If →a=α^i+β^j+3^k,→b=−β^i−α^j−^k and →c=^i−2^j−^k
such that →a⋅→b=1 and →b⋅→c=−3, then 13((→a×→b)⋅→c) is equal to
A
06
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
6
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
6.0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
6.00
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
→a=α^i+β^j+3^k→b=−β^i−α^j−^k→c=^i−2^j−^k
Now, →a⋅→b=1⇒−2αβ−3=1⇒αβ=−2⋯(1)
Also, →b⋅→c=−3⇒−β+2α+1=−3⇒2α−β=−4
Using equation (1), we get 2α+2α=−4⇒α2+2α+1=0⇒α=−1⇒β=2
Therefore, 13((→a×→b)⋅→c)=∣∣
∣∣−123−21−11−2−1∣∣
∣∣=−1(−1−2)−2(2+1)+3(4−1)=3−6+9=6