Question

Consider three vectors ${\left(\overrightarrow{V}\right)}_1=\left(sin\theta \right)\hat{i}+\left(cos\theta \right)\hat{j}+\left(a-3\right)\hat{k}$ , ${\left(\overrightarrow{V}\right)}_2=\left(sin\theta +cos\theta \right)\hat{i}+\left(cos\theta -sin\theta \right)\hat{j}+\left(b-4\right)\hat{k}$ and ${\left(\overrightarrow{V}\right)}_3=\left(cos\theta \right)\hat{i}+\left(sin\theta \right)\hat{j}+\left(c-5\right)\hat{k}$ . If the resultant of ${\overrightarrow{V}}_1,{\overrightarrow{V}}_2$ and ${\overrightarrow{V}}_3$ is equal to $\lambda \hat{i}$ , where $\theta \in \left[-\pi ,\pi \right]$ and $a,b,c\in N$ , then the number of quadruplets $\left(a,b,c,\theta \right)$ are

• A. $55$
• B. $110$
• C. $91$
• D. $182$

Answer 1

Answer: (B)

Resultant of ${\overrightarrow{V}}_1,{\overrightarrow{V}}_2,{\overrightarrow{V}}_3$ is ${\overrightarrow{V}}_1+{\overrightarrow{V}}_2+{\overrightarrow{V}}_3=\lambda \hat{i}$
$\Rightarrow \left(2sin\theta +2cos\theta \right)\hat{i}+\left(2cos\theta -2sin\theta \right)\hat{j}$ $+\left(a+b+c-12\right)\hat{k}=\lambda \hat{i}$
On comparison, we get,
$2sin\theta +2cos\theta =\lambda$
$2cos\theta -2sin\theta =0\Rightarrow tan\theta =1\Rightarrow$ number of values of $\theta$ in $\left[-\pi ,\pi \right]$ is $2$
Also, $a+b+c=12\Rightarrow$ number of solutions ${=}^{11}{C}_2=55$
Hence, number of quadruplets $=55\times 2=110$