An angle between the plane, x+y+z=5 and the line of intersection of the planes, 3x+4y+z−1=0 and 5x+8y+2z+14=0, is :
A
sin−1(√317)
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B
cos−1(√317)
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C
cos−1(3√17)
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D
sin−1(3√17)
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Solution
The correct option is Asin−1(√317) Given planes: 3x+4y+z−1=0 and 5x+8y+2z+14=0
The normal to the planes will be, →n1=3^i+4^j+^k→n2=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+4√3√17⇒θ=sin−1√317