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Conduction Law

Let V⃗=2i⃗+...

Question

Let $\to V = 2 \to i + \to j - \to k$ and $\to W = \to i + 3 \to k .$ If $\to U$ is a unit vector then the 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 C $\sqrt{59}$
$[\to U \to V - \to W] = \to U \cdot (\to V \times - \to W)$
$= \to U . [(2^i +^j -^k) \times (^i + 3^k)]$
$= \to U . (3^i - 7^j -^k)$
The maximum value of $\to a . \to b$ is $a b$ .
So, similarly, the maximum value of the above expression will be $\sqrt{59}$

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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

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Let →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] is

The correct option is C √59[→U→V−→W]=→U⋅(→V×−→W)=→U.[(2^i+^j−^k)×(^i+3^k)]=→U.(3^i−7^j−^k)The maximum value of →a.→b is ab.So, similarly, the maximum value of the above expression will be √59

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

The correct option is C $\sqrt{59}$
$[\to U \to V - \to W] = \to U \cdot (\to V \times - \to W)$
$= \to U . [(2^i +^j -^k) \times (^i + 3^k)]$
$= \to U . (3^i - 7^j -^k)$
The maximum value of $\to a . \to b$ is $a b$ .
So, similarly, the maximum value of the above expression will be $\sqrt{59}$