Question

Find the ratio in which the plane 3x+4y-5z=1 divides the line joining the points(-2,4,-6) and (3,-5,8).

Answer 1

Answer:

Therefore, the plane 3x + 4y - 5z - 1 = 0 divides the line segment joining (-2, 4, -6) and (3, -5, 8) in the ratio 3:4

Step-by-step explanation:

Let P(x, y, z) be a point in the given plane which divides the line segment joining A(-2, 4, -6) and B(3, -5, 8) in the ratio 1: k

ii) So by section formula,P(x, y, z) = $(\frac{-2k+3}{k+1} , \frac{4k-5}{k+1}, \frac{-6k+8}{k+1} )$

iii) Substituting this in plane equation,   3x + 4y - 5z = 1

$3(\frac{-2k+3}{k+1}) + 4(\frac{4k-5}{k+1}) - 5(\frac{-6k+8}{k+1}) - 1 = 0$

Solving the above, k = 4/3

So ratio of division is 1:(4/3) = 3:4

Therefore, the plane 3x + 4y - 5z - 1 = 0 divides the line segment joining (-2, 4, -6) and (3, -5, 8) in the ratio 3:4

Answer 2

Answer:

Therefore, the plane 3x + 4y - 5z - 1 = 0 divides the line segment joining (-2, 4, -6) and (3, -5, 8) in the ratio 3:4

Step-by-step explanation:

Let P(x, y, z) be a point in the given plane which divides the line segment joining A(-2, 4, -6) and B(3, -5, 8) in the ratio 1: k

ii) So by section formula,P(x, y, z) = (\frac{-2k+3}{k+1} , \frac{4k-5}{k+1}, \frac{-6k+8}{k+1} )(

k+1

−2k+3

,

k+1

4k−5

,

k+1

−6k+8

)

iii) Substituting this in plane equation, 3x + 4y - 5z = 1

3(\frac{-2k+3}{k+1}) + 4(\frac{4k-5}{k+1}) - 5(\frac{-6k+8}{k+1}) - 1 = 03(

k+1

−2k+3

)+4(

k+1

4k−5

)−5(

k+1

−6k+8

)−1=0

Solving the above, k = 4/3

So ratio of division is 1:(4/3) = 3:4

Therefore, the plane 3x + 4y - 5z - 1 = 0 divides the line segment joining (-2, 4, -6) and (3, -5, 8) in the ratio 3:4