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Question
Show that the line whose vector equation is `vecr = (2hati - 2hatj + 3hatk) + λ(hati - hatj + 4hatk)` is parallel to the plane whose vector equation is `vecr.(hati + 5hatj + hatk) = 5`. Also find the distance between them.
Solution
P.V. of the point on the line `2hati - 2hatj + 3hatk`
Direction vector `hati - hatj + 4hatk`
Given plane `vecr.(hati + 5hatj + hatk) - 5 = 0`
∴ Normal vector of the plane `hati + 5hatj + hatk`
If straight line and plane are parallel to each other
`\implies` Direction vector of line and normal vector of the plane are perpendicular to each other
Now we get `(hati - hatj + 4hatk)(hati + 5hatj + hatk) = 0`
∵ Since the dot product is zero, they are perpendicular to each other.
`\implies` Straight line and plane are parallel to each other.
Required distance = `((2hati - 2hatj + 3hatk)*(hati + 5hatj + hatk) - 5)/sqrt(1 + 25 + 1)`
= `10/sqrt(27)`
= `10/(3sqrt(3))` units.
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