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The length of the perpendicular drawn fr...

The length of the perpendicular drawn from the point `(2,1,4)` to the plane containing the lines
`vec(r) = (hati +hatj) +lambda (hati +2 hatj - hatk)` and `vec(r) = (hati +hatj) +mu (-hati +hatj - 2hatk)` is:

A

3

B

`(1)/(3)`

C

`sqrt(3)`

D

`(1)/(sqrt(3))`

Text Solution

Verified by Experts

The correct Answer is:
A
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The length of the perpendicular drawn from the point (2, 1, 4) to the plane containing the lines r=(hati+hatj)+lambda(hati+2hatj-hatk)" and "r=(hati+hatj)+mu(-hati+hatj-2hatk) is

The shortest distance between the lines vecr = (-hati - hatj) + lambda(2hati - hatk) and vecr = (2hati - hatj) + mu(hati + hatj -hatk) is

Find the shortest distance betwee the lines : vec(r) = (hati + 2 hatj + hatk ) + lambda ( hati - hatj + hatk) and vec(r) = 2 hati - hatj - hakt + mu (2 hati + hatj + 2 hatk) .

Find the shortest distance between the lines bar(r) = (4hati - hatj) + lamda(hati + 2hatj - 3hatk) and bar(r) = (hati - hatj + 2hatk) + mu(hati + 4hatj - 5hatk).

The shortest distance between the lines vecr = (2hati - hatj) + lambda(2hati + hatj - 3hatk) vecr = (hati - hatj + 2hatk) + lambda(2hati + hatj - 5hatk)

Find the shortest distance between lines: vec(r) = 6 hati + 2 hatj + 2 hatk + lambda ( hati - 2 hatj + 2 hatk) and vec(r) = -4 hati - hatk + mu (3 hati - 2 hatj - 2 hatk) .

Find the vector equation of the plane in which the lines vecr=hati+hatj+lambda(hati+2hatj-hatk) and vecr=(hati+hatj)+mu(-hati+hatj-2hatk) lie.

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