The angle subtended by the vector A-4- i -3- j -12- k with the x-axis is

The correct option is B cos ^ 1 ( 4 / 13 ) The angle substended by the vector → #160; A with the x axis is cosθ #160; = 4 /√( 4 ^ 2 + 3 ^ ...

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Applications of Cross Product

The angle sub...

Question

The angle subtended by the vector $A = 4 \land i + 3 \land j + 12 \land k$ with the x-axis is

{s i n}^{- 1} (\frac{3}{13})

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{s i n}^{- 1} (\frac{4}{13})

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{c o s}^{- 1} (\frac{4}{13})

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{c o s}^{- 1} (\frac{3}{13})

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Find the value of sin 13x + sin x if cos 12x = cos 14 x

F_{1} = - 10^k, F_{2} = u (\frac{4}{13}^i - \frac{12}{13}^j + \frac{3}{13}^k), F_{3} = v (- \frac{4}{13}^i - \frac{12}{13}^j + \frac{3}{13}^k)

F_{4} = w (cos θ^i + sin θ^j)

F_{1} = - 10 k, F_{2} = U [\frac{4}{13} i - \frac{12}{13} j + \frac{3}{13} k],

F_{3} = V [- \frac{4}{13} i - \frac{12}{13} j + \frac{3}{13} k],

F_{4} = W [cos θ i + sin θ j]

θ .

\tan 4 π - \cos \frac{3 π}{2} - \sin \frac{5 π}{6} \cos \frac{2 π}{3} = \frac{1}{4}

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\sin \frac{13 π}{3} \sin \frac{2 π}{3} + \cos \frac{4 π}{3} \sin \frac{13 π}{6} = \frac{1}{2}

\sin \frac{10 π}{3} \cos \frac{13 π}{6} + \cos \frac{8 π}{3} \sin \frac{5 π}{6} = - 1

\tan \frac{5 π}{4} \cot \frac{9 π}{4} + \tan \frac{17 π}{4} \cot \frac{15 π}{4} = 0

{cos}^{- 1} (\frac{3}{5}) - {sin}^{- 1} (\frac{4}{5}) = {cos}^{- 1} x

x =

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