Question

Let $a$, $b$ and $c$ be the distinct non-negative numbers. If the vectors $a\stackrel{\rightharpoonup }{i}+a\stackrel{\rightharpoonup }{j}+c\stackrel{\rightharpoonup }{k}$,$\stackrel{\rightharpoonup }{i}+\stackrel{\rightharpoonup }{k}$and $c\stackrel{\rightharpoonup }{i}+c\stackrel{\rightharpoonup }{j}+b\stackrel{\rightharpoonup }{k}$ lie in a plane, then $c$ is?

• A. the harmonic mean of $a\text{and}b$
• B. equal to zero
• C. the arithmetic mean of $a\text{and}b$
• D. the geometric mean of $a\text{and}b$

Answer 1

The correct option is D

the geometric mean of $a\text{and}b$

Explanation for correct option:

Given the vectors $a\stackrel{\rightharpoonup }{i}+a\stackrel{\rightharpoonup }{j}+c\stackrel{\rightharpoonup }{k}$,$\stackrel{\rightharpoonup }{i}+\stackrel{\rightharpoonup }{k}$and $c\stackrel{\rightharpoonup }{i}+c\stackrel{\rightharpoonup }{j}+b\stackrel{\rightharpoonup }{k}$ lie in a plane. Three vectors are coplanar.

Therefore $\left[\left(a\stackrel{\rightharpoonup }{i}+a\stackrel{\rightharpoonup }{j}+c\stackrel{\rightharpoonup }{k}\right)\left(\stackrel{\rightharpoonup }{i}+\stackrel{\rightharpoonup }{k}\right)\left(c\stackrel{\rightharpoonup }{i}+c\stackrel{\rightharpoonup }{j}+b\stackrel{\rightharpoonup }{k}\right)\right]\ne 0$

$$\begin{gathered} \Rightarrow \left|\begin{array}{ccc}a& a& c\\ 1& 0& 1\\ c& c& b\end{array}\right|=0 \\ a(0-c)-a(b-c)+c(c-0)=0 \\ -ac+ac-ab+c^2=0 \\ {c}^2=ab \end{gathered}$$

The geometric mean of $a\mathrm{and}b$is $\sqrt{ab}$.

Therefore, $c$ is the geometric mean of $a\mathrm{and}b$

Hence, option (D) is the correct answer