(a) For A(3,-1,2), B(1,-1,2) and C(4,-2,1) vector area of triangle is:
21(AB×AC)AB=(1−3)i+(−1+1)j+(2−2)k=−2iAC=(4−3)i+(−2+1)j+(1−2)k=i−j−k(AB×AC)=∣∣i−21j0−1k0−1∣∣=0i−2j+2k=−2j+2kVector area of the triangle ABC:
21(AB×AC)=−j+kArea of the triangle ABC:
Area=∣21(AB×AC)∣=(−1)2+12=2
(b)
Given that
r=(1+t2)i^+(4t−3)j^+(2t2−6t)k^
Now , Differentiating it with respect to t:
r′=2ti^+4j^+(4t−6)k^
Now, Unit tangent vector,
r′^=∣r′∣r′⇒4t2+16+(4t−6)22ti^+4j^+(4t−6)k^
⇒(20t2−12t+52)2ti^+4j^+(4t−6)k^⇒(5t2−3t+13)ti^+2j^+(2t−3)k^
at t=1
the unit tangent vector = 15i^+2j^+(−1)k^=151i^+152j^−151k^
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