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Vectors · Question 91 of 125

Question 91

HardOriginal question

a⃗+b⃗+c⃗=0⃗\vec{a} + \vec{b} + \vec{c} = \vec{0}, with ∣a⃗∣=2\lvert \vec{a} \rvert = 2, ∣b⃗∣=3\lvert \vec{b} \rvert = 3 and ∣c⃗∣=4\lvert \vec{c} \rvert = 4. What is a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}?

Options for question 91

Show the answer and explanation
CorrectNot quiteAnswer B

Correct answer: B

Explanation

  1. Square a⃗+b⃗+c⃗=0⃗\vec{a} + \vec{b} + \vec{c} = \vec{0}: ∣a⃗∣2+∣b⃗∣2+∣c⃗∣2+2(a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗)=0\lvert \vec{a} \rvert^{2} + \lvert \vec{b} \rvert^{2} + \lvert \vec{c} \rvert^{2} + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0.

  2. 4+9+16=294 + 9 + 16 = 29, so the sum is −292-\frac{29}{2}.

Worth noting

292\frac{29}{2} loses the sign when the squares move to the other side.

Concept
Three vectors that add to zero
Typical time
120 s
Your time
—
The rule this testsThe scalar product
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Ten questions from Vectors, at your own pace.

Practise this chapter →

Ten questions from Vectors, at your own pace.