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Question

If the acute angle between the line r=^i+2^j+λ(4^i3^k) and xyplane is α and the acute angle between the planes x+2y=0 and 2x+y=0 is β, then (cos2α+sin2β) equals

A
01
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B
1.00
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C
1
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D
1.0
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Solution

Line r=^i+2^j+λ(4^i3^k)
Vector along the direction of line =4^i3^k
Normal unit vector of xyplane = ^k
Angle between line and plane is
sinα=∣ ∣ ∣(4^i3^k)^k42+321∣ ∣ ∣=35
cosα=45

Angle between planes x+2y=0 and 2x+y=0 is
cosβ=(^i+2^j)(2^i+^j)55=2+25=45
sinβ=35
Now, cos2α+sin2β=4252+3252=2525=1

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