Question

The vectors $2{\mathbf{\hat i}} - m{\mathbf{\hat j}} + 3m{\mathbf{\hat k}}$ and $\left( {1 + m} \right){\mathbf{\hat i}} - 2m{\mathbf{\hat j}} + {\mathbf{\hat k}}$ include an acute angle for
(a) all real $m$
(b) $m < \- 2$ or $m > - \dfrac{1}{2}$
(c) $m = - \dfrac{1}{2}$
(d) $m \in \left[ { - 2, - \dfrac{1}{2}} \right]$

Answer 1

Here, we need to find which of the given options are correct. Let ${\mathbf{v}} = 2{\mathbf{\hat i}} - m{\mathbf{\hat j}} + 3m{\mathbf{\hat k}}$ and ${\mathbf{w}} = \left( {1 + m} \right){\mathbf{\hat i}} - 2m{\mathbf{\hat j}} + {\mathbf{\hat k}}$. We will find the magnitude of the two vectors, and their dot product. Then, using these, we will find the angle between the two vectors. The cosine of any acute angle is greater than 0. Using this, we will form an inequation, and solve it to find the correct values of $m$.
Formula used: The dot product of two vectors ${\mathbf{v}} = {a_1}{\mathbf{\hat i}} + {b_1}{\mathbf{\hat j}} + {c_1}{\mathbf{\hat k}}$ and ${\mathbf{w}} = {a_2}{\mathbf{\hat i}} + {b_2}{\mathbf{\hat j}} + {c_2}{\mathbf{\hat k}}$ is given by ${\mathbf{v}} \cdot {\mathbf{w}} = {a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}$.
The magnitude of a vector ${\mathbf{v}} = {a_1}{\mathbf{\hat i}} + {b_1}{\mathbf{\hat j}} + {c_1}{\mathbf{\hat k}}$ is given by $\left\| {\mathbf{v}} \right\| = \sqrt {{{\left( {{a_1}} \right)}^2} + {{\left( {{b_1}} \right)}^2} + {{\left( {{c_1}} \right)}^2}}$.
The angle $\theta$ between two vectors non-zero vectors $u$ and $v$ is given by $\cos \theta = \dfrac{{u \cdot v}}{{\left\| u \right\|{\text{ }}\left\| v \right\|}}$.
If $\left( {x + a} \right)\left( {x + b} \right) > 0$, where $a > b$, then $x + a < 0$ or $x + b > 0$.

