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Scalar Triple Product

V⃗ = 2 î+ ĵ -...

Question

$\to V = 2^i +^j -^k$ 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

-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
$\sqrt{59}$

Given $\to V = 2^i +^j -^k$ and $\to W =^i + 3^k$
$[\to U \to V \to W] = \to U . [(2^i +^j -^k) \times (^i + 3^k)] \to U . (3^i - 7^j -^k) = | \to U | | 3^i - 7^j -^k | cos θ$
which is maximum, if angle between $\to U$ and $3^i - 7^j -^k$ is 0 and maximum value
$= | 3^i - 7^j -^k | = \sqrt{59}$

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$\to V = 2^i +^j -^k$ 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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→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 Given →V=2^i+^j−^k and →W=^i+3^k [→U →V →W]=→U.[(2^i+^j−^k)×(^i+3^k)]→U.(3^i−7^j−^k)=|→U||3^i−7^j−^k| cos θ which is maximum, if angle between →U and 3^i−7^j−^k is 0 and maximum value =|3^i−7^j−^k|=√59

$\to V = 2^i +^j -^k$ 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