# Vectors Set-2

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Example – 3

If $$\vec a,\,\vec b,\,\vec c$$ are three non-coplanar unit vectors equally inclined to one another at an angle $$\theta$$ such that

$\vec a \times \vec b + \vec b \times \vec c = p\vec a + q\vec b + r\vec c,$

find $$\vec p,\,\vec q,\,\vec r$$ in terms of $$\theta$$ Also, show that

${p^2} + \frac{{{q^2}}}{{\cos {{\theta }}}} + {r^2} = 2$

Solution: Note that

$\vec a \cdot \vec a = \vec b \cdot \vec b = \vec c \cdot \vec c = 1;\,\,\,\,\,\vec a \cdot \vec b = \vec b \cdot \vec c = \vec c \cdot \vec a = \cos \theta$

Also

\begin{align} &\;\qquad{\left[ {\vec a\,\,\,\,\,\vec b\,\,\,\,\,\vec c} \right]^2} = \left| {\begin{array}{*{20}{c}} {\vec a \cdot \vec a}&{\vec a \cdot \vec b}&{\vec a \cdot \vec c} \\ {\vec b \cdot \vec a}&{\vec b \cdot \vec b}&{\vec b \cdot \vec c} \\ {\vec c \cdot \vec a}&{\vec c \cdot \vec b}&{\vec c \cdot \vec c} \end{array}} \right|\qquad\qquad\left( \begin{gathered}{\text{We'll need this value }} \hfill \\{\text{later; that's why we }} \hfill \\{\text{are evaluating it here}} \hfill \\ \end{gathered} \right) \hfill \\\\&\qquad\qquad\qquad\;\,\;{\text{ = }}\left| {\begin{array}{*{20}{c}}{\text{1}}&{{\text{cos}}\theta }&{{\text{cos}}\theta } \\ {{\text{cos}}\theta }&{\text{1}}&{{\text{cos}}\theta } \\ {{\text{cos}}\theta }&{{\text{cos}}\theta }&1 \end{array}} \right| \hfill \\\\&\qquad\qquad\qquad\;\,\;= 1 - 3{\cos ^2}\theta + 2{\cos ^3}\theta \hfill \\\\&\qquad\qquad\qquad\;\,\;= {\left( {1 - \cos \theta } \right)^2}\left( {1 + 2\cos \theta } \right) \hfill \\\\& \Rightarrow \quad \left[ {\vec a\,\,\,\,\,\vec b\,\,\,\,\,\vec c} \right] = \left( {1 - \cos \theta } \right)\sqrt {1 + 2\cos \theta } \hfill \\ \end{align}

The given relation is

$\vec a \times \vec b + \vec b \times \vec c = p\vec a + q\vec b + r\vec c$

Taking the dot product on both sides successively with $$\vec a,\,\vec b,\,\vec c,$$ we get the following system of equations.

\begin{align}&p + q\cos \theta + r\cos \theta = \left[ {\vec a\,\,\,\,\,\vec b\,\,\,\,\,\vec c} \right] \hfill \\& p\cos \theta + q + r\cos \theta = 0 \hfill \\&p\cos \theta + q\cos \theta + r = \left[ {\vec a\,\,\,\,\,\vec b\,\,\,\,\,\vec c} \right] \hfill \\ \end{align}

Using Cramer’s rule (or otherwise, by elimination), p, q, r can be evaluated. For example,

\begin{align} &p\; = \frac{{\left| {\begin{array}{*{20}{c}}{\left[ {\vec a\,\,\,\,\,\vec b\,\,\,\,\,\vec c} \right]}&{\cos \theta }&{\cos \theta } \\ 0&1&{\cos \theta } \\ {\left[ {\vec a\,\,\,\,\,\vec b\,\,\,\,\,\vec c} \right]}&{\cos \theta }&1 \end{array}} \right|}}{{\left| {\begin{array}{*{20}{c}} 1&{\cos \theta }&{\cos \theta } \\ {\cos \theta }&1&{\cos \theta } \\ {\cos \theta }&{\cos \theta }&1 \end{array}} \right|}} \hfill \\\\&\;\, = \frac{{\left[ {\vec a\,\,\,\,\,\vec b\,\,\,\,\,\vec c} \right]\left( {1 - \cos \theta } \right)}}{{{{\left[ {\vec a\,\,\,\,\,\vec b\,\,\,\,\,\vec c} \right]}^2}}} \hfill \\\\&\;\, = \frac{1}{{\sqrt {1 + 2\cos \theta } }} \hfill \\\end{align}

