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Question
Solve the following:
Find the distance of the point `3hat"i" + 3hat"j" + hat"k"` from the plane `bar"r".(2hat"i" + 3hat"j" + 6hat"k")` = 21.
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Solution
The distance of the point `"A"(bara)` from the plane `bar"r".bar"n" = p "is given by" d = |bar"a".bar"n" - p|/|bar"n"|` ...(1)
Here, `bar"a" = 3hat"i" + 3hat"j" + hat"k", bar"n" = 2hat"i" + 3hat"j" + 6hat"k"`, p = 21
∴ `bar"a".bar"n" = (3hat"i" + 3hat"j" + hat"k").(2hat"i" + 3hat"j" + 6hat"k")`
= (3)(2) + (3)(3) + (1)(–6)
= 6 + 9 – 6
= 9
Also, `|bar"n"| = sqrt(3^2 + 3^2 + (-6)^2) = sqrt(-12)` = 0
∴ from (1), the required distance
= `|- 12 - 21|/(12)`
= 0 units.
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