Let a=2^−^+3k^, b=3^−5^+k^ and c be a vector such that a×c=c×b and (a+c)⋅(b+c)=168. Then the maximum value of ∣c∣2 is:
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Step-by-Step Solution
Step 1: Simplify the cross product equation
The given cross product equation is a×c=c×b. We know that the cross product is anti-commutative, meaning c×b=−(b×c). Substituting this into the equation and rearranging terms, we get (a+b)×c=0. This implies that c is parallel to (a+b).
Step 2: Calculate a+b and express c
First, we calculate the sum of vectors a and b. Since c is parallel to (a+b), we can express c as a scalar multiple of (a+b), where λ is a scalar.
Step 3: Expand the dot product equation
We are given the dot product equation (a+c)⋅(b+c)=168. Expanding this equation, we get a⋅b+a⋅c+c⋅b+c⋅c=168. We can rewrite the middle terms as (a+b)⋅c and c⋅c is simply ∣c∣2.
Step 4: Calculate dot products and magnitudes
We calculate the dot product a⋅b. Then, we use the fact that c=λ(a+b) to simplify (a+b)⋅c to λ∣a+b∣2. We also calculate the magnitude squared of (a+b) and express ∣c∣2 in terms of λ.
Step 5: Substitute into the expanded dot product equation
Substitute the calculated values into the expanded dot product equation. This results in a quadratic equation in terms of λ. Solving this quadratic equation by factoring gives us two possible values for λ: 1 and −2.
Step 6: Find the maximum value of ∣c∣2
We use the expression for ∣c∣2=77λ2 and substitute the two values of λ we found. For λ=1, ∣c∣2=77. For λ=−2, ∣c∣2=308. The maximum value is the larger of these two.