# Examples On Scalar Triple Product Of Vectors Set-2

Go back to  'Vectors and 3-D Geometry'

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

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

Vectors
grade 11 | Questions Set 1
Vectors