Question

The set of values of x for which the angle between the vectors $\overrightarrow{a}=x\hat{i}-3\hat{j}-\hat{k}$ and $\overrightarrow{b}=2x\hat{i}+x\hat{j}-\hat{k}$ is acute and the angle between the vector $\overrightarrow{b}$ and the axis of ordinates is obtuse, is:

• A. $1<x<2$
• B. $x>2$
• C. $x<1$
• D. $x<0$

Answer 1

Answer: (D)

$x<0$

Angle is given by, $\cos \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{ab}$
Thus, $\cos \theta =\frac{(x\hat{i}-3\hat{j}-\hat{k}).(2x\hat{i}+x\hat{j}-\hat{k})}{\sqrt{x^2+3^2+(-1{)}^2}.\sqrt{(2x{)}^2+x^2+(-1{)}^2}}$
which gives, $\cos \theta =\frac{2x^2-3x+1}{\sqrt{x^2+10}.\sqrt{5x^2+1}}$

For angle to be acute, its cosine needs to be positive.

Denominator is always positive. Thus, numerator has to be positive.
Thus, $2x^2-3x+1>0$ or $x$ can be anything but should not lie between $\frac{1}{2}$ and $1$.
Also, angle between b and y-axis is obtuse. Thus, $(2x\hat{i}+x\hat{j}-\hat{k}).\hat{j}$ should be negative (because denominator of $\cos \theta$ is magnitude which is always positive.)
Thus, $x<0$
Taking intersection of the two solution sets we get x<0 as final solution set.