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Applications of Dot Product

a⃗,b⃗,c⃗ are ...

Question

$\to a, \to b, \to c$ are mutually perpendicular vectors of equal magnitude then angle between $\to a + \to b + \to c$ and $\to a$ is

A

{cos}^{- 1} ∣ ∣ ∣ \frac{1}{3} ∣ ∣ ∣

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B

{cos}^{- 1} ∣ ∣ ∣ \frac{1}{\sqrt{3}} ∣ ∣ ∣

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C

π - {cos}^{- 1} ∣ ∣ ∣ \frac{1}{\sqrt{3}} ∣ ∣ ∣

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D

{tan}^{- 1} \sqrt{2}

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Solution

The correct options are
B ${cos}^{- 1} ∣ ∣ ∣ \frac{1}{\sqrt{3}} ∣ ∣ ∣$
D ${tan}^{- 1} \sqrt{2}$
$\to a, \to b, \to c$ are vectors of equal magnitude and mutually perpendicular to each other then angle between $\to a + \to b + \to c$ & $\to a$ is given by
$cos θ = \frac{(\to a + \to b + \to c) . \to a}{∣ ∣ ∣ \to a + \to b + \to c ∣ ∣ ∣ ∣ ∣ \to a ∣ ∣} = \frac{(\to a + \to b + \to c) . \to a}{\sqrt{^{2} +^{2} +^{2}} ∣ ∣ \to a ∣ ∣} = \frac{∣ ∣ \to a ∣ ∣}{\sqrt{{\to a}^{2} +^{2} +^{2}}} = \frac{1}{\sqrt{3}}$
$\Rightarrow θ = {cos}^{- 1} (\frac{1}{\sqrt{3}}) = {tan}^{1} \sqrt{2}$
Hence, options B and D.

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