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प्रश्न
Find the distance of the point (–1, –5, – 10) from the point of intersection of the line `vecr = 2hati - hatj + 2hatk + lambda(3hati + 4hatj + 2hatk)` and the plane `vecr * (hati - hatj + hatk)` = 5.
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उत्तर
We have `vecr = 2hati - hatj + 2hatk + lambda(3hati + 4hatj + 2hatk)` and `vecr * (hati - hatj + hatk)` = 5.
Solving these two equations
We get `[(2hati - hatj + 2hatk) + lambda(3hati + 4hatj + 2hatk)]*(hati - hatj + hatk)` = 5
Which gives `lambda` = 0
Therefore, the point of intersection of line and the plane is (2, 1, 2) − and the other given point is (– 1, – 5, – 10).
Hence the distance between these two points is `sqrt([2 - (-1)^2] + [-1 + 5]^2 + [2 - (-10)]^2`
i.e. 13
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