Question

Question 3D vector!


Let line x-2/3 = y+1/2= z-1/-1 intersect the curve xy=c²,z=0 then find 'c'.


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Answer 1

Answer:

C = +-√5

Step-by-step explanation:

Given :- z = 0 , since ,

x-2/3 = y+1/2 = z-1/-1 ..

x- 3 /3 =y+1/2 = 1

x = 3+2 = 5 ,

y = 1

but ,xy = c²

since, c = +-√5 Answer

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Hope it helps !

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Answer 2

Given the equation of line

$\frac{x - 2}{3} = \frac{y + 1}{ 2} = \frac{z - 1}{ - 1} = p(let)$

X = 3p +2

y = 2p -1

z = -p + 1

so p' (3p + 2,2p - 1,1 -p )

Now given the line intersect the given curve so point p' satisfy the equation

for given equation 1

Z = 0

1 - p= 0

p = 1

for given equation 2

xy = c^2

(3p + 2) (2p - 1) = c^2

Now on putting p = 1 we get

c^2 = (3 + 2)(2 - 1)

c = sqrt5