The angle which a vector -2k- makes with x-axis isA- 60-B- 120-C- 150-D- tan -1-1-2

The correct option is B 120^oLet A = î ĵ + √(2)k̂ and vector representing y axis be B = ĵ. We know that cosβ = A·B|A||B| = 1√((1)^2+ ( 1)^2+(√(2))^2) = ...

The angle whi...

Question

The angle which a vector $^i -^j + \sqrt{2}^k$ makes with x-axis is

60^{o}

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120^{o}

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150^{o}

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{tan}^{- 1} (- \frac{1}{2})

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Solution

The correct option is B $120^{o}$
Let $\to A =^i -^j + \sqrt{2}^k$ and vector representing y - axis be $\to B =^j$ .
We know that $cos β = \frac{\to A \cdot \to B}{| \to A | | \to B |} = \frac{- 1}{\sqrt{(1)^{2} + (- 1)^{2} + (\sqrt{2})^{2}}} = - \frac{1}{2}$
$∴ β = 120^{o}$

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The correct option is B 120oLet →A=^i−^j+√2^k and vector representing y - axis be →B=^j. We know that cosβ=→A⋅→B|→A||→B|=−1√(1)2+(−1)2+(√2)2=−12 ∴ β=120o

The angle which a vector $^i -^j + \sqrt{2}^k$ makes with x-axis is

The correct option is B $120^{o}$
Let $\to A =^i -^j + \sqrt{2}^k$ and vector representing y - axis be $\to B =^j$ .
We know that $cos β = \frac{\to A \cdot \to B}{| \to A | | \to B |} = \frac{- 1}{\sqrt{(1)^{2} + (- 1)^{2} + (\sqrt{2})^{2}}} = - \frac{1}{2}$
$∴ β = 120^{o}$