Answer:
Option (b) is correct.
A⃗ – 2B⃗ + 3C⃗ = ( 2i + j ) – 2( 3i - k ) + 3( 6i - k )
⇒2i + j − 6j + 2k + 18i − 6k
⇒20i – 5j – 4k
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Verified answer
Answer:
Option (b) is correct.
Given:
To find:
Solution:
A⃗ – 2B⃗ + 3C⃗ = ( 2i + j ) – 2( 3i - k ) + 3( 6i - k )
⇒2i + j − 6j + 2k + 18i − 6k
⇒20i – 5j – 4k