Let $\to b = -^i + 4^j + 6^k, \to c = 2^i - 7^j - 10^k$ . If a be a unit vector and the scalr triple product $a, b, c$ has the greatest value, then $\to a$ is

\frac{1}{\sqrt{3}} (^i +^j +^k)

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\frac{1}{\sqrt{5}} (\sqrt{2}^i -^j - \sqrt{2}^k)

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\frac{1}{3} (2^i + 2^j -^k)

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\frac{1}{3} (2^i + 2^j +^k)

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Solution

The correct option is C $\frac{1}{3} (2^i + 2^j -^k)$
Let $\to a = p^i + q^j + r^k$
$\to b = -^i + 4^j + 6^k, \to c = 2^i - 7^j - 10^k$
$[\to a \to b \to c] = ∣ ∣ ∣ ∣ \begin{matrix} p & q & r - 1 & 4 & 6 2 & - 7 & - 10 \end{matrix} ∣ ∣ ∣ ∣$
$= p ∣ ∣ ∣ \begin{matrix} 4 & 6 - 7 & - 10 \end{matrix} ∣ ∣ ∣ - q ∣ ∣ ∣ \begin{matrix} - 1 & 6 2 & - 10 \end{matrix} ∣ ∣ ∣ + r ∣ ∣ ∣ \begin{matrix} - 1 & 4 2 & - 7 \end{matrix} ∣ ∣ ∣$
$= 2 p + 2 q - r$
Since, $p^{2} + q^{2} + r^{2} = 1$ , we can write by cylindrical (spherical ) co ordinates
$p = sin θ cos ϕ$
$q = sin θ sin ϕ$
$r = cos θ$
$∴ S = 2 sin θ cos ϕ + 2 sin θ sin ϕ - cos ϕ$
$= 2 sin θ (sin ϕ + cos ϕ) - cos θ$
Now, S is maximum (given) then $sin ϕ + cos ϕ$ is also maximum.
$∴ sin ϕ + cos ϕ = \sqrt{2} sin (\frac{π}{4} + ϕ) \leq \sqrt{2}$
At $ϕ = 45 °$
Also $S \leq 2 \sqrt{2} sin θ - cos θ$
$a cos θ + b sin θ \leq \sqrt{a^{2} + b^{2}}$
When, $cos θ = \frac{a}{\sqrt{a^{2} + b^{2}}}, sin θ = \frac{b}{\sqrt{a^{2} + b^{2}}}$
$∴ S_{m a x} = \sqrt{{(2 \sqrt{2})}^{2} + 12} = 3$
when $cos θ = \frac{- 1}{3}, sin θ = \frac{2 \sqrt{2}}{3}$
$∴ p = \frac{2}{3}, q = \frac{2}{3}, r = \frac{- 1}{3}$
$∴ \to a = \frac{1}{3} (2^i + 2^j -^k)$

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Let →b=−^i+4^j+6^k,→c=2^i−7^j−10^k. If a be a unit vector and the scalr triple product a,b,c has the greatest value, then →a is

Let $\to b = -^i + 4^j + 6^k, \to c = 2^i - 7^j - 10^k$ . If a be a unit vector and the scalr triple product $a, b, c$ has the greatest value, then $\to a$ is