Vectors each have magnitude L. When drawn with their tails at the same point, the angle between them is 30 . The magnitude of is: A) L²/2 B) L² D) 2L² E) none of these

Question

Quizplus · Accepted Answer

Let the two vectors be represented by A and B respectively. Then, the angle between them is given by:

cosθ = (A•B) / (|A| |B|)

where |A| = |B| = L (as given in the question), and θ = 30°.

Substituting the values, we get:

cos30° = (A•B) / (L × L)

√3 / 2 = (A•B) / L^2

Therefore, A•B = (√3 / 2) L^2.

Now, consider the third vector C whose tail is at the same point as A and B. Then, the angle between A and C is 150° (since the angle between A and B is 30°). Using the same formula as above, we get:

cos150° = (A•C) / (L × L)

-1/2 = (A•C) / L^2

Therefore, A•C = -(1/2) L^2.

Now, let us consider the vector D = C - A, i.e., the vector going from the tail of A to the head of C. Since A and C have the same magnitude, |D| is simply the length of the diagonal of an equilateral triangle of side L (as A, C, and D form an equilateral triangle). Using the Pythagorean theorem, we get:

|D|^2 = L^2 + L^2 - 2L^2 cos60°

|D|^2 = 3L^2 - L^2

|D|^2 = 2L^2

|D| = L√2

Therefore, the magnitude of vector D is L√2.

Now, we can use the fact that D = C - A, i.e., A + D = C. Since we know A and D (from the above calculations), we can easily find C. We get:

C = A + D

|C| = √[(Ax + Dx)^2 + (Ay + Dy)^2]

|C| = √[(L + L√2 cos30°)^2 + (L√2 sin30°)^2]

|C| = √[L^2 + 2L^2√2 cos30° + 2L^3 cos^230° + 2L^2]

|C| = √[4L^2 + 2L^2√6 + 2L^2]

|C| = L√(6 + 3√2)

Therefore, the magnitude of vector C is L√(6 + 3√2), which is option A.

Vectors Each Have Magnitude L