Similarly,

$q = \frac{{ - 2\cos \theta }}{{\sqrt {1 + 2\cos \theta } }},\,\,\,\,\,r = \frac{1}{{\sqrt {1 + 2\cos \theta } }}$

These values confirm that

${p^2} + \frac{{{q^2}}}{{\cos \theta }} + {r^2} = 2$

Example – 4

Let V be the volume of the parallelopiped formed by the vectors

\begin{align}& \vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k \hfill \\\\& \vec b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k \hfill \\\\&\vec c = {c_1}\hat i + {c_2}\hat j + {c_3}\hat k \hfill \\ \end{align}

If $${a_r},\,{b_r},\,{c_r},\,r = 1,\,2,\,3$$ are non-negative numbers such that $$\sum\limits_{r = 1}^3 {\left( {{a_r} + {b_r} + {c_r}} \right) = 3L,}$$ show that $$V \leqslant {L^3}.$$

Solution: We have,

\begin{align}& V = \left| {\left[ {\vec a\,\,\,\,\,\vec b\,\,\,\,\,\vec c} \right]} \right| \hfill \\\\&\;\;\, = \left| {\,\left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}&{{a_3}} \\ {{b_1}}&{{b_2}}&{{b_3}} \\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}} \right|\,} \right| \hfill \\\\&\;\;\,= \left| {{a_1}{b_2}{c_3} + {a_2}{b_3}{c_1} + {a_3}{b_1}{c_2} - \left( {{a_1}{b_3}{c_2} + {a_2}{b_1}{c_3} + {a_3}{b_2}{c_1}} \right)} \right| \hfill \\\\&\;\;\, \leqslant {a_1}{b_2}{c_3} + {a_2}{b_3}{c_1} + {a_3}{b_1}{c_2}\qquad\qquad\qquad...\left( 1 \right) \hfill \\ \end{align}

The last step is justified since all the $${a'_i}s,\,\,{b'_i}s\,\,and\,\,{c'_i}s$$ are positive.

Now, by the AM-GM in equality, we have

\begin{align}&27\left( {{a_1}{b_2}{c_3}} \right) \leqslant {\left( {{a_1} + {b_2} + {c_2}} \right)^3} \hfill \\\\&27\left( {{a_2}{b_3}{c_1}} \right) \leqslant {\left( {{a_2} + {b_3} + {c_1}} \right)^3} \hfill \\\\& 27\left( {{a_3}{b_1}{c_2}} \right) \leqslant {\left( {{a_3} + {b_1} + {c_2}} \right)^3} \hfill \\ \end{align}

Adding these three inequalities, we have

$27\left( {{a_1}{b_2}{c_3} + {a_2}{b_3}{c_1} + {a_3}{b_1}{c_2}} \right) \leqslant {\left( {{a_1} + {b_2} + {c_3}} \right)^3} + {\left( {{a_2} + {b_3} + {c_1}} \right)^3} + {\left( {{a_3} + {b_1} + {c_2}}\right)^3}\qquad\qquad\qquad...\left( 2 \right)$

The right hand side is of the form

${x^3} + {y^3} + {z^3}$

where $$x,y,z \geqslant$$ so that

${x^3} + {y^3} + {z^3} \leqslant {\left( {x + y + z} \right)^3}\qquad\qquad\qquad...\left( 3 \right)$

