Let the position vectors of the points A- B and C be a- b- and c- respectively- Let Q be the point of intersection of the medians of the - ABC- Then- Q-A-Q-B-Q-C- is equal to

The correct option is E 0⃗ Since, #160; O is the intersecting point of all the medians of Δ ABC .Hence, Q= a+b+c 3 Now, Q⃗A⃗=a a+b+ ...

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Byju's Answer

Standard XII

Mathematics

Addition of Vectors

Let the posit...

Question

Let the position vectors of the points $A, B$ and $C$ be $\to a, \to b$ and $\to c$ , respectively. Let $Q$ be the point of intersection of the medians of the $Δ A B C$ . Then, $\to Q A + \to Q B + \to Q C$ is equal to

\frac{\to a + \to b + \to c}{2}

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2 \to a + \to b + \to c

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\to a + \to b + \to c

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\frac{\to a + \to b + \to c}{3}

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\to 0

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Solution

The correct option is E $\to 0$
Since, $O$ is the intersecting point of all the medians of $Δ A B C$ .
Hence, $Q = \frac{a + b + c}{3}$
Now, $\to Q A = a - \frac{a + b + c}{3}$
$= \frac{2 a - b - c}{3}$
Similarly, $\to Q B = \frac{2 b - a - c}{3}$
and $\to Q C = \frac{2 c - a - b}{3}$
Hence, $\to Q A + \to Q B + \to Q C = \frac{2 a - b - c}{3} + \frac{2 b - a - c}{3} + \frac{2 c - a - b}{3} = 0$

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Let the position vectors of the points A,B and C be →a,→b and →c, respectively. Let Q be the point of intersection of the medians of the ΔABC. Then, →QA+→QB+→QC is equal to

The correct option is E →0Since, O is the intersecting point of all the medians of ΔABC.Hence, Q=a+b+c3Now, →QA=a−a+b+c3 =2a−b−c3Similarly, →QB=2b−a−c3and →QC=2c−a−b3Hence, →QA+→QB+→QC=2a−b−c3+2b−a−c3+2c−a−b3=0

Let the position vectors of the points $A, B$ and $C$ be $\to a, \to b$ and $\to c$ , respectively. Let $Q$ be the point of intersection of the medians of the $Δ A B C$ . Then, $\to Q A + \to Q B + \to Q C$ is equal to