⎡⎢
⎢⎣→OA×→OB=^i−^j+^k→OB×→OC=^i→OC×→OA=−^i+^j⎤⎥
⎥⎦ where 'O' is the the origin.
Then ∣∣→CA×→CB∣∣ has the value equal to
A
1/2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1/√2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
√2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C√2 The concept here is "the sum of all normal vectors in a tetrahedron is zero" →OA×→OB+→OB×→OC+→OC×→OA+→CB×→CA=0 →CB×→CA=−((^i−^j+^k)+(^i)+(−^i+^j)) |→CB×→CA|= Magnitude is √2