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Vectors · Question 113 of 125

Question 113

MediumOriginal question

(λi^+2j^−k^)×(i^−j^+k^)=i^−3j^+μk^(\lambda\hat{i} + 2\hat{j} - \hat{k}) \times (\hat{i} - \hat{j} + \hat{k}) = \hat{i} - 3\hat{j} + \mu\hat{k}. What is λ+μ\lambda + \mu?

Options for question 113

Show the answer and explanation
CorrectNot quiteAnswer B

Correct answer: B

Explanation

  1. i^(2⋅1−(−1)(−1))−j^(λ⋅1−(−1)⋅1)+k^(λ(−1)−2⋅1)=i^−(λ+1)j^−(λ+2)k^\hat{i}(2 \cdot 1 - (-1)(-1)) - \hat{j}(\lambda \cdot 1 - (-1) \cdot 1) + \hat{k}(\lambda(-1) - 2 \cdot 1) = \hat{i} - (\lambda + 1)\hat{j} - (\lambda + 2)\hat{k}.

  2. Matching: λ+1=3\lambda + 1 = 3, so λ=2\lambda = 2, and μ=−(λ+2)=−4\mu = -(\lambda + 2) = -4: λ+μ=−2\lambda + \mu = -2.

Worth noting

6 works out the k^\hat{k} part the wrong way round, 2⋅1−λ(−1)2 \cdot 1 - \lambda(-1), which gives μ=4\mu = 4.

Concept
Computing a vector product
Typical time
90 s
Your time
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The rule this testsThe vector product
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Ten questions from Vectors, at your own pace.

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Ten questions from Vectors, at your own pace.