Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
1630 Upvotes
Verified Answer
The correct answer is:
$\frac{409}{136}$
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 $\mathbf{a} \times \mathbf{b}]$
$\Rightarrow \quad|\mathbf{a}|^2|\mathbf{b}| \sin \theta=|\mathbf{c}|$
$\Rightarrow \quad \sin \theta=\frac{|\mathbf{c}|}{|\mathbf{a}|^2|\mathbf{b}|}=\frac{8}{1 \times 17}=\frac{8}{17}$
$\therefore \quad \cos \theta=\sqrt{1-\left(\frac{8}{17}\right)^2}=\frac{15}{17}$
$\cos \theta+\operatorname{cosec} \theta=\frac{15}{17}+\frac{17}{8}=\frac{120+289}{136}=\frac{409}{136}$
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 $\mathbf{a} \times \mathbf{b}]$
$\Rightarrow \quad|\mathbf{a}|^2|\mathbf{b}| \sin \theta=|\mathbf{c}|$
$\Rightarrow \quad \sin \theta=\frac{|\mathbf{c}|}{|\mathbf{a}|^2|\mathbf{b}|}=\frac{8}{1 \times 17}=\frac{8}{17}$
$\therefore \quad \cos \theta=\sqrt{1-\left(\frac{8}{17}\right)^2}=\frac{15}{17}$
$\cos \theta+\operatorname{cosec} \theta=\frac{15}{17}+\frac{17}{8}=\frac{120+289}{136}=\frac{409}{136}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.