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Vector Algebra

Question
CBSEENMA12035746

If straight a with rightwards arrow on top space and space straight b with rightwards arrow on top are two vectors such that open vertical bar straight a with rightwards arrow on top plus straight b with rightwards arrow on top close vertical bar equals open vertical bar straight a with rightwards arrow on top close vertical bar comma then prove that vector 2 straight a with rightwards arrow on top plus straight b with rightwards arrow on top is perpendicular to vector straight b with rightwards arrow on top.

Solution
open vertical bar straight a with rightwards arrow on top plus straight b with rightwards arrow on top close vertical bar equals open vertical bar straight a with rightwards arrow on top close vertical bar
space rightwards double arrow space space open vertical bar straight a with rightwards arrow on top plus straight b with rightwards arrow on top close vertical bar squared space equals space open vertical bar straight a with rightwards arrow on top close vertical bar squared
space rightwards double arrow space space space open vertical bar straight a with rightwards arrow on top close vertical bar squared plus 2 straight a with rightwards arrow on top. straight b with rightwards arrow on top plus open vertical bar straight b with rightwards arrow on top close vertical bar squared space equals space open vertical bar straight a with rightwards arrow on top close vertical bar squared
space rightwards double arrow 2 straight a with rightwards arrow on top. straight b with rightwards arrow on top plus open vertical bar straight b with rightwards arrow on top close vertical bar squared space equals space 0 space space space space... left parenthesis 1 right parenthesis
Now comma space open parentheses 2 straight a with rightwards arrow on top plus straight b with rightwards arrow on top close parentheses. space left parenthesis straight b with rightwards arrow on top right parenthesis space equals space 2 straight a with rightwards arrow on top. straight b with rightwards arrow on top plus straight b with rightwards arrow on top. straight b with rightwards arrow on top space equals space 2 straight a with rightwards arrow on top. straight b with rightwards arrow on top plus open vertical bar straight b with rightwards arrow on top close vertical bar squared space equals 0       [Using (1)]
We know that if the dot product of two vectors is zero, then either of the two vectors is zero or the vectors are perpendicular to each other.
Thus, 2 straight a with rightwards arrow on top plus straight b with rightwards arrow on top is perpendicular to straight b with rightwards arrow on top.