Question

Given A=(1,7) and B=(2,3), the unit vector along 2A+7B is:

A) 2/√77(1,7)+7/√77(2,3)
B) 2/√77(2,3)+7/√77(1,7)
C) 1√53(2,3)+7/√53(1,7)
D) 1/√53(1,7)+7/√53(2,3)

Answer 1

The vector \(2A + 7B\) is obtained by multiplying each component of A by 2 and each component of B by 7, and then adding the corresponding components. Let's calculate it:

\[ 2A + 7B = 2(1,7) + 7(2,3) = (2 + 14, 14 + 21) = (16, 35) \]

Now, to find the unit vector along \(2A + 7B\), divide each component by the magnitude of \(2A + 7B\):

\[ \text{Magnitude of } (16, 35) = \sqrt{16^2 + 35^2} = \sqrt{256 + 1225} = \sqrt{1481} \]

So, the unit vector is:

\[ \left(\frac{16}{\sqrt{1481}}, \frac{35}{\sqrt{1481}}\right) \]

To simplify this expression, you can multiply both components by \(\frac{1}{\sqrt{53}}\) to get the unit vector:

\[ \left(\frac{16}{\sqrt{53 \cdot 28}}, \frac{35}{\sqrt{53 \cdot 28}}\right) \]

This simplifies to:

\[ \left(\frac{2}{\sqrt{77}}, \frac{7}{\sqrt{77}}\right) \]

Comparing this with the given options, it matches with option B:

\[ \frac{2}{\sqrt{77}}(2,3) + \frac{7}{\sqrt{77}}(1,7) \]

So, the correct answer is B.

Answer 2

The unit vector along 2A+7B is found by first calculating the vector 2A+7B, then finding its magnitude and dividing each component of 2A+7B by its magnitude. None of the given options match the resulting simplified unit vector.

Step-by-step explanation:

To find the unit vector along 2A+7B, we first determine the vector by multiplying the given vectors A and B by the scalars 2 and 7, respectively, and then add them together. So, if A = (1,7) and B = (2,3), then 2A+7B = 2(1,7) + 7(2,3) = (2,14) + (14,21) = (16,35).

Next, we calculate the magnitude of 2A+7B using the formula for the magnitude of a vector, which is the square root of the sum of the squares of its components: |2A+7B| = √(16² + 35²) = √(256 + 1225) = √1481 = 38.47 (approx).

Finally, to get the unit vector, we divide each component of 2A+7B by its magnitude: unit vector = (16/38.47, 35/38.47). Simplifying this we get approximately (0.416, 0.909). However, none of the options provided match this result, suggesting either a typo in the options or in the computation.