Question

Find the ratio in which the plane x-2y+3z=17 divides the line joining points (-2,4,7)&(3,-5,8)

Answer 1

Answer:

Step-by-step explanation:

Given Find the ratio in which the plane x-2y+3z=17 divides the line joining points (-2,4,7) and (3,-5,8)

Let m = k and n = 1

It is in the form (mx2 + nx1 / m + n , my2 + ny1 / m + n, mz2 + nz1 / m + n)

So here x1 = -2, y1 = 4 and z1 = 7  x2 = 3, y2 = -5, z2 = 8

Substituting the values we get

 3k - 2/ x+ 1, -5k + 4 / k + 1, 8k + 7 / k + 1

From the given equation x - 2y + 3z = 17 we get

3k - 2/k+1 - 2 (-5k + 4/k + 1) + 3 (8k + 7 / k + 1) = 17

3k - 2 + 10k - 8 + 24k + 21 = 17k + 17

20k = 6

k = 3/10

So the ratio is 3 : 10

Answer 2

Answer:

The given points cuts the plane in 3:10 ratio.

Step-by-step explanation:

A plane x-2y+3z=17

Let point P(-2,4,7) and Q(3,-5,8) cut the plane in k:1 ratio.

Section formula:  $(x,y,z)\rightarrow (\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n},\dfrac{mz_2+nz_1}{m+n})$

Using section formula to find the coordinate of intersection of line and plane.

$x=\dfrac{3k-2}{k+1}$

$y=\dfrac{-5k+4}{k+1}$

$z=\dfrac{8k+7}{k+1}$

This point must be satisfy the given equation of plane.

Now, we will plug in the value of x, y and z into x+2y+3z=17 and solve for k

$\dfrac{3k-2}{k+1}+\dfrac{-10k+8}{k+1}+\dfrac{24k+21}{k+1}=17$

$3k-2+10k-8+24k+21=17k+17$

$37k+11=17k+17$

$20k=6$

$k=\dfrac{3}{10}$

Hence, The given points cuts the plane in 3:10 ratio.