Let $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ be the three vectors. If $|\mathbf{a}|=1$, $|\mathbf{b}|=17$…

Let $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ be the three vectors. If $|\mathbf{a}|=1$, $|\mathbf{b}|=17$ and $|\mathbf{c}|=8$ and the angle between $\mathbf{a}$ and $\mathbf{b}$ is $\theta$ and $\mathbf{a} \times(\mathbf{a} \times \mathbf{b})-\mathbf{c}=0$, then $\cos \theta+\operatorname{cosec} \theta$ is equal to
  1. $\frac{409}{136}$
  2. $\frac{309}{136}$
  3. $\frac{419}{126}$
  4. $\frac{409}{126}$

Solution

Given, $|\mathbf{a}|=1,|\mathbf{b}|=17$ and $|\mathbf{c}|=8$ Angle between $\mathbf{a}$ and $\mathbf{b}$ is $\theta$. $\mathbf{a} \times(\mathbf{a} \times \mathbf{b})-\mathbf{c}=0$ $\begin{array}{lc}\Rightarrow & \mathbf{a} \times(\mathbf{a} \times \mathbf{b})=\mathbf{c} \\ \Rightarrow & |\mathbf{a} \times(\mathbf{a} \times \mathbf{b})|=|\mathbf{c}|\end{array}$ $\Rightarrow|\mathbf{a} \| \mathbf{a}||\mathbf{b}| \sin \theta \cdot \sin 90^{\circ}=|\mathbf{c}|$ $[\because \mathbf{a}$ is perpendicular to $\mathbf{a} \times \mathbf{b}]$ $\Rightarrow \quad|\mathbf{a}|^2|\mathbf{b}| \sin \theta=|\mathbf{c}|$ $\Rightarrow \quad \sin \theta=\frac{|\mathbf{c}|}{|\mathbf{a}|^2|\mathbf{b}|}=\frac{8}{1 \times 17}=\frac{8}{17}$ $\therefore \quad \cos \theta=\sqrt{1-\left(\frac{8}{17}\right)^2}=\frac{15}{17}$ $\cos \theta+\operatorname{cosec} \theta=\frac{15}{17}+\frac{17}{8}=\frac{120+289}{136}=\frac{409}{136}$

Asked in: AP EAMCET 2021 (24 Aug Shift 2)

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