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! :]