Equation of a Line Passing through Two Given Points
If the acute ...
Question
If the acute angle between the line →r=^i+2^j+λ(4^i−3^k) and xy−plane 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^i−3^k)
Vector along the direction of line =4^i−3^k
Normal unit vector of xy−plane = ^k
Angle between line and plane is sinα=∣∣
∣
∣∣(4^i−3^k)⋅^k√42+32⋅1∣∣
∣
∣∣=35 ⇒cosα=45
Angle between planes x+2y=0 and 2x+y=0 is cosβ=(^i+2^j)⋅(2^i+^j)√5√5=2+25=45 ⇒sinβ=35
Now, cos2α+sin2β=4252+3252=2525=1