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Relative to the origin O, the position vectors of two points A and B are a and b respectively. b is a unit vector and the magnitude of a is twice that of b. The angle between a and b is 60°. Show that [a×[ob + (1-o)a] =√k, where k is a constant to be determined.​

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Answer:

|a × [ob + (1 - o)a]| = √(7 - 8o(a · o) - 8(a · o)^2)

where k = 7 - 8o(a · o) - 8(a · o)^2.

Explanation:

Given the position vectors of two points A and B as a and b respectively, where b is a unit vector, and the magnitude of a is twice that of b, we are asked to show that:

|a × [ob + (1-o)a]| = √k,

where k is a constant to be determined.

We can begin by expanding the vector inside the cross product:

ob + (1 - o)a = ob + a - oa

Since b is a unit vector, we can write:

ob = b - o

Substituting this into the previous equation, we get:

ob + (1 - o)a = b - o + a - oa = b + (1 - o)a - oa

Next, we can use the vector cross product formula:

|a × b| = |a||b|sinθ

where θ is the angle between a and b.

We are given that the angle between a and b is 60°, so we can substitute this value into the formula:

|a × b| = |a||b|sin60° = (2|b|)(1)(√3/2) = √3

Now we can calculate the cross product of a and the vector we just derived:

a × [ob + (1 - o)a] = a × (b + (1 - o)a - oa)

= a × (b + a - oa)

= a × b + a × a - a × oa

Since b is a unit vector, we know that a × b is a vector perpendicular to both a and b, and therefore perpendicular to the plane containing a and b. The vector a × a is 0 since the cross product of a vector with itself is 0. Finally, we can use the vector triple product to simplify a × oa:

a × oa = (a · a)o - (a · o)a = |a|^2 o - (a · o)a

Since |a| is twice |b|, we have:

|a|^2 = 4|b|^2 = 4

Substituting this back in, we get:

a × oa = 4o - (a · o)a

Putting it all together, we have:

a × [ob + (1 - o)a] = a × b + 4o - (a · o)a

Now we can take the magnitude squared of both sides:

|a × [ob + (1 - o)a]|^2 = (a × b + 4o - (a · o)a) · (a × b + 4o - (a · o)a)

Expanding the dot product, we get:

|a × [ob + (1 - o)a]|^2 = |a × b|^2 + 16o^2 + |a|^2(o · o) - 8o(a · o)b + 8(a · o)(a × b) - 2(a · o)^2|a|^2

Substituting the values we derived earlier, we get:

|a × [ob + (1 - o)a]|^2 = 3 + 16o^2 + 4(o · o) - 8o(a · o) + 0 - 2(a · o)^2(4)

= 7 - 8o(a · o) - 8(a · o)^2

Now we need to find the value of k such that the left-hand side equals k:

|a × [ob + (1 - o)a]|^2 = k

Using the vector triple product again, we can simplify the left-hand side as:

|a × [ob + (1 - o)a]|^2 = |a|^2|ob + (1 - o)a|^2 - ((a · [ob + (1 - o)a])^2)

Since we know that the magnitude of a is twice that of b, we have:

|a|^2 = 4|b|^2 = 4

Substituting this back in, we get:

|a × [ob + (1 - o)a]|^2 = 4|ob + (1 - o)a|^2 - ((a · [ob + (1 - o)a])^2)

Now we can substitute the expanded expression for ob + (1 - o)a:

|a × [ob + (1 - o)a]|^2 = 4|b + (1 - o)a|^2 - ((a · [b + (1 - o)a - oa])^2)

= 4|b|^2 + 8|b|(1 - o)(a · b) + 4(1 - o)^2|a|^2 - ((a · b + (1 - o)(a · b) - (a · o)(a · b))^2)

= 4 + 8(1 - o)(a · b) + 4(1 - o)^2(4) - ((a · b + (1 - o)(a · b) - (a · o)(a · b))^2)

= 28 - 8o(a · b) - 8(a · o)^2

Substituting this back into the previous equation, we get:

28 - 8o(a · b) - 8(a · o)^2 = k

Therefore, we have:

|a × [ob + (1 - o)a]| = √(28 - 8o(a · b) - 8(a · o)^2) and

k = 28 - 8o(a · b) - 8(a · o)^2

Hope this helps! Sorry if it's wrong! If you need more help, ask me! :]

User Kriggs
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