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Question
Let, →a=2^i+23^j+22^k and →b=^i+22^j+2^k, the vector component of ^a perpendicular to the direction of ^b is
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Solution
Given →a=2^i+23^j+22^k→b=^i+22^j+2^kvector component of →a parallel to →b is =→a.^b=(2^i+23^j+22^k).→b∣∣→a∣∣ =(2^i+23^j+22^k).^i+22^j+2^k∣∣^i+22^j+2^k∣∣ =2+506+44√1+484+4=552√489Vector component of →a along →b is→a∥=(→a.^b).^b=552√489.^i+22^j+2^k√1+484+4 =552√489.^i+22^j+2^k√489 =552(^i+22^j+2^k)489∴→a=→a∥+→a⊥ since →a⊥ is a perpendicular component of →bWe have →a⊥=→a−→a∥ =2^i+23^j+22^k−552(^i+22^j+2^k)489 =(2−552489)^i+(23−552×22489)^j+(22−552×2489)^k =426489^i−896489^j+9654489^k
Given →a=2^i+23^j+22^k
→b=^i+22^j+2^k
vector component of →a parallel to →b is
=→a.^b=(2^i+23^j+22^k).→b∣∣→a∣∣
=(2^i+23^j+22^k).^i+22^j+2^k∣∣^i+22^j+2^k∣∣
=2+506+44√1+484+4=552√489
Vector component of →a along →b is
→a∥=(→a.^b).^b=552√489.^i+22^j+2^k√1+484+4
=552√489.^i+22^j+2^k√489
=552(^i+22^j+2^k)489
∴→a=→a∥+→a⊥ since →a⊥ is a perpendicular component of →b
We have →a⊥=→a−→a∥
=2^i+23^j+22^k−552(^i+22^j+2^k)489
=(2−552489)^i+(23−552×22489)^j+(22−552×2489)^k
=426489^i−896489^j+9654489^k
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