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Question
Let →a=^i+^j+^k,→b=−^i+^j+^k,→c=^i−^j+^k. The line of intersection of the plane determined by →a and →b and the plane determined by →c and →d is parallel to
Let →a=^i+^j+^k,→b=−^i+^j+^k,→c=^i−^j+^k. The line of intersection of the plane determined by →a and →b and the plane determined by →c and →d is parallel to
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Solution
The correct option is C x−axis
The normal to the plane determined by →a,→b is along →a×→b
=∣∣
∣
∣∣^i^j^k111−111∣∣
∣
∣∣
=^i(1−1)−^j(1+1)+^k(1+1)=−2^j+2^k=−2(^j−^j)
The normal to the plane determined by →c,→d is along →c×→d
=∣∣
∣
∣∣^i^j^k1−111−11∣∣
∣
∣∣
=^i(1−1)−^j(−1−1)+^k(1+1)=2^j+2^k=2(^j+^k)
The line of intersection of the two planes is along (→a×→b).(→c×→d)
=∣∣
∣
∣∣^i^j^k0−22022∣∣
∣
∣∣
=^i(−4−4)−^j(0−0)+^k(0−0)
=−8^i Which is along x− axis
The normal to the plane determined by →a,→b is along →a×→b
=∣∣ ∣ ∣∣^i^j^k111−111∣∣ ∣ ∣∣
=^i(1−1)−^j(1+1)+^k(1+1)=−2^j+2^k=−2(^j−^j)
The normal to the plane determined by →c,→d is along →c×→d
=∣∣ ∣ ∣∣^i^j^k1−111−11∣∣ ∣ ∣∣
=^i(1−1)−^j(−1−1)+^k(1+1)=2^j+2^k=2(^j+^k)
The line of intersection of the two planes is along (→a×→b).(→c×→d)
=∣∣ ∣ ∣∣^i^j^k0−22022∣∣ ∣ ∣∣
=^i(−4−4)−^j(0−0)+^k(0−0)
=−8^i Which is along x− axis
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