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Vector Addition

If a⃗, b⃗ a...

Question

If $\to a, \to b$ and $\to c$ are mutually perpendicular vectors of equal magnitudes, If the angles which the vector $2 \to a + \to b + 2 \to c$ makes with the vectors $\to a$ is ${cos}^{- 1} \sqrt{\frac{2}{m}}$ . Find $m$

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Solution

$\to a . \to b = \to b . \to c = \to c . \to a = 0$

$| \to a | = | \to b | = | \to c | = k$

$\to c . (2 \to a + \to b + 2 \to c) = | \to c | | 2 \to a + \to b + 2 \to c | cos θ$

Where $θ$ is the angle between $\to c$ and $2 \to a + \to b + 2 \to c$

$2 \to a . \to c + \to b . \to c + 2 \to c . \to c = k | 2 \to a + \to b + 2 \to c | cos θ$ $[\to a . \to b = \to b . \to a]$

$2 | \to c |^{2} = k [2 \to a + \to b + 2 \to c] cos θ$

$2 k^{2} = k [2 \to a + \to b + 2 \to c] cos θ$

$cos θ = \frac{2 k}{| 2 \to a + \to b + 2 \to c |}$

$θ = {cos}^{- 1} \frac{2 k}{| 2 \to a + \to b + 2 \to c |}$

${cos}^{- 1} \sqrt{\frac{2}{m}} = {cos}^{- 1} \frac{2 k}{| 2 \to a + \to b + 2 \to c |}$

$\sqrt{\frac{2}{m}} = \frac{2 k}{| 2 \to a + \to b + 2 \to c |}$ ---- (i)

Now,

$\begin{matrix} {∣ ∣ 2 \to a + \to b + 2 \to c ∣ ∣}^{2} = (2 \to a + \to b + 2 \to c) . (2 \to a + \to b + 2 \to c) = 4. \to a . \to a + 2 \to a . \to b + 4. \to a . \to c + 2. \to a . \to b + \to b . \to b + 2. \to b . \to c + 4. \to a . \to c + 2. \to b . \to c + 4. \to c . \to c = 4 {| \to a |}^{2} + {∣ ∣ \to b ∣ ∣}^{2} + 4 {| \to c |}^{2} = 4 k^{2} + k^{2} + 4 k^{2} = 9 k^{2} ∣ ∣ 2 \to a + \to b + 2 \to c ∣ ∣ = 3 k \end{matrix}$

Putting these values in (i)

$\sqrt{\frac{2}{m}} = \frac{2 k}{3 k}$

$\sqrt{\frac{2}{m}} = \frac{2}{3}$

$\frac{2}{m} = \frac{4}{9}$

$m = \frac{9}{2}$

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