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Dot Product of Two Vectors

Let V=2î+ĵ−k̂...

Question

Let $\to V = 2^i +^j -^k$ 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 $\sqrt{59}$
$\to V \times - \to W = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 2 & 1 & - 1 1 & 0 & 3 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 3 i - 7^j -^k$
$\Rightarrow [\to U \to V - \to W] = \to U \cdot (3^i - 7^j -^k) = | U | | 3^i - 7^j -^k | cos θ$ $\Rightarrow \sqrt{59} cos θ \leq \sqrt{59}$
The maximum value $\sqrt{59}$ is attained when $θ = 0^{\circ}$

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Dot Product of Two Vectors

Standard XII Mathematics

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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→V×−→W=∣∣ ∣ ∣∣^i^j^k21−1103∣∣ ∣ ∣∣=3i−7^j−^k ⇒[→U →V −→W]=→U⋅(3^i−7^j−^k)=|U||3^i−7^j−^k|cosθ ⇒√59cosθ≤√59 The maximum value √59 is attained when θ=0∘

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