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A uni-modular tangent vector on the curve x=t2+2,y=4t5,z=2t26t at t=2 is
(A) 13(2i+2j+k)(B13(ijk)(C13(2i+j+k)(D23(i+j+k)

Answer
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Hint: We know that if f(x,y,z,t)=0 represents a curve then the tangent vector of f(x,y,z,t)=0 is represented by dxdti+dydtj+dzdtk. From the question, we were given the equation of curve is x=t2+2,y=4t5,z=2t26t. Now we have to calculate dxdt,dydt,dzdt. From the values of dxdt,dydt,dzdt we can get the vector dxdti+dydtj+dzdtk. Now we have to substitute t=2. This will give us the tangent vector of f(x,y,z,t)=0 at t=2. We know that the unit vector of ai+bj+ck is (aa2+b2+c2)i+(ba2+b2+c2)j+(ca2+b2+c2)k. Now by using this concept, we can find the uni-modular tangent vector of curve x=t2+2,y=4t5,z=2t26t at t=2.

Complete step-by-step solution:
Before solving the question, we should know that if f(x,y,z,t)=0 represents a curve then the tangent vector of f(x,y,z,t)=0 is represented by dxdti+dydtj+dzdtk.
From the question, we were given the equation of curve is x=t2+2,y=4t5,z=2t26t.
Let us consider
x=t2+2......(1)y=4t5.......(2)z=2t26t......(3)
Now let us differentiate equation (1) on both sides, then we get
dxdt=ddt(t2+2)dxdt=2t.....(4)
Now let us differentiate equation (2) on both sides, then we get
dydt=ddt(4t5)dydt=4.....(5)
Now let us differentiate equation (3) on both sides, then we get
dzdt=ddt(2t26t)dzdt=4t6.....(6)
From equation (4), equation (5) and equation (6), we can get the tangent vector of the curve x=t2+2,y=4t5,z=2t26t.
Now we have to find the equation of the tangent vector.
dxdti+dydtj+dzdtk=(2t)i+4j+(4t6)k.....(7)
Now we have to find the equation of the tangent at t=2.
So, now we have to substitute t=2 in equation (7), then we get
dxdti+dydtj+dzdtk=(2(2))i+4j+(4(2)6)kdxdti+dydtj+dzdtk=4i+4j+2k.....(8)
Now we have to find the unimodular vector of equation (8).
We know that the unit vector of ai+bj+ck is (aa2+b2+c2)i+(ba2+b2+c2)j+(ca2+b2+c2)k.
Now let us compare 4i+4j+2k with ai+bj+ck. Then we get
a=4......(9)b=4......(10)c=2.......(11)
Now from equation (9), equation (10) and equation (11) we have to calculate the value of a2+b2+c2.
a2+b2+c2=42+42+22a2+b2+c2=36a2+b2+c2=6.....(12)
Now we have to find the unit tangent vector.
From equation (9), equation (10), equation (11) and equation (12), we get
(aa2+b2+c2)i+(ba2+b2+c2)j+(ca2+b2+c2)k=(46)i+(46)j+(26)k(aa2+b2+c2)i+(ba2+b2+c2)j+(ca2+b2+c2)k=4i+4j+2k6(aa2+b2+c2)i+(ba2+b2+c2)j+(ca2+b2+c2)k=2i+2j+k3......(13)
From equation we can say that uni-modular tangent vector on the curve x=t2+2,y=4t5,z=2t26t at t=2 is 2i+2j+k3.
Hence, option A is correct.

Note: Students may do the solution up to equation (8) and they may conclude that 4i+4j+2k is the tangent vector. They may think that this is the correct answer but this will give us the wrong answer. The reason is that in the question it was given that to find the uni-modular vector. We know that 4i+4j+2k will not represents a uni -modular vector. So, students should follow the question properly and then proceed through the solution.

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