Question

The set of values of x for which, the angle between $\vec a = x\hat i - 3\hat j - \hat k$ and $\vec b = 2x\hat i + x\hat j - \hat k$ is an acute angle and the angle between the axis of ordinate and vector $\vec b$ is obtuse.

Answer 1

Hint: For any acute angle $\theta $, $\cos \theta $ is positive and for any obtuse angle $\theta $, $\cos \theta $ is negative.

Complete step by step solution:
An acute angle is an angle whose measure is less than 90 degrees. For any quadrant system the acute angle lies in the first quadrant. An obtuse angle is an angle whose measure is greater than 90 degrees. For any quadrant system the obtuse angle lies in the second quadrant.
Now when we talk about the sign convention for the sine and cosines, we see that they have different signs in different quadrants. Now for acute angles as the angle lies in the first quadrant, the cosine of the angle is positive.
So let us take the dot product of the two vectors $\vec a$ and $\vec b$.
Now, as we know that the angle is acute thee cosine is positive;
So, $\vec a \cdot \vec b = \left| {\vec a} \right| \cdot \left| {\vec b} \right|\cos \theta $.
Therefore, $\cos \theta = \dfrac{{\vec a \cdot \vec b}}{{\left| {\vec a} \right|\left| {\vec b} \right|}}$
And we know that for acute angle $\cos \theta $ is positive;
So, $\dfrac{{\vec a \cdot \vec b}}{{\left| {\vec a} \right|\left| {\vec b} \right|}} > 0$
So, $\vec a \cdot \vec b = 2{x^2} - 3x + 1$
And $\left| {\vec a} \right| = \sqrt {{x^2} + 10} $, $\left| {\vec b} \right| = \sqrt {5{x^2} + 1} $ and $\left| {\vec a} \right| \cdot \left| {\vec b} \right| = \sqrt {({x^2} + 10)(5{x^2} + 1)} $
Substituting these values we get;
$\dfrac{{\vec a \cdot \vec b}}{{\left| {\vec a} \right| \cdot \left| {\vec b} \right|}} = \dfrac{{2{x^2} - 3x + 1}}{{\sqrt {({x^2} + 10)(5{x^2} + 1)} }} > 0$
Solving the above inequality for x we get;
As the denominator always exists because the quantity inside the root is always greater zero, we will consider the numerator to find the range of x,
So, $2{x^2} - 3x + 1 > 0$
$(x - 1)(2x - 1) > 0$
Thus, by wavy curve method;
Thus, $x \in \left( { - \infty ,\dfrac{1}{2}} \right) \cup \left( {1,\infty } \right)$
Also, the second case the axis of orinate and vector b is an obtuse angle;
So, let the unit vector along the axis be $\hat j$.
Hence, by taking dot product, we get;
$\vec b \cdot \hat i = \left| {\vec b} \right|\left| {\hat j} \right|\cos \theta '$
Hence, $\dfrac{{\vec b \cdot \hat j}}{{\left| {\vec b} \right| \cdot \left| {\hat j} \right|}} < 0$ since $\cos \theta ' < 0$ for $\theta '$ is obtuse.
$\vec b \cdot \hat j = x$.
So, $\dfrac{x}{{\sqrt {5{x^2} + 1} }} < 0$
Now, the numerator is always true for all x so we will consider the denominator;
Now, case: 1- for $x > 0$;
$\sqrt {5{x^2} + 1} < 0$
$5{x^2} + 1 < 0$
Thus, ${x^2} < \dfrac{{ - 1}}{5}$
But this is not possible so case 1 is not true.

Case: 2- for $x < 0$;
$\sqrt {5{x^2} + 1} > 0$
Hence here the case is true.
Now let us combine all we have;
We have; $x \in \left( { - \infty ,\dfrac{1}{2}} \right) \cup \left( {1,\infty } \right)$
And $x < 0$
So we take the intersection of the two sets and we get;
$x \in \left( {0,\dfrac{1}{2}} \right) \cup \left( {1,\infty } \right)$
Therefore for the above range of x the condition given in the question is true.

Note: 1. Dot product is proportional to the cosine of the angle made by the vectors.
2. For such problems we do not need the vector formats, as we are working here with scalar calculus.
3. So, we need to make use of the dot product concept.