Using (3) in (2), we have

$27\left( {{a_1}{b_2}{c_3} + {a_2}{b_3}{c_1} + {a_3}{b_1}{c_2}} \right) \leqslant {\left( {{a_1} + {a_2} + {a_3} + {b_1} + {b_2} + {b_3} + {c_1} + {c_2} + {c_3}} \right)^3} = 27{L^3}\qquad\qquad\qquad...\left( 4 \right)$

From (1) and (4), it follows that

$V \leqslant {L^3}$

Example – 5

Let  $$\vec u,\,\vec v,\,\vec w$$ be three non-coplanar unit vectors and $$\alpha ,\,\beta ,\,\gamma$$ be the angles between $$\vec u\,\,and\,\,{\text{ }}\vec v,\,\vec v\,\,and\,\,\vec w\,\,and\,\,\vec w\,\,and\,\,\vec u$$ respectively. Let $$\vec x,\,\vec y,\,\vec z$$ be the unit vectors along the bisectors of the angles $$\alpha ,\,\beta ,\,\gamma$$ respectively. Prove that

$\left[ {\vec x \times \vec y\,\,\,\vec y \times \vec z\,\,\,\vec z \times \vec x} \right] = \frac{1}{{16}}{\left[ {\vec u\,\,\,\vec v\,\,\,\vec w} \right]^2}{\sec ^2}\frac{\alpha }{2}{\sec ^2}\frac{\beta }{2}{\sec ^2}\frac{\gamma }{2}$

Solution: Since $$\vec u,\,\vec v,\,\vec w$$ are unit vectors, we have

$\left| {\vec u + \vec v} \right| = 2\cos \frac{\alpha }{2},\,\,\,\,\,\left| {\vec v + \vec w} \right| = 2\cos \frac{\beta }{2},\,\,\,\,\,\left| {\vec w + \vec u} \right| = 2\cos \frac{\gamma }{2}$

Now, since  $$\vec x,\,\vec y,\,\vec z$$ are unit vectors along the bisectors of $$\alpha ,\,\beta ,\,\gamma ,$$ we have

$\vec{x}=\frac{\vec{u}+\vec{v}}{2\cos {}^{\alpha }\!\!\diagup\!\!{}_{2}\;},\,\,\,\,\,\vec{y}=\frac{\vec{v}+\vec{w}}{2\cos {}^{\beta }\!\!\diagup\!\!{}_{2}\;},\,\,\,\,\,\vec{z}=\frac{\vec{w}+\vec{u}}{2\cos {}^{\gamma }\!\!\diagup\!\!{}_{2}\;}$

\begin{align}& LHS=\left[ \vec{x}\times \vec{y}\,\,\,\,\,\vec{y}\times \vec{z}\,\,\,\,\,\vec{z}\times \vec{x} \right]={{\left[ \vec{x}\,\,\,\,\,\vec{y}\,\,\,\,\,\vec{z} \right]}^{2}} \\\\& \qquad\;\;\;\qquad\qquad\qquad\qquad\qquad=\frac{{{\left[ \vec{u}+\vec{v}\,\,\,\,\,\vec{v}+\vec{w}\,\,\,\,\,\vec{w}+\vec{u} \right]}^{2}}}{64{{\cos }^{2}}\frac{\alpha }{2}\,\,{{\cos }^{2}}\frac{\beta }{2}\,\,{{\cos }^{2}}\frac{\gamma }{2}} \\\\ & \qquad\;\;\;\qquad\qquad\qquad\qquad\qquad=\frac{{{\left( 2\left[ \vec{u}\,\,\,\,\,\vec{v}\,\,\,\,\,\vec{w} \right] \right)}^{2}}}{64{{\cos }^{2}}\frac{\alpha }{2}\,\,{{\cos }^{2}}\frac{\beta }{2}\,\,{{\cos }^{2}}\frac{\gamma }{2}} \\\\ &\qquad\;\;\;\qquad\qquad\qquad\qquad\qquad =RHS \\ \end{align}

This proves the stated assertion.

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