Question

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Answer 1

Given Question

A lines makes an angle α, β, γ with x - axis, y - axis and z - axis respectively, then cos2α + cos2β + cos2γ is

(a) 2

(b) 1

(c) - 2

(d) - 1

$\large\underline{\sf{Solution-}}$

Given that lines makes an angle α, β, γ with x - axis, y - axis and z - axis respectively.

So, By definition of direction cosines,

$\rm :\longmapsto\:l = cos \alpha$

$\rm :\longmapsto\:m = cos \beta$

$\rm :\longmapsto\:n = cos \gamma$

So,

$\rm :\longmapsto\: {l}^{2} + {m}^{2} + {n}^{2} = 1$

$\rm :\longmapsto\: {cos}^{2} \alpha + {cos}^{2} \beta + {cos}^{2} \gamma = 1$

On multiply by 2 on both sides we get

$\rm :\longmapsto\: 2{cos}^{2} \alpha + 2{cos}^{2} \beta + 2 {cos}^{2} \gamma = 2$

can be further rewritten as

$\rm :\longmapsto\: 2{cos}^{2} \alpha - 1 + 1 + 2{cos}^{2} \beta - 1 + 1 + 2 {cos}^{2} \gamma - 1 + 1 = 2$

$\rm :\longmapsto\: (2{cos}^{2} \alpha - 1)+ (2{cos}^{2} \beta - 1)+ (2 {cos}^{2} \gamma - 1) + 3= 2$

$\rm :\longmapsto\:cos2 \alpha + cos2 \beta + cos2 \gamma + 3= 2$

$\red{ \bigg\{ \sf \: \because \: cos2x = {2cos}^{2}x - 1 \bigg\}}$

$\rm :\longmapsto\:cos2 \alpha + cos2 \beta + cos2 \gamma= 2 - 3$

$\rm :\longmapsto\:cos2 \alpha + cos2 \beta + cos2 \gamma= - 1$

Hence,

$\bf\implies \:\boxed{\tt{ \: cos2 \alpha + cos2 \beta + cos2 \gamma = - 1 \: }}$

So, option (d) is correct.

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MORE TO KNOW

Direction cosines of a line segment is defined as the cosines of the angle which a line makes with the positive direction of respective axis.

The scalar components of unit vector always give direction cosines.

The scalar components of a vector gives direction ratios.

Answer 2

Question:-

A lines makes an angle α, β, γ with x - axis, y - axis and z - axis respectively, then cos2α + cos2β + cos2γ is

(a) 2

(b) 1

(c) - 2

(d) - 1

Given:-

A line makes angles α, β, γ with x - axis, y - axis and z - axis respectively.

To Find:-

cos2α + cos2β + cos2γ is equal to = ?

Solution:-

As we know direction Cosines of a line are l,m,n.

So, we can write l = cos²α, m = cos²β, n = cos²γ.

Now using Property,

$\sf \red \because \red {l {}^{2} + m {}^{2} + n {}^{2} }$

Putting values of l,m,n we get,

cos²α + cos²β + cos²γ = 1.

We know that,

$\sf \blue{cos {}^{2} \theta = \frac{1 +cos {}^{2} \theta }{2} }$

Thus, we write about equations as

$\sf (\frac{1 + cos {}^{2} \alpha }{2} )+ ( \frac{1 + cos {}^{2} \beta }{2} ) +( \frac{1 + cos {}^{2} \gamma }{2} )$

1 + cos²α + cos²β + cos²γ = 2.

3 + cos²α + cos²β + cos²γ = 2.

cos²α + cos²β + cos²γ = 2 – 3.

cos²α + cos²β + cos²γ = 1.

Answer:-

$\sf \red{\: cos {}^{2} α + cos {}^{2} β + cos {}^{2} γ \: is} \: \\ \sf \red{ \: equal \: to \: = 1.}$

Hope you have satisfied.⚘