Lf -2- - cos 2 - cos-sin- cos-sin- sin 2 - - cos 2 - cos-sin- cos-sin- sin 2 - -

The correct option is C 0 Given, #160; θ =π #160;/ 2 +ϕ Consider , [ cos ^ 2 θ #160; #160; amp; cos #160;θsin ...

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Perpendicular Distance of a Point from a Line

lf θ=π/2+ϕ,...

Question

lf $θ = \frac{π}{2} + ϕ, [\begin{matrix} {cos}^{2} θ & cos θ sin θ cos θ sin θ & {sin}^{2} θ \end{matrix}] \times [\begin{matrix} {cos}^{2} ϕ & cos ϕ sin ϕ cos ϕ sin ϕ & {sin}^{2} ϕ \end{matrix}] =$

0

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2 I

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Solution

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Similar questions

A B = 0

A = [\begin{matrix} {cos}^{2} θ & cos θ sin θ cos θ sin θ & {sin}^{2} θ \end{matrix}]

B = [\begin{matrix} {cos}^{2} ϕ & cos ϕ sin ϕ cos ϕ sin ϕ & {sin}^{2} ϕ \end{matrix}]

θ - ϕ

θ - ϕ = π / 2

[\begin{matrix} {cos}^{2} θ & cos θ sin θ cos θ sin θ & {sin}^{2} θ \end{matrix}] [\begin{matrix} {cos}^{2} φ & cos φ sin φ cos φ sin φ & {sin}^{2} φ \end{matrix}] = 0

A = [\begin{matrix} {cos}^{2} θ & cos θ sin θ cos θ sin θ & {sin}^{2} θ \end{matrix}]; B = [\begin{matrix} {cos}^{2} ϕ & cos ϕ sin ϕ cos ϕ sin ϕ & {sin}^{2} ϕ \end{matrix}]

A B

θ

ϕ

\frac{π}{2}

3 {sin}^{2} θ + 2 {sin}^{2} ϕ = 1

3 sin 2 θ = 2 sin 2 ϕ

0 < θ < \frac{π}{2}

0 < ϕ < \frac{π}{2}

θ + 2 ϕ

θ

θ = 0

\frac{π}{2}

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + {sin}^{2} θ & {cos}^{2} θ & 4 sin 4 θ {sin}^{2} θ & 1 + {cos}^{2} θ & 4 sin 4 θ {sin}^{2} θ & {cos}^{2} θ & 1 + 4 sin 4 θ \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0

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