Question

find the point of intersection of the line x+3/4=y+4/3=z-8/-5 with the sphere x^2+y^2+z^2+2x-10y=23

Answer 1

SOLUTION

TO DETERMINE

The point of intersection of the line

$\displaystyle \sf{ \frac{x + 3}{4} = \frac{y + 4}{3} = \frac{z - 8}{ - 5} }$

with the sphere

$\displaystyle \sf{ {x}^{2} + {y}^{2} + {z}^{2} + 2x - 10y = 23 }$

EVALUATION

Here the given equation of the line is

$\displaystyle \sf{ \frac{x + 3}{4} = \frac{y + 4}{3} = \frac{z - 8}{ - 5} = t \: (say)}$

Thus any point on the above line is of the form ( 4t - 3 , 3t - 4 , - 5t + 8 )

Now the given equation of the sphere is

$\displaystyle \sf{ {x}^{2} + {y}^{2} + {z}^{2} + 2x - 10y = 23 }$

$\displaystyle \sf{ \implies {x}^{2} +2x + {y}^{2} - 10y + {z}^{2} = 23 }$

$\displaystyle \sf{ \implies {x}^{2} +2x + 1+ {y}^{2} - 10y +25 + {z}^{2} = 23 + 1 + 25 }$

$\displaystyle \sf{ \implies {(x + 1)}^{2} + {(y - 5)}^{2} + {z}^{2} = 49 }$

Now we put the point ( 4t - 3 , 3t - 4 , - 5t + 8 ) we get

$\displaystyle \sf{ {(4t - 3 + 1)}^{2} + {(3t - 4- 5)}^{2} + {( - 5t + 8)}^{2} = 49 }$

$\displaystyle \sf{ \implies {(4t - 2)}^{2} + {(3t - 9)}^{2} + {( 5t - 8)}^{2} = 49 }$

$\displaystyle \sf{ \implies 16 {t}^{2} - 16t + 4 + 9 {t}^{2} - 54t + 81 + 25 {t}^{2} - 80t + 64 = 49 }$

$\displaystyle \sf{ \implies 50 {t}^{2} - 150t + 100 = 0 }$

$\displaystyle \sf{ \implies {t}^{2} - 3t + 2 = 0 }$

$\displaystyle \sf{ \implies (t - 1)(t - 2) = 0 }$

$\displaystyle \sf{ \implies t = 1 \: , \: 2}$

Putting t = 1 we get the point of intersection as ( 1 , - 1 , 3 )

Putting t = 2 we get the point of intersection as ( 5 , 2 , - 2 )

FINAL ANSWER

Hence the required point of intersection of the line and the sphere is

• ( 1 , - 1 , 3 )

• ( 5 , 2 , - 2 )

━━━━━━━━━━━━━━━━

Learn more from Brainly :-

$1.$ a butterfly is moving in a straight line in the space.

let this path be denoted by a line l whose equation is x-2/2=2-y...

https://brainly.in/question/30868719

2. Find the distance of point p(3 4 5) from the yz plane

https://brainly.in/question/8355915