wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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(112,012,012)=^i^j^k2
q=SQ(012,112,012)=^i+^j^k2
r=SR(012,012,112)=^i^j+^k2
t=ST(112,112,112)=^i+^j+^k2
(p×q)=12×12∣ ∣ ∣^i^j^k111111∣ ∣ ∣=14(2^i+2^j)
(p×q)=^i+^j2
(r×t)=12×12∣ ∣ ∣^i^j^k111111∣ ∣ ∣=14(2^i+2^j)
(r×t)=^i+^j2
|(p×q)×(r×t)|=12×12∣ ∣ ∣^i^j^k110110∣ ∣ ∣=2^k4=(12)2=12=0.5

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon