Given, →a=^i+^j+^k and →b=^i+2^j+3^k
⇒→a+→b=2^i+3^j+4^k and →a−→b=−^j−2^k
A vector which is perpendicular to both (→a+→b) and (→a−→b) is given by,
(→a+→b)×(→a−→b)=∣∣
∣
∣∣^i^j^k2340−1−2∣∣
∣
∣∣
=−2^i+4^j−2^k (=→c, say)
Now, |→c|=√4+16+4=√24=2√6
Therefore, the required unit vector is,
→c|→c|=−1√6^i+2√6^j−1√6^k