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प्रश्न
Read the following passage and answer the questions given below.
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Two motorcycles A and B are running at the speed more than the allowed speed on the roads represented by the lines `vecr = λ(hati + 2hatj - hatk)` and `vecr = (3hati + 3hatj) + μ(2hati + hatj + hatk)` respectively.
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Based on the above information, answer the following questions:
- Find the shortest distance between the given lines.
- Find the point at which the motorcycles may collide.
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उत्तर
a. Given, lines are: `vecr = λ(hati + 2hatj - hatk)` and `vecr = (3hati + 3hatj) + μ(2hati + hatj + hatk)`
We know that, shortest distance between the lines
`vecr_1 = veca + λb_1` and `vecr = veca_2 + λvecb_1` is d = `(|(veca_2 - veca_1).(vecb_1 xx vecb_2)|)/(|vecb_1 xx vecb_2|)`
Here, `veca_1 = 0, veca_2 = (3hati + 3hatj)`
`vecb_1 = hati + 2hatj - hatk`
and `vecb_2 = 2hati + hatj + hatk`
∴ `veca_2 - veca_1 = (3hati + 3hatj) - 0 = 3hati + 3hatj`
`vecb_1 xx vecb_2 = |(hati, hatj, hatk),(1, 2, -1),(2, 1, 1)|`
= `hati(2 + 1) - hatj(1 + 2) + hatk(1 - 4)`
= `3hati - 3hatj - 3hatk`
and `|vecb_1 xx vecb_2| = sqrt(3^2 + (-3)^2 + (-3)^2)`
= `sqrt(9 + 9 + 9)`
= `3sqrt(3)`
Also, `(veca_2 - veca_1).(vecb_1 xx vecb_2) = (3hati + 3hatj).(3hati - 3hatj - 3hatk)`
= 9 – 9
= 0
d = `0/(3sqrt(3))` = 0
Thus, distance between lines is 0.
b. We have, `vecr = λ(hati + 2hatj - hatk)` ...(i)
and `vecr = 3hati + 3hatj + μ(2hati + hatj + hatk)`
or `vecr = (3 + 2μ)hati + (3 + μ)hatj + μhatk` ...(ii)
Now, from equation (i) and equation (ii), we get
`λ(hati + 2hatj - hatk) = (3 + 2μ)hati + (3 + μ)hatj + μhatk`
On comparing both sides, we get
3 + 2µ = λ, 3 + µ = 2λ and µ = –λ
On solving for values of λ and µ, we get
λ = 1 and µ = –1
From equation (i), we get `vecr = hati + 2hatj - hatk`
`xhati + yhatj + zhatk = hati + 2hatj - hatk`
So, required point is (1, 2, –1).
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