→a1=^i+2^j+^k;→b1=^i−^j+^k
→a2=2^i−^j−^k;→b2=2^i+^j+2^k
(→a2−→a1)=^i−3^j−2^k
→b1×→b2=∣∣
∣
∣∣^i^j^k1−11212∣∣
∣
∣∣=−3^i+3^k⇒∣∣→b1×→b2∣∣=√18
(→b1×→b2).((→a2−→a1))=(−3^i+3^k).(^i−3^j−2^k)=−3−6=−9
d=∣∣
∣
∣∣(→b1×→b2).((→a2−→a1))(→b1×→b2)∣∣
∣
∣∣=∣∣∣−9√18∣∣∣=3√2