Question

Consider three vectors $\left(\overset{ \rightarrow }{V}\right)_{1}=\left(sin \theta \right)\hat{i}+\left(cos ⁡ \theta \right)\hat{j}+\left(a - 3\right)\hat{k}$ , $\left(\overset{ \rightarrow }{V}\right)_{2}=\left(sin \theta + cos ⁡ \theta \right)\hat{i}+\left(cos ⁡ \theta - sin ⁡ \theta \right)\hat{j}+\left(b - 4\right)\hat{k}$ and $\left(\overset{ \rightarrow }{V}\right)_{3}=\left(cos \theta \right)\hat{i}+\left(sin ⁡ \theta \right)\hat{j}+\left(c - 5\right)\hat{k}$ . If the resultant of $\overset{ \rightarrow }{V}_{1},\overset{ \rightarrow }{V}_{2}$ and $\overset{ \rightarrow }{V}_{3}$ 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 $\overset{ \rightarrow }{V}_{1},\overset{ \rightarrow }{V}_{2},\overset{ \rightarrow }{V}_{3}$ is $\overset{ \rightarrow }{V}_{1}+\overset{ \rightarrow }{V}_{2}+\overset{ \rightarrow }{V}_{3}=\lambda \hat{i}$
$\Rightarrow \left(2 sin \theta + 2 cos ⁡ \theta \right)\hat{i}+\left(2 cos ⁡ \theta - 2 sin ⁡ \theta \right)\hat{j}$ $+\left(a + b + c - 12\right)\hat{k}=\lambda \hat{i}$
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 $=55\times 2=110$