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Question

An angle between the plane, x+y+z=5 and the line of intersection of the planes, 3x+4y+z1=0 and 5x+8y+2z+14=0, is :

A
sin1(317)
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B
cos1(317)
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C
cos1(317)
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D
sin1(317)
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Solution

The correct option is A sin1(317)
Given planes:
3x+4y+z1=0 and 5x+8y+2z+14=0
The normal to the planes will be,
n1=3^i+4^j+^kn2=5^i+8^j+2^k
The direction ratios of the line of intersection will be,
n1×n2=∣ ∣ ∣^i^j^k341582∣ ∣ ∣n1×n2=^j+4^k
Normal vector to plane x+y+z=5
n3=^i+^j+^k
The angle between plane and the line of intersection will be,
sinθ=n3(n1×n2)|n3||n1×n2|sinθ=1+4317θ=sin1317

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