The position vector of the center of the circle - r- -5-r- i-j-k- -3- 3 is

The correct option is A √( 3 )(#160;i+j⃗+k⃗) The equation of ON is r=λ(#160;i+j⃗+k⃗) #160; ... (1) Since it passes through the origin a ...

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Standard XII

Mathematics

Standard Equation of Parabola

The position ...

Question

The position vector of the center of the circle $| \to r | = 5, \to r . (\to i + \to j + \to k) = 3 \sqrt{3}$ is

\sqrt{3} (\to i + \to j + \to k)

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\to i + \to j + \to k

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3 (\to i + \to j + \to k)

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None of these

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Solution

The correct option is A $\sqrt{3} (\to i + \to j + \to k)$
The equation of $O N$ is $r = λ (\to i + \to j + \to k)$ ... $(1)$
Since it passes through the origin and is parallel to the vector $(\to i + \to j + \to k)$ , any point on it is $λ (\to i + \to j + \to k)$ .
If this point lies on the plane $r . (\to i + \to j + \to k) = 3 \sqrt{3}$ ,
then $λ (i + j + k) . (\to i + \to j + \to k) = 3 \sqrt{3}$
$\Rightarrow λ (1 + 1 + 1) = 3 \sqrt{3}$
$\Rightarrow λ = \sqrt{3}$
Putting the value of $λ$ in $(1)$ , we get the position vector $N$
i.e. center of the circle as $\sqrt{3} (\to i + \to j + \to k)$

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The position vector of the center of the circle |→r|=5,→r.(→i+→j+→k)=3√3 is

The position vector of the center of the circle $| \to r | = 5, \to r . (\to i + \to j + \to k) = 3 \sqrt{3}$ is