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+2 votes
183k views
in Mathematics by (31.5k points)
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Show that the lines intersect and hence find the point of intersection : 

vector r = (3i + 2j - 4k) + λ (i + 2j + 2k)

and vector r = (5i - 2j) + μ (3i + 2j + 6k).

2 Answers

+1 vote
by (15.0k points)
selected by
 
Best answer

Let the lines are \(M:\vec r = 3\hat i+ 2\hat j - 4\hat k + \lambda(\hat i + 2\hat j + 2\hat k)\) and \(N: \vec r = 5\hat i - 2\hat j + \,u(3\hat i + 2\hat j + 6 \hat k)\)

Coordinates of any random point on M are P(3 + λ, 2 + 2λ, −4 + 2λ) and on N are Q(5 + 3μ, −2 + 2μ, 6μ)

If the lines M and N intersect then, they must have a common point on them i.e., P and Q must coincide for some values of λ and μ

Now, 3 + λ = 5 + 3μ    -----(1)

2 + 2λ = −2 + 2μ-   ----(2)

−4 + 2λ = 6μ    -----(3)

Solving (1) and (2),

we get λ = −4, μ = −2

Substitute the values in equation 3,

−4 + 2(−4) = 6(−2)

−4 = −12 + 8

−4 = −4

So, the given lines intersect each other

Now, point of intersection is P(−1, −6, −12).

+1 vote
by (323k points)

The position vectors of arbitrary points on the given lines are :

3i + 2j - 4k + λ (i + 2j + 2k)

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