1
Question
Consider the cube in the first octant with sides OP,OQ and OR of length 1, along the x-axis, y-axis and z-axis, respectively, where O(0,0,0) is the origin. Let S(12,12,12) be the centre of the cube and T be the vertex of the cube opposite to the origin O such that S lies on the diagonal OT.If →p=−→SP,→q=−−→SQ,→r=−−→SR and →t=−→ST, then the value of |(→p×→q)×(→r×→t)| is
Open in App
Solution
→p=−→SP(1−12,0−12,0−12)=^i−^j−^k2
→q=−−→SQ(0−12,1−12,0−12)=−^i+^j−^k2
→r=−−→SR(0−12,0−12,1−12)=−^i−^j+^k2
→t=−→ST(1−12,1−12,1−12)=^i+^j+^k2
(→p×→q)=12×12∣∣
∣
∣∣^i^j^k1−1−1−11−1∣∣
∣
∣∣=14(2^i+2^j)
(→p×→q)=^i+^j2
(→r×→t)=12×12∣∣
∣
∣∣^i^j^k−1−11111∣∣
∣
∣∣=14(−2^i+2^j)
(→r×→t)=−^i+^j2
|(→p×→q)×(→r×→t)|=12×12∣∣
∣
∣∣^i^j^k110−110∣∣
∣
∣∣=∣∣2^k4∣∣=√(12)2=12=0.5
→p=−→SP(1−12,0−12,0−12)=^i−^j−^k2→q=−−→SQ(0−12,1−12,0−12)=−^i+^j−^k2
→r=−−→SR(0−12,0−12,1−12)=−^i−^j+^k2
→t=−→ST(1−12,1−12,1−12)=^i+^j+^k2
(→p×→q)=12×12∣∣ ∣ ∣∣^i^j^k1−1−1−11−1∣∣ ∣ ∣∣=14(2^i+2^j)
(→p×→q)=^i+^j2
(→r×→t)=12×12∣∣ ∣ ∣∣^i^j^k−1−11111∣∣ ∣ ∣∣=14(−2^i+2^j)
(→r×→t)=−^i+^j2
|(→p×→q)×(→r×→t)|=12×12∣∣ ∣ ∣∣^i^j^k110−110∣∣ ∣ ∣∣=∣∣2^k4∣∣=√(12)2=12=0.5
Suggest Corrections
0
Similar questions