Question

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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.​

Answer 1

$\mathfrak\red{answer}$

Point of intersection is P(1,-2,7).

P externally divides the line segment AB in the ratio 2:1

step-by-step explanation:

The equation of the plane passing through three given points can be given by

$|x - 2 \: \: \ \: y - 2 \: \: \: z - 1| \: \: \: \: \: \: \: \: \\ |x - 3 \: \: \: \: y - 0 \: \: \: z - 1| = 0 \\ | x - 4 \: \: \: \: y - 1 \: \: \: z - 0| \: \: \: \: \: \: \: \:$

Performing elementary row operations R2=>R1-R2 and R3=>R1-R3,

we get

=》

$|x - 2 \: \: \ \: y - 2 \: \: \: z - 1| \: \: \: \: \: \: \: \: \\ |3 - 2 \: \: \: \: 0 - 2 \: \: \: \: \: 0 \: \: \: \: \: \: | = 0 \\ | x - 4 \: \: \: \: y - 1 \: \: \: z - 0| \: \: \: \: \: \: \: \:$

=》

$|x - 2 \: \: \ \: y - 2 \: \: \: z - 1| \: \: \: \: \: \: \: \: \\ |1 \: \: \: \: \: \: - 2 \: \: \: \: \: \: \: \: \: \: \: 0 \: \: \: \: \: \: \: \: | = 0 \\ | 2 \: \: \: \: \: \: \: - 3 \: \: \: \: \: \: - 1 \: \: \: \: \: \: \: | \: \: \: \: \: \: \: \: \:$

Solving the above determinant, we get

=》(x-2)(2-0)-(y-2)(-1-0)+(z-1)(-3+4)=0

=》(2x-4)+(y-2)+(z-1)=0

=》2x+y+z-7=0

Therefore, the equation of the plane is

2x+y+z-7=0

Now, the equation of the line passing through two given points is

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

$= > \frac{x - 3}{ - 1} = \frac{y + 4}{1} = \frac{z + 5}{6} = \alpha$

$= > x = ( - \alpha + 3) \\ y = ( \alpha - 4) \\ z = (6 \alpha - 5)$

At the point of intersection, these points satisfy the equation of the plane

2x+y+z-7=0.

Putting the values of x, y and z in the equation of the plane, we get the value of a.

$2( - \alpha + 3) + ( \alpha - 4) - (6 \alpha - 5) + 7 = 0$

$= > 2 \alpha + 6 + \alpha - 4 + 6 \alpha - 5 - 7 = 0$

$5 \alpha = 10 \\ \alpha = 2$

Thus, the point of intersection is P(1, -2, 7).

Now, let P divide the line AB in the ratio m:

n.

By the section formula, we have

$1 = \frac{2m + 3n}{m + n} \: \: \: \: \: \: \\ = > m + 2n = 0 \\ = > m = - 2n \: \: \: \: \\ = > \frac{m}{n} = > \frac{ - 2}{1}$

Hence, P externally divides the line segment AB in the ratio 2:1

Answer 2

Answer:

Point of intersection is P(1,-2,7).

P externally divides the line segment AB in the ratio 2:1

Step-by-step explanation:

The equation of the plane passing through three given points can be given by

=》| x-2 y-2 z-1 |

| 3-2 y-0 z-1 | => 0

| x-4 y-1 z-0 |

Performing elementary row operations R2=>R1-R2 and R3=>R1-R3,

we get

=》| x-2 y-2 z-1 |

| 3-2 y-0 z-1 | => 0

| x-4 y-1 z-0 |

=》x-2 y-2 z-1

1 - 1 0 = 0

2 - 3 - 1 = 0

Solving the above determinant, we get

=》(x-2)(2-0)-(y-2)(-1-0)+(z-1)(-3+4)=0

=》(2x-4)+(y-2)+(z-1)=0

=》2x+y+z-7=0

Therefore, the equation of the plane is

2x+y+z-7=0

Now, the equation of the line passing through two given points is

x-3 = y+4 = z+5 = α

2-3 -3+4 1+5

=>

x = (-α+3)

y = (α-4)

z = (6α-5)

At the point of intersection, these points satisfy the equation of the plane

2x+y+z-7=0.

Putting the values of x, y and z in the equation of the plane, we get the value of a.

2α+6+α-4+6α-5-7=0

5α=10

α = 2

Thus, the point of intersection is P(1, -2, 7).

Now, let P divide the line AB in the ratio m:

n.

By the section formula, we have

1 = 2m+3n

m+n

= m+2n =0

m= -2n

m = -2

n 1

Hence, P externally divides the line segment AB in the ratio 2:1