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प्रश्न
Find the distance of the point (–1, –5, – 10) from the point of intersection of the line `vec"r" = 2hat"i" - hat"j" + 2hat"k" + lambda(3hat"i" + 4hat"j" + 2hat"k")` and the plane `vec"r" * (hat"i" - hat"j" + hat"k")` = 5
योग
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उत्तर
We have `vec"r" = 2hat"i" - hat"j" + 2hat"k" + lambda(3hat"i" + 4hat"j" + 2hat"k")` and `vec"r" * (hat"i" - hat"j" + hat"k")` = 5
Solving these two equations,
We get `[(2hat"i" - hat"j" + 2hat"k") + lambda(3hat"i" + 4hat"j" + 2hat"k")]*(hat"i" - hat"j" + hat"k")` = 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). H
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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