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Question
Find the distance of the point (1, –2, 0) from the point of the line `vecr = 4hati + 2hatj + 7hatk + λ(3hati + 4hatj + 2hatk)` and the point `vecr.(hati - hatj + hatk)` = 10.
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Solution
Given line is `vecr = 4hati + 2hatj + 7hatk + λ(3hati + 4hatj + 2hatk)` ...(i)
and plane is `vecr.(hati - hatj + hatk)` = 10 ...(ii)
From (i) and (ii), we get
`4hati + 2hatj + 7hatk + λ(3hati + 4hatj + 2hatk).(hati - hatj + hatk)` = 10
⇒ `(4 + 3λ)hati + (2 + 4λ)hatj + (7 + 2hatk)(hati - hatj + hatk)` = 10
⇒ (4 + 3λ).1 + (2 + 4λ)(–1) + (7 + 2λ).1 = 10
⇒ 4 + 3λ – 2 – 4λ + 7 + 2λ = 10
⇒ λ + 9 = 10
⇒ λ = 1
Now, from (i), we get
`vecr = (4hati + 2hatj + 7hatk) + 1(3hati + 4hatj + 2hatk)`
or, `vecr = 7hati + 6hatj + 9hatk`
or, `xhati + yhatj + zhatk = 7hati + 6hatj + 9hatk`
∴ Required point is (7, 6, 9)
Now, distance between (7, 6, 9) and (1, –2, 9) is
d = `sqrt((7 - 1)^2 + (6 + 2)^2 + (9 - 9)^2)`
= `sqrt(6^2 + 8^2 + 0^2)`
= `sqrt(36 + 64)`
= `sqrt(100)`
= 10 units
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