If - - - be the direction angles of a vector and cos-14-15- cos-1-3 then cos y-

The correct option is C cos^2 α+cos^2 β+cos^2 γ=1 ⇒ cos γ=√(1 (14/15 )^2 (1/3 )^2)=√(8/9 (196/225 ))=±2/15. ...

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Standard XII

Mathematics

Applications of Cross Product

If α, β, γ be...

Question

If $α, β, γ$ be the direction angles of a vector and $c o s α = \frac{14}{15}, c o s β = \frac{1}{3} t h e n c o s γ =$

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Solution

The correct option is A

$c o s^{2} α + c o s^{2} β + c o s^{2} γ = 1 \Rightarrow c o s γ = \sqrt{1 - {(\frac{14}{15})}^{2} - {(\frac{1}{3})}^{2}} = \sqrt{\frac{8}{9} - (\frac{196}{225})} = \pm \frac{2}{15} .$

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If α,β,γ be the direction angles of a vector and cosα=1415,cosβ=13 then cos γ=

The correct option is A cos2 α+cos2 β+cos2 γ=1⇒cos γ=√1−(1415)2−(13)2=√89−(196225)=±215.

If $α, β, γ$ be the direction angles of a vector and $c o s α = \frac{14}{15}, c o s β = \frac{1}{3} t h e n c o s γ =$

The correct option is A

$c o s^{2} α + c o s^{2} β + c o s^{2} γ = 1 \Rightarrow c o s γ = \sqrt{1 - {(\frac{14}{15})}^{2} - {(\frac{1}{3})}^{2}} = \sqrt{\frac{8}{9} - (\frac{196}{225})} = \pm \frac{2}{15} .$