A- B- C and D are four points in a plane with Position Vectors a- b- c- and d- respectively such that a-d- - b-c- - b-d- - c-a- -0- then for the triangle ABC- D is its

The correct option is B OrthocentreHere in triangle ABC #160; (a⃗ d⃗)·(b⃗ c⃗)=(b⃗ d⃗)·(c⃗ a⃗)=0 a⃗·b⃗ a⃗·c⃗ b⃗·d⃗+c⃗·d⃗=b⃗·c⃗ a⃗·b⃗ c⃗·d⃗+a⃗·d⃗=0 he ...

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Byju's Answer

Standard XII

Mathematics

Applications of Dot Product

A, B, C and ...

Question

$A, B, C$ and $D$ are four points in a plane with Position Vectors $\to a, \to b, \to c$ and $\to d$ respectively such that $(\to a - \to d) . (\to b - \to c) = (\to b - \to d) . (\to c - \to a) = 0,$ then for the triangle $A B C, D$ is its

Incentre

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Circumcentre

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Orthocentre

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Centroid

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Solution

The correct option is B Orthocentre
Here in triangle ABC
$(\to a - \to d) \cdot (\to b - \to c) = (\to b - \to d) \cdot (\to c - \to a) = 0$
$\to a \cdot \to b - \to a \cdot \to c - \to b \cdot \to d + \to c \cdot \to d = \to b \cdot \to c - \to a \cdot \to b - \to c \cdot \to d + \to a \cdot \to d = 0$
here the given expression can be only when the dot product of above vector in eq be zero
SO d should orthocentre of ABC due to which DOT PRODUCT will zero because of perpendicular

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A,B,C and D are four points in a plane with Position Vectors →a,→b,→c and →d respectively such that (→a−→d).(→b−→c)=(→b−→d).(→c−→a)=0, then for the triangle ABC,D is its

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