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Question
If a=i−j+k,a⋅b=0,a×b=c,, where c=−2i−j+k, then b is equal to
If a=i−j+k,a⋅b=0,a×b=c,, where c=−2i−j+k, then b is equal to
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Solution
The correct option is B (0,1,1)
→a=^i−^j+^k,→c=−2^i−^j+^k
Let →b=x^i+y^j+z^k
→a×→b=→c
⇒∣∣
∣
∣∣^i^j^k1−11xyz∣∣
∣
∣∣=−2^i−^j+^k
⇒^i(−z−y)−^j(z−x)+^k(y+x)=−2^i−^j+^k
Therefore,
−z−y=−2 ...(1)
z−x=1 ...(2)
x+y=1 ...(3)
It is given, →a⋅→b=0
⇒x−y+z=0...(4)
Substracting (1) from (3)
x+z+2y=3⇒x+z=3−2y
put in equation (4)
3−2y−y=0⇒y=1
From equation (3), x=0
From equation (2), z=1
Hence, (x,y,z)≡(0,1,1)
→a=^i−^j+^k,→c=−2^i−^j+^k
Let →b=x^i+y^j+z^k
→a×→b=→c
⇒∣∣ ∣ ∣∣^i^j^k1−11xyz∣∣ ∣ ∣∣=−2^i−^j+^k
⇒^i(−z−y)−^j(z−x)+^k(y+x)=−2^i−^j+^k
Therefore,
−z−y=−2 ...(1)
z−x=1 ...(2)
x+y=1 ...(3)
It is given, →a⋅→b=0
⇒x−y+z=0...(4)
Substracting (1) from (3)
x+z+2y=3⇒x+z=3−2y
put in equation (4)
3−2y−y=0⇒y=1
From equation (3), x=0
From equation (2), z=1
Hence, (x,y,z)≡(0,1,1)
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