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Vector Algebra question

2019 · 9 Apr · Shift 2 · Q31
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Vector Algebra question

2019 · 9 Apr · Shift 2 · Q31

JEE MainMathematicsVector AlgebraMCQ+4 / −1
If a unit vector a→\overrightarrow aa makes angles π\piπ/3 with i^\widehat ii, π\piπ/ 4 with j^\widehat jj​ and θ∈\theta \inθ∈(0, π\piπ) with k^\widehat kk, then a value of θ\thetaθ is :-
  1. A
    5π6{{5\pi } \over {6}}65π​
  2. B
    5π12{{5\pi } \over {12}}125π​
  3. C
    2π3{{2\pi } \over {3}}32π​
  4. D
    π4{{\pi } \over {4}}4π​
View written solutionFree

Correct answer: C

  1. Use direction cosines of a unit vector

For a unit vector making angles α,β,γ\alpha,\beta,\gammaα,β,γ with the coordinate axes, its direction cosines satisfy

cos⁡2α+cos⁡2β+cos⁡2γ=1.\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1.cos2α+cos2β+cos2γ=1.

Here,

α=π3,β=π4,γ=θ.\alpha = \frac{\pi}{3}, \qquad \beta = \frac{\pi}{4}, \qquad \gamma = \theta.α=3π​,β=4π​,γ=θ.

So,

cos⁡2π3+cos⁡2π4+cos⁡2θ=1.\cos^2\frac{\pi}{3} + \cos^2\frac{\pi}{4} + \cos^2\theta = 1.cos23π​+cos24π​+cos2θ=1.
  1. Substitute the known values
cos⁡π3=12  ⟹  cos⁡2π3=14\cos\frac{\pi}{3} = \frac12 \implies \cos^2\frac{\pi}{3} = \frac14cos3π​=21​⟹cos23π​=41​

and

cos⁡π4=12  ⟹  cos⁡2π4=12.\cos\frac{\pi}{4} = \frac{1}{\sqrt2} \implies \cos^2\frac{\pi}{4} = \frac12.cos4π​=2​1​⟹cos24π​=21​.

Thus,

14+12+cos⁡2θ=1.\frac14 + \frac12 + \cos^2\theta = 1.41​+21​+cos2θ=1. 34+cos⁡2θ=1\frac34 + \cos^2\theta = 143​+cos2θ=1 cos⁡2θ=14.\cos^2\theta = \frac14.cos2θ=41​.
  1. Find θ\thetaθ in (0,π)(0,\pi)(0,π)

From

cos⁡2θ=14,\cos^2\theta = \frac14,cos2θ=41​,

we get

cos⁡θ=±12.\cos\theta = \pm \frac12.cosθ=±21​.

In the interval (0,π)(0,\pi)(0,π), this gives

θ=π3or2π3.\theta = \frac{\pi}{3} \quad \text{or} \quad \frac{2\pi}{3}.θ=3π​or32π​.
  1. Match with the options

Among the given options, only

2π3\frac{2\pi}{3}32π​

is present.

So the correct option is C.

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