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 $[¯ U ¯ V ¯ W]$ is

\sqrt{6}

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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 B $\sqrt{59}$
we have $[¯ U ¯ V ¯ W] = ¯ U \cdot (¯ V \times ¯ W)$
$= ∣ ∣ ¯ U ∣ ∣ ∣ ∣ ¯ V \times ¯ W ∣ ∣ cos ϕ$

$ϕ$ is the angle between $¯ U$ and $¯ V \times ¯ W$

$\Rightarrow ∣ ∣ ¯ V \times ¯ W ∣ ∣ cos ϕ \leq ∣ ∣ ¯ V ¯ W ∣ ∣$

$∴$ max. value of $[¯ U ¯ V ¯ W] = ∣ ∣ ¯ V \times ¯ W ∣ ∣ = \sqrt{59}$

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and

\to w =^i + 3^k

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

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Q. Let

¯ u =^i +^j, ¯ v =^i -^j, ¯ w =^i + 2^j + 3^k .

^n

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equals

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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] is

The correct option is B √59we have [¯U¯V¯W]=¯U⋅(¯V×¯W)=∣∣¯U∣∣∣∣¯V×¯W∣∣cosϕϕ is the angle between ¯U and ¯V×¯W⇒∣∣¯V×¯W∣∣cosϕ≤∣∣¯V¯W∣∣∴ max. value of [¯U¯V¯W]=∣∣¯V×¯W∣∣=√59

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 $[¯ U ¯ V ¯ W]$ is