If a- b- c are different real numbers and a-b-ck-b-c-ak- c-a-bk- are position vectors of three non-collinear points A- B C then

The correct options are A centroid of triangle ABC is a+b+c/3 ( î+ĵ+k̂ ) D triangle ABC is an equilateral triangleGiven position #160;vectors: #160; ...

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Standard XII

Physics

Relative Velocity in 2D

If a, b, c ar...

Question

If a, b, c are different real numbers and $a^i + b^j + c^k; b^i + c^j + a^k$ & $c^i + a^j + b^k$ are position vectors of three non-collinear points A, B & C then

centroid of triangle ABC is

\frac{a + b + c}{3} (^i +^j +^k)

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^i +^j +^k

is equally inclined to the three vectors

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perpendicular from the origin to the plane of triangle ABC meet at centroid

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triangle ABC is an equilateral triangle

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Solution

The correct options are
A centroid of triangle ABC is $\frac{a + b + c}{3} (^i +^j +^k)$
D triangle ABC is an equilateral triangle
Given position vectors: $[\to A = a^i + b^j + c^k]$ ; $[\to B = b^i + c^j + a^k]$ ; $[\to C = c^i + a^j + b^k]$

Use formula for centroid of triangle when three position vectors A, B and C are
given $[\frac{\to A + \to B + \to C}{3}]$

$\frac{(a^i + b^j + c^k) + (b^i + c^j + a^k) + (c^i + a^j + b^k)}{3}$

$\frac{^i (a + b + c) +^j (a + b + c) + \to k (a + b + c)}{3}$

$\frac{(a + b + c)}{3} (^i +^j + \to k) .$

Also, it is an equilateral triangle if
$| - - \to A B | = | - - \to B C | = | - - \to A C |$

Let O be the position vector where $O = (0^i + 0^j + 0 \to k)$ then;
$- - \to O A = a^i + b^j + c^k$

$- - \to O B = b^i + c^j + a^k$

$- - \to O C = c^i + a^j + b^k$

Find $| - - \to A B |$ ;

$- - \to A B = - - \to O B - - - \to O A$

$(b^i + c^j + a^k) - (a^i + b^j + c^k)$

$(b - a)^i + (c - b)^i + (a - c)^i$

$| - - \to A B | = - - \to O B - - - \to O A$

$| - - \to A B | = \sqrt{{(b - a)}^{2} + {(c - b)}^{2} + {(a - c)}^{2}}]$

$- - \to A B | = \sqrt{a^{2} + b^{2} - 2 a b + c^{2} + b^{2} - 2 b c + a^{2} + c^{2} - 2 a c}$

$| - - \to A B | = \sqrt{2 (a^{2} + b^{2} + c^{2}) - 2 a b - 2 b c - 2 a c}$

Similarly, $| - - \to A C | = - - \to O C - - - \to O A]$

$| - - \to A C | = \sqrt{{(c - a)}^{2} + {(a - b)}^{2} + {(b - c)}^{2}}$

$| - - \to A C | = \sqrt{c^{2} + a^{2} - 2 a c + a^{2} + b^{2} - 2 a b + b^{2} + c^{2} - 2 b c}$

$| - - \to A C | = \sqrt{2 (a^{2} + b^{2} + c^{2}) - 2 a b - 2 b c - 2 a c}$

Similarly, $| - - \to B C | = \sqrt{2 (a^{2} + b^{2} + c^{2}) - 2 a b - 2 b c - 2 a c}$

Hence, the given position vectors forms and equilateral triangle ABC and the centroid of the triangle is
given as $\frac{(a + b + c)}{3} (^i +^j + \to k) .$
Option (A) and (D) both are correct.

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If a, b, c are different real numbers and a^i+b^j+c^k;b^i+c^j+a^k & c^i+a^j+b^k are position vectors of three non-collinear points A, B & C then

If a, b, c are different real numbers and $a^i + b^j + c^k; b^i + c^j + a^k$ & $c^i + a^j + b^k$ are position vectors of three non-collinear points A, B & C then