Solution:

We are given that a vector Q with magnitude 8 is added to a vector P which lies along the x-axis. Let the magnitude of P be denoted by p.

Step 1: Finding the Resultant of the Two Vectors

As the resultant of the vectors lies along the y-axis, it can be represented as R = k * j where j is the unit vector along the y-axis and k is a scalar. Let the vector Q be represented as Q = q * u where u is the unit vector in the direction of Q and q is the magnitude of Q.

Now, as vector addition is commutative, the vector sum of P and Q can be written as:

P + Q = P + q * u

The resultant of the two vectors is along the y-axis which implies that the vector sum of P and Q has no x-component. Therefore, we have:

P + q * u = k * j

As u lies in the x-y plane, it has no y-component. Therefore, the above equation can be written as:

P + q * u = k * (0 * i + 1 * j + 0 * k)

Comparing the x and y-components, we get:

P + q * u = 0 * i + k * j + 0 * k

P = -q * u + k * j ...(1)

Step 2: Finding the Magnitude of the Resultant

The magnitude of the resultant vector is given to be twice that of P. Therefore, we have:

|R| = 2|P|

As R = k * j, we have:

|R| = |k * j| = |k| * |j| = |k|

Therefore, we can write:

|k| = 2|P|

Substituting the value of P from equation (1), we get:

|k| = 2|-q * u + k * j|

Squaring both sides, we get:

k^2 = 4q^2 + 4k^2

Solving for k, we get:

k = ±2sqrt(q^2 + k^2)

As k is a scalar, it cannot be negative. Therefore, we have:

k = 2sqrt(q^2 + k^2)

Substituting the value of k in equation (1), we get:

P = -q * u + 2sqrt(q^2 + k^2) * j

Step 3: Finding the Magnitude of P

We are given that the magnitude of Q is 8. Therefore, we have:

|Q| = q = 8

Substituting the value of q in the equation for P, we get:

P = -8u + 2sqrt(64 + k^2) * j

Taking the magnitude of both sides, we get:

|P| = |-8u + 2sqrt(64 + k^2) * j|

Squaring both sides, we get:

|P|^2 = 64 + 4k^2

Substituting the value of k in the above equation, we get:

|P|^2 = 64 + 4(2sqrt(64 + k^2))^2

Simplifying, we get:

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8/√5