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SSSSS [86.1K]
3 years ago
15

The points A, B and C have position vectors OA = 6,2, OB = 3,4 and OC = 12, -2 respectively.

Mathematics
1 answer:
kari74 [83]3 years ago
5 0

Answer:(a) BA = (-3,2)   (b) BC = (-9,6)

Step-by-step explanation: (a) Vector BA = position vector of B - position vector of A = vector OB - vector OA = (3-6) i + (4-2) j = -3i+2j [i, j are unit vectors along x and y-axis respectively]. Also vector BA=(-3,2).                     (b) Vector BC = position vector of B - position vector of C = vector OB - vector OC = (3-12) i + (4-(-2)) j = -9i+6j [i, j are unit vectors along x and y-axis respectively]. Also vector BA=(-9,6).

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Applying the differentiation rule, it can be obtained that:

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\frac{d}{dt}(e^{mt})=me^{mt}.

Given that r(t)=(e^t,e^{2t}).

So, differentiating r(t)=(e^t,e^{2t}) with respect to t, we get: r'(t)=\left(\frac{d}{dt}(e^t),\frac{d}{dt}(e^{2t})\right).

So, using the above formula \frac{d}{dt}(e^{mt})=me^{mt}, we get: r'(t)=(e^t,2e^{2t}).

Now, substituting t=t_0=0 in r(t)=(e^t,e^{2t}) and r'(t)=(e^t,2e^{2t}), we obtain:

r(t_0=0)=(e^0,e^{2\times 0})=(1,1) and r'(t_0=0)=(e^0,2e^{2\times 0})=(1,2).

Therefore, applying the differentiation rule, we get:

r'(t)=(e^t,2e^{2t}), r(t_0)=(1,1) and r'(t_0)=(1,2).

To know about the differentiation rule, refer:

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