Given α→$$=3i^+j^+2k^$$ and β→$$=i^$$−$$2j^$$−$$4k^$$ are the position vectors of the points A and B. Then the distance of the point −$$i^+j^+k^$$ from the plane passing through B and perpendicular to AB is

Question

Given α→$$=3i^+j^+2k^$$ and β→$$=i^$$−$$2j^$$−$$4k^$$ are the  position vectors of the points A and B. Then the distance   of the point −$$i^+j^+k^$$ from the plane passing through B and perpendicular to AB is

Infinity Learn · Accepted Answer

AB→$$=$$β→−α→$$=$$−$$2i^$$−$$3j^$$−$$6k^Equation$$ of the plane passing through B and perpendicular to AB is  (r$$→−OB→$$)⋅AB→$$=0r$$→⋅(2i^+3j^+6k^)+28=0Hence$$, the required distance from r→$$=$$−$$i^+j^+k^=($$−$$i^+j^+k^)⋅$$(2i^+3j^+6k^)+28$$|$$2i^+3j^+6k^$$|$$=$$−$$2+3+6+287=5$$ units

$Given \vec{α} = 3 \hat{i} + \hat{j} + 2 \hat{k} and \vec{β} = \hat{i} - 2 \hat{j} - 4 \hat{k} are the position vectors of the points A and B . Then the distance$
$of the point - \hat{i} + \hat{j} + \hat{k} from the plane passing through B and perpendicular to AB is$

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Given α→=3i^+j^+2k^ and β→=i^−2j^−4k^ are the position vectors of the points A and B. Then the distance of the point −i^+j^+k^ from the plane passing through B and perpendicular to AB is

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$Given \vec{α} = 3 \hat{i} + \hat{j} + 2 \hat{k} and \vec{β} = \hat{i} - 2 \hat{j} - 4 \hat{k} are the position vectors of the points A and B . Then the distance$
$of the point - \hat{i} + \hat{j} + \hat{k} from the plane passing through B and perpendicular to AB is$