Examples On Scalar Triple Product Of Vectors Set-2

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Example – 30

If the vector

\[\begin{align}&  \vec \alpha  = a\hat i + \hat j + \hat k & ,a \ne 1 \hfill \\\\& \vec \beta  = \hat i + b\hat j + \hat k & ,b \ne 1 \hfill \\\\& \vec \gamma  = \hat i + \hat j + c\hat k & ,c \ne 1 \hfill \\ \end{align} \]

are coplanar, prove that

\[\frac{1}{{1 - a}} + \frac{1}{{1 - b}} + \frac{1}{{1 - c}} = 1\]

Solution: The coplanarity of the three vectors means that their STP must be zero:

\[ \Rightarrow  \quad \left| {\;\begin{align}&a&1&&1 \\ &1&b&&1 \\ &1&1&&c \end{align}\;} \right|\; = 0\]

\[\qquad\qquad\; \Rightarrow  \quad a(bc - 1) + (1 - c) + (1 - b) = 0\]

\[ \qquad\qquad\quad\;\;\;\Rightarrow  \quad a + b + c = abc + 2\qquad\qquad\quad...(1)\]

We now have

\[\begin{align}&\frac{1}{{1 - a}} + \frac{1}{{1 - b}} + \frac{1}{{1 - c}} \hfill \\\\& = \frac{{(1 - b)(1 - c) + (1 - a)(1 - c) + (1 - a)(1 - b)}}{{(1 - a)(1 - b)(1 - c)}} \hfill \\\\& = \frac{{3 - 2(a + b + c) + (ab + bc + ac)}}{{1 - (a + b + c) + (ab + bc + ac) - abc}} \hfill \\\\&  = \frac{{3 - 2(a + b + c) + (ab + bc + ac)}}{{1 - (a + b + c) + (ab + bc + ac) + 2 - (a + b + c)}}\qquad \left\{ {Using{\text{ }}\left( 1 \right){\text{ }}for{\text{ }}the{\text{ }}value{\text{ }}of\;abc} \right\} \hfill \\\\&  = 1 \hfill \\ \end{align} \]

Example – 31

Let \(\vec a,\;\,\vec b,\;\,\vec c\) be three non-zero vectors such that \(\vec c\) is a unit vector perpendicular to both \(\vec a\;{\text{and}}\;\vec b\). If the angle between \(\vec a\;{\text{and}}\;\vec b\) is \(\begin{align}\frac{\pi }{6}\end{align}\), prove that

\[{[\vec a\;\;\vec b\;\;\vec c]^2} = \frac{1}{4}{\left| {\vec a} \right|^2}\;\;{\left| {\vec b} \right|^2}\]

Solution: Since \(\vec c\) is perpendicular to both \(\vec a\;{\text{and}}\;\vec b,\,\,\vec c\) must be parallel to \(\vec a\;\; \times \;\vec b,\) i.e, the angle between \(\vec c\) and  \((\vec a \times \vec b)\) must be 0. Thus,

\[\begin{align}&[\vec a\;\,\vec b\;\,\vec c] = [\vec c\;\,\vec a\;\,\vec b] \hfill \\\\& \qquad\quad= \vec c \cdot (\vec a \times \vec b) \hfill \\\\& \qquad\quad = \left| {\vec c} \right|\;\left| {\vec a \times \vec b} \right|\cos 0 \hfill \\\\& \qquad\quad = 1 \cdot \left| {\vec a} \right|\;\left| {\vec b} \right|\sin \frac{\pi }{6} \cdot 1 \hfill \\\\& \qquad\quad= \frac{1}{2}\left| {\vec a} \right|\;\left| {\vec b} \right|\qquad\qquad\qquad...\left( 1 \right) \hfill \\ \end{align} \]

Squaring both sides of (1) proves the stated assertion.

Example – 32

For three arbitrary vectors  \(\vec a,\;\;\vec b,\;\,\vec c,\)  prove that

\[{[\vec a\;\;\vec b\;\vec c]^2} = \left| {\begin{align}&{\vec a \cdot \vec a}&{\vec a \cdot \vec b}&&{\vec a \cdot \vec c} \\ &  {\vec b \cdot \vec a}&{\vec b \cdot \vec b}&&{\vec b \cdot \vec c} \\ & {\vec c \cdot \vec a}&{\vec c \cdot \vec b}&&{\vec c \cdot \vec c} \end{align}} \right|\]

Solution: The relation is most easily proved by assuming  \(\vec a,\,\,\vec b,\,\,\vec c\) in rectangular form:

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

There’s no loss of generality in this assumption.

