i - 2k

The cross product of AB and n is normal to the required plane.

$\vec{n}=AB\times n'$

$\vec{n}=<2,-3,-1>\times <1,0,-2>$

$\vec{n}=<6,3,3>$

The normal vector to required plane, n = 3i + j + k

Passing point of plane, A(2,1,2)

Equation of required plane:-

$<x-2,y-1,z-2>\cdot<3,1,1>=0$

$3x+y+z=9$

A line passing through (3,4,1) and (5,1,6) crosses the plane.

Equation of line:-

$\dfrac{x-3}{5-3}=\dfrac{y-4}{1-4}=\dfrac{z-1}{6-1}=t$

$x=2t+3$

$y=-3t+4$

$z=5t+1$

Substitute the value of x, y and z into plane equation.

The point of intersection of line and plane is,

3(2t+3)+(-3t+4)+(5t+1)=9

6t+9-3t+4+5t+1=9

8t=-5

$t=-\dfrac{5}{8}$

The coordinate of point of intersection,

$x=2\times -\dfrac{5}{8}+3=\dfrac{7}{4}$

$y=-3\times -\dfrac{5}{8}+4=\dfrac{47}{8}$

$z=5\times -\dfrac{5}{8}+1=-\dfrac{17}{8}$

Hence, The coordinate of the point of intersection of line and plane $(\dfrac{7}{4},\dfrac{47}{8},-\dfrac{17}{8})$

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