Let $\to v = 2^i +^j -^k$ and $\to w =^i + 3^k$ . If $\to u$ is a unit vector, then maximum value of the scalar triple product $[\to u, \to v, \to w]$ is

- 1

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\sqrt{10} + \sqrt{6}

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\sqrt{59}

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\sqrt{60}

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Solution

The correct option is A $\sqrt{59}$
$[\to u \to v \to w] = ∣ ∣ \to u . (\to v \times \to w) ∣ ∣$
$= ∣ ∣ \to u . (3^i - 7^j -^k) ∣ ∣$
$= ∣ ∣ \to u ∣ ∣ \sqrt{59} cos θ$
$∴$ Maximum $[\to u, \to v, \to w] = \sqrt{59}$
$(∴ ∣ ∣ \to u ∣ ∣ = 1, cos θ \leq 1)$

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Similar questions

$\to V = 2^i +^j -^k$ and $\to W =^i + 3^k$ . If $\to U$ is a unit vector, then the maximum value of the scalar triple product $[\to U \to V \to W]$ is

Q. Let

\to U = 2^i +^j -^k, - \to W =^i + 3^k

, if

\to V

is a unit vector,then the maximum value of

[\to U, \to V, - \to W]

Q. If

¯ V = 2^i +^j -^k

and

¯ W =^i + 3^k

. If

¯ U

is a unit vector then the maximum value of the scalar triple product

[¯ U ¯ V ¯ W]

$\to V = 2^i +^j -^k$ and $\to W =^i + 3^k$ . If $\to U$ is a unit vector, then the maximum value of the scalar triple product $[\to U \to V \to W]$ is

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Let →v=2^i+^j−^k and →w=^i+3^k. If →u is a unit vector, then maximum value of the scalar triple product [→u,→v,→w] is

The correct option is A √59[→u→v→w]=∣∣→u.(→v×→w)∣∣=∣∣→u.(3^i−7^j−^k)∣∣=∣∣→u∣∣√59cosθ∴ Maximum [→u,→v,→w]=√59(∴∣∣→u∣∣=1,cosθ≤1)

Let $\to v = 2^i +^j -^k$ and $\to w =^i + 3^k$ . If $\to u$ is a unit vector, then maximum value of the scalar triple product $[\to u, \to v, \to w]$ is

The correct option is A $\sqrt{59}$
$[\to u \to v \to w] = ∣ ∣ \to u . (\to v \times \to w) ∣ ∣$
$= ∣ ∣ \to u . (3^i - 7^j -^k) ∣ ∣$
$= ∣ ∣ \to u ∣ ∣ \sqrt{59} cos θ$
$∴$ Maximum $[\to u, \to v, \to w] = \sqrt{59}$
$(∴ ∣ ∣ \to u ∣ ∣ = 1, cos θ \leq 1)$