First slide

Algebra of vectors

Question

If vector $\vec{b} = (\tan α, - 1, 2 \sqrt{\sin α / 2}) and \vec{c} = (\tan α, \tan α, - \frac{3}{\sqrt{\sin α / 2}})$ are orthogonal and vector $\vec{a} = (1, 3, \sin 2 α)$ makes an obtuse angle with the z-axis, then the value of $α$ is

Moderate

A

$α = (4 n + 1) π + \tan^{- 1} 2$
B

$α = (4 n + 1) π - \tan^{- 1} 2$
C

$α = (4 n + 2) π + \tan^{- 1} 2$
D

$α = (4 n + 2) π - \tan^{- 1} 2$

Solution

Since $\vec{a} = (1, 3, \sin 2 α)$ makes an obtuse angle with the z-axis, its z-component is negative. Thus,
$\begin{array}{ll} - 1 \leq \sin 2 α < 0 - - - - (i) \\ But \vec{b} \cdot \vec{c} = 0 \\ \tan^{2} α - \tan α - 6 = 0 \\ ∴ (\tan α - 3) (\tan α + 2) = 0 \\ \Rightarrow \tan α = 3, - 2 \end{array}$
Now, tan $α$ = 3. Therefore,
$\sin 2 α = \frac{2 \tan α}{1 + \tan^{2} α} = \frac{6}{1 + 9} = \frac{3}{5}$ (not possible as sin2 $α$ < 0)
Now, if tan $α$ = -2
$\begin{array}{l} \Rightarrow \sin 2 α = \frac{2 \tan α}{1 + \tan^{2} α} = \frac{- 4}{1 + 4} = \frac{- 4}{5} \\ \Rightarrow \tan 2 α > 0 \end{array}$
Hence, 2 $α$ is in the third quadrant. Also meaningful. If $0 < \sin α / 2 < 1$ then
$\begin{aligned} α = (4 n + 1) π - \tan^{- 1} 2 \\ and α = (4 n + 2) π - \tan^{- 1} 2 \end{aligned}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App