Now,

\[\begin{align} {[\vec a\;\;\vec b\;\;\vec c]^2}&= {\left| {\;\begin{array}{*{20}{c}}
  {{a_1}}&{{a_2}}&{{a_3}} \\ 
  {{b_1}}&{{b_2}}&{{b_3}} \\ 
  {{c_1}}&{{c_2}}&{{c_3}} 
\end{array}\;} \right|^2} \\\\
 & = \left| {\;\begin{array}{*{20}{c}}
  {{a_1}}&{{a_2}}&{{a_3}} \\ 
  {{b_1}}&{{b_2}}&{{b_3}} \\ 
  {{c_1}}&{{c_2}}&{{c_3}} 
\end{array}\;} \right|\;\left| {\;\begin{array}{*{20}{c}}
  {{a_1}}&{{a_2}}&{{a_3}} \\ 
  {{b_1}}&{{b_2}}&{{b_3}} \\ 
  {{c_1}}&{{c_2}}&{{c_3}} 
\end{array}\;} \right| \\\\
 &= \left| {\begin{array}{*{20}{c}}
  {a_1^2 + a_2^2 + a_3^2}&{{a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3}}&{{a_1}{c_1} + {a_2}{c_2} + {a_3}{c_3}} \\ 
  {{b_1}{a_1} + {b_2}{a_2} + {b_3}{a_3}}&{b_1^2 + b_2^2 + b_3^2}&{{b_1}{c_1} + {b_2}{c_2} + {b_3}{c_3}} \\ 
  {{c_1}{a_1} + {c_2}{a_2} + {c_3}{a_3}}&{{c_1}{b_1} + {c_2}{b_2} + {c_3}{b_3}}&{c_1^2 + c_2^2 + c_3^2} 
\end{array}} \right| \\\\
  &= \left| {{\mkern 1mu} \begin{array}{*{20}{c}}
  {\vec a \cdot \vec a}&{\vec a \cdot \vec b}&{\vec a \cdot \vec c} \\ 
  {\vec b \cdot \vec a}&{\vec b \cdot \vec b}&{\vec b \cdot \vec c} \\ 
  {\vec c \cdot \vec a}&{\vec c \cdot \vec b}&{\vec c \cdot \vec c} 
\end{array}{\mkern 1mu} } \right| \\ 
\end{align} \]

TRY YOURSELF - IV

Q. 1 Find the volume of the parallelopiped with sides given by \(( - 3\hat i + 7\hat j + 5\hat k),\;\;( - 5\hat i + 7\hat j - 3\hat k)\) and \((7\hat i - 5\hat j - 3\hat k)\)

Q. 2 Find the value of \(\lambda \)  for which the points

\[\begin{align}&A(3\hat i - 2\hat j - \hat k),\;B(2\hat i + 3\hat j - 4\hat k),\;C( - \hat i + \hat j + 2\hat k)\,and\\\\&D(4\hat i + 5\hat j + \lambda \hat k)\,\,are{\text{ }}coplanar. \\ \end{align} \]

Q. 3 For three vectors \(\vec a,\;\vec b,\;\vec c\) prove that

\[[\vec a + \vec b + \vec c\;\;\vec a + \vec b\;\;\vec a + \vec c] =  - [\vec a\;\;\vec b\;\;\vec c]\]

Q. 4 If the vectors \(\vec \alpha  = a\hat i + a\hat j + c\hat k,\;\;\vec \beta  = \hat i + \hat k\,\,and\,\,\vec r = c\hat i + c\hat j + b\hat k\) are coplanar, prove that c is the geometric mean of a and b.

Q. 5 Let \(\vec a,\;\;\vec b,\;\;\vec c,\;\,\vec l,\;\;\vec m,\;\;\vec n\) be arbitrary vectors. Prove that

\[[\vec a\;\;\vec b\;\;\vec c]\;[\vec l\;\;\vec m\;\;\vec n] = \left| {\begin{array}{*{20}{c}}
  {\vec a \cdot \vec l}&{\vec b \cdot \vec l}&{\vec c \cdot \vec l} \\ 
  {\vec a \cdot \vec m}&{\vec b \cdot \vec m}&{\vec c \cdot \vec m} \\ 
  {\vec a \cdot \vec n}&{\vec b \cdot \vec n}&{\vec c \cdot \vec n} 
\end{array}} \right|\]

Q. 6 Let \(\vec r = l(\vec b \times \vec c) + m(\vec c \times \vec a) + n(\vec a \times \vec b)\) . Prove that

\[\vec r \cdot (\vec a + \vec b + \vec c) = l + m + n\]

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Vectors
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Vectors
grade 11 | Questions Set 2
Vectors
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Three Dimensional Geometry
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