Let a be a non-zero vector parallel to the line of intersection of the two planes described by i^+j^,i^+k^ and i^−j^,j^−k^. If θ is the angle between the vector a and the vector b=2i^−2j^+k^ and a⋅b=6, then the ordered pair (θ,∣a×b∣) is equal to :
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Official solution
We have, a is non-zero vector parallel to the line of intersection of the two planes described by i^+j^,i^+k^ and i^−j^,j^−k^.
Let n1 and n2 are the normal vector to the plane i^+j^,i^+k^ and i^−j^,j^−k^, respectively.
n1=∣i^j^k^110101∣=i^−j^−k^n2=∣i^j^k^1−1010−1∣=i^+j^+k^a=λ∣n1×n2∣
[ ∵a is parallel to line of intersection of both planes]
=λ∣i^j^k^1−1−1111∣=λ(−2j^+2k^)a⋅b=6[Given]λ(0+4+2)=6λ=1∴a=−2j^+2k^cosθ=∣a∣∣b∣a⋅b=4+44+4+16=21∴θ=4πa×b=∣i^j^k^0−222−21∣=2i^+4j^+4k^∣a×b∣=4+16+16=6(θ,∣a×b∣)=(4π,6)