Complete step-by-step answer:
We will use the formula for angle between two vectors to find the angle between the two vectors.
Let ${\mathbf{v}} = 2{\mathbf{\hat i}} - m{\mathbf{\hat j}} + 3m{\mathbf{\hat k}}$ and ${\mathbf{w}} = \left( {1 + m} \right){\mathbf{\hat i}} - 2m{\mathbf{\hat j}} + {\mathbf{\hat k}}$.
First, we will find the dot product of the two vectors.
The dot product of two vectors ${\mathbf{v}} = {a_1}{\mathbf{\hat i}} + {b_1}{\mathbf{\hat j}} + {c_1}{\mathbf{\hat k}}$ and ${\mathbf{w}} = {a_2}{\mathbf{\hat i}} + {b_2}{\mathbf{\hat j}} + {c_2}{\mathbf{\hat k}}$ is given by ${\mathbf{v}} \cdot {\mathbf{w}} = {a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}$.
Substitute ${a_1} = 2$, ${b_1} = - m$, ${c_1} = 3m$, ${a_2} = 1 + m$, ${b_2} = - 2m$, and ${c_2} = 1$ in the formula to find the dot product of ${\mathbf{v}} = 2{\mathbf{\hat i}} - m{\mathbf{\hat j}} + 3m{\mathbf{\hat k}}$ and ${\mathbf{w}} = \left( {1 + m} \right){\mathbf{\hat i}} - 2m{\mathbf{\hat j}} + {\mathbf{\hat k}}$, we get
$\Rightarrow {\mathbf{v}} \cdot {\mathbf{w}} = 2\left( {1 + m} \right) + \left( { - m} \right)\left( { - 2m} \right) + \left( {3m} \right)\left( 1 \right)$
Multiplying the terms in the expression, we get
$\Rightarrow {\mathbf{v}} \cdot {\mathbf{w}} = 2 + 2m + 2{m^2} + 3m$
Adding the like terms, we get
$\Rightarrow {\mathbf{v}} \cdot {\mathbf{w}} = 2{m^2} + 5m + 2$
Now, we will find the magnitudes of the two vectors.
The magnitude of a vector ${\mathbf{v}} = {a_1}{\mathbf{\hat i}} + {b_1}{\mathbf{\hat j}} + {c_1}{\mathbf{\hat k}}$ is given by $\left\| {\mathbf{v}} \right\| = \sqrt {{{\left( {{a_1}} \right)}^2} + {{\left( {{b_1}} \right)}^2} + {{\left( {{c_1}} \right)}^2}}$.
Substituting ${a_1} = 2$, ${b_1} = - m$, ${c_1} = 3m$ in the formula, we get the magnitude of ${\mathbf{v}} = 2{\mathbf{\hat i}} - m{\mathbf{\hat j}} + 3m{\mathbf{\hat k}}$ as
$\Rightarrow \left\| {\mathbf{v}} \right\| = \sqrt {{2^2} + {{\left( { - m} \right)}^2} + {{\left( {3m} \right)}^2}}$
Applying the exponents on the bases, we get
$\Rightarrow \left\| {\mathbf{v}} \right\| = \sqrt {4 + {m^2} + 9{m^2}}$
Adding the like terms, we get
$\Rightarrow \left\| {\mathbf{v}} \right\| = \sqrt {4 + 10{m^2}}$
Substituting ${a_2} = 1 + m$, ${b_2} = - 2m$, and ${c_2} = 1$ in the formula for magnitude, we get the magnitude of ${\mathbf{w}} = \left( {1 + m} \right){\mathbf{\hat i}} - 2m{\mathbf{\hat j}} + {\mathbf{\hat k}}$ as
$\Rightarrow \left\| {\mathbf{w}} \right\| = \sqrt {{{\left( {1 + m} \right)}^2} + {{\left( { - 2m} \right)}^2} + {{\left( 1 \right)}^2}}$
Applying the exponents on the bases, we get
$\Rightarrow \left\| {\mathbf{w}} \right\| = \sqrt {1 + {m^2} + 2m + 4{m^2} + 1}$
Adding the like terms, we get
$\Rightarrow \left\| {\mathbf{w}} \right\| = \sqrt {2m + 5{m^2} + 2}$
Now, we will find the angle between the two vectors.
The angle $\theta$ between two vectors non-zero vectors $u$ and $v$ is given by $\cos \theta = \dfrac{{u \cdot v}}{{\left\| u \right\|{\text{ }}\left\| v \right\|}}$.
Therefore, the angle between the two vectors ${\mathbf{v}} = 2{\mathbf{\hat i}} - m{\mathbf{\hat j}} + 3m{\mathbf{\hat k}}$ and ${\mathbf{w}} = \left( {1 + m} \right){\mathbf{\hat i}} - 2m{\mathbf{\hat j}} + {\mathbf{\hat k}}$ is given by
$\Rightarrow \cos \theta = \dfrac{{{\mathbf{v}} \cdot {\mathbf{w}}}}{{\left\| {\mathbf{v}} \right\|{\text{ }}\left\| {\mathbf{w}} \right\|}}$
Substituting ${\mathbf{v}} \cdot {\mathbf{w}} = 2{m^2} + 5m + 2$, $\left\| {\mathbf{v}} \right\| = \sqrt {4 + 10{m^2}}$, and $\left\| {\mathbf{w}} \right\| = \sqrt {2m + 5{m^2} + 2}$ in the expression, we get
$\Rightarrow \cos \theta = \dfrac{{2{m^2} + 5m + 2}}{{\sqrt {4 + 10{m^2}} \sqrt {2{m^2} + 5m + 2} }}$
It is given that the angle between the two vectors is an acute angle.
We know that the cosine of an acute angle is always greater than 0.
Therefore, we can write
$\Rightarrow \cos \theta > 0$
Thus, we get
$\Rightarrow \dfrac{{2{m^2} + 5m + 2}}{{\sqrt {4 + 10{m^2}} \sqrt {2{m^2} + 5m + 2} }} > 0$
Multiplying both sides of the equation by $\sqrt {4 + 10{m^2}} \sqrt {2{m^2} + 5m + 2}$, we get
$\Rightarrow 2{m^2} + 5m + 2 > 0$
The left side of the inequation is a quadratic polynomial.
We will factorise the quadratic polynomial.
\Rightarrow 2{m^2} + 4m + m + 2 > 0 \\\ \Rightarrow 2m\left( {m + 2} \right) + 1\left( {m + 2} \right) > 0 \\\ \Rightarrow \left( {2m + 1} \right)\left( {m + 2} \right) > 0 \\\
Factoring out 2, we get
\Rightarrow 2\left( {m + \dfrac{1}{2}} \right)\left( {m + 2} \right) > 0 \\\ \Rightarrow \left( {m + \dfrac{1}{2}} \right)\left( {m + 2} \right) > 0 \\\
Now, we know that if $\left( {x + a} \right)\left( {x + b} \right) > 0$, where $a > b$, then $x + a < 0$ or $x + b > 0$.
Therefore, since $\left( {m + \dfrac{1}{2}} \right)\left( {m + 2} \right) > 0$, we get
$\Rightarrow m + 2 < 0$ or $m + \dfrac{1}{2} > 0$
Simplifying the two expressions, we get
$\therefore m < \- 2$ or $m > - \dfrac{1}{2}$
Therefore, we get $m > - \dfrac{1}{2}$ or $m < \- 2$.
Thus, the correct option is option (b).

Note: The cosine of an acute angle is always greater than 0 because the cosine of an acute angle lies in the range $\left( {0,1} \right)$.
We multiplied both sides of the equation $\dfrac{{2{m^2} + 5m + 2}}{{\sqrt {4 + 10{m^2}} \sqrt {2{m^2} + 5m + 2} }} > 0$ by $\sqrt {4 + 10{m^2}} \sqrt {2{m^2} + 5m + 2}$ without changing the sign. This is because since $\sqrt {4 + 10{m^2}} \sqrt {2{m^2} + 5m + 2}$ is positive, we can multiply both sides by $\sqrt {4 + 10{m^2}} \sqrt {2{m^2} + 5m + 2}$ without changing the sign of inequality.