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

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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 $\sqrt{59}$
$V \times W = ∣ ∣ ∣ ∣ \begin{matrix} i & j & k 2 & 1 & - 1 1 & 0 & 3 \end{matrix} ∣ ∣ ∣ ∣ = 3 i - 7 j - k$

$\Rightarrow [U V W] = U \cdot (3 i - 7 j - k) = | U | | 3 i - 7 j - k | cos θ$
$\Rightarrow \sqrt{59} cos θ \leq \sqrt{59}$

The maximum value $\sqrt{59}$ is attained when $θ = 0$

i.e., taking $U = \frac{V \times W}{| V \times W |}$

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

The correct option is C √59V×W=∣∣ ∣∣ijk21−1103∣∣ ∣∣=3i−7j−k⇒[UVW]=U⋅(3i−7j−k)=|U||3i−7j−k|cosθ⇒√59cosθ≤√59The maximum value √59 is attained when θ=0 i.e., taking U=V×W|V×W|

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