Question

If the relation between ‘x’, ‘y’ and ‘z’ is defined as $x=y\cos \dfrac{2\pi }{3}=z\cos \dfrac{4\pi }{3}$, then find the value of xy + yz + zx?
(a) –1
(b) 0
(c) 1
(d) 2

Answer 1

Hint: We start solving the problem by converting variables ‘y’ and ‘z’ in terms of ‘x’. Once, we convert every variable in terms of ‘x’, we start calculating the terms xy, yz and zx separately. We add them to each other later to get the required result.

Complete step by step answer:
We have given the relation between ‘x’, ‘y’ and ‘z’ as $x=y\cos \dfrac{2\pi }{3}=z\cos \dfrac{4\pi }{3}$. We need to find the value of xy + yz + zx.
Let us first find all the variables in terms of ‘x’. We first convert ‘y’ in terms of ‘x’.
$\Rightarrow $ \[x=y\cos \dfrac{2\pi }{3}\].
$\Rightarrow $ $x=y\times \left( \dfrac{-1}{2} \right)$.
$\Rightarrow $ $\dfrac{x}{\dfrac{-1}{2}}=y$.
$\Rightarrow $ y= –2x ---(1).
$\Rightarrow $ $x=z\cos \dfrac{4\pi }{3}$.
$\Rightarrow $ $x=z\times \dfrac{-1}{2}$.
$\Rightarrow $ $z=\dfrac{x}{\dfrac{-1}{2}}$.
$\Rightarrow $ z=-2x ---(2).
We first find the values of xy, yz, zx and then add all together.
$\Rightarrow $ xy = x.(–2x).
$\Rightarrow $ $xy=-2{{x}^{2}}$ ---(3).
$\Rightarrow $ yz = (–2x).(–2x).
$\Rightarrow $ $yz=4{{x}^{2}}$ ---(4).
$\Rightarrow $ zx = (–2x).x.
$\Rightarrow $ $zx=-2{{x}^{2}}$ ---(5).
Now we find the value of xy + yz + zx by using the obtained values from (3), (4) and (5).
$\Rightarrow $ $xy+yz+zx=-2{{x}^{2}}+4{{x}^{2}}-2{{x}^{2}}$.
$\Rightarrow $ xy + yz + zx = 0.
∴ The value of xy + yz + zx is 0.

So, the correct answer is “Option B”.

Note: We should not confuse with the values of cosine function while doing the problem. We should not make any calculation mistakes while solving the problem. Similarly, we can expect problems to find the xyz, x + y + z.