12th NCERT Vector Algebra Exercise 10.5 Questions 7
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Question (1)

Find $\left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right]$ if $\overrightarrow a = \widehat i - 2\widehat j + 3\widehat k$, $\overrightarrow b = 2\widehat i - 3\widehat j + \widehat k$   and   $\overrightarrow c = 3\widehat i + \widehat j - 2\widehat k$

Solution

$\left[ {\overrightarrow a \,\,\overrightarrow b \,\overrightarrow c } \right] = \left| {\begin{array}{*{20}{c}}1&{ - 2}&3\\2&{ - 3}&1\\3&1&{ - 2}\end{array}} \right|$ $\left[ {\overrightarrow a \,\,\overrightarrow b \,\overrightarrow c } \right] = 1\left( 5 \right) + 2\left( { - 4 - 3} \right) + 3\left( {2 + 9} \right)$ $\left[ {\overrightarrow a \,\,\overrightarrow b \,\overrightarrow c } \right] = 5 - 14 - 33 = 24$

Question (2)

Show that the vectors $\overrightarrow a = \widehat i - 2\widehat j + 3\widehat k$ , $\overrightarrow b = - 2\widehat i + 3\widehat j - 4\widehat k$, and $\overrightarrow c = \widehat i - 3\widehat j + 5\widehat k$   are coplanar

Solution

$\left[ {\overrightarrow a \,\,\overrightarrow b \,\overrightarrow c } \right] = \left| {\begin{array}{*{20}{c}}1&{ - 2}&3\\{ - 2}&3&{ - 4}\\1&{ - 3}&5\end{array}} \right|$ $\left[ {\overrightarrow a \,\,\overrightarrow b \,\overrightarrow c } \right] = 1\left( {15 - 12} \right) + 2\left( { - 10 + 4} \right) + 3\left( {6 - 3} \right)$ $\left[ {\overrightarrow a \,\,\overrightarrow b \,\overrightarrow c } \right] = 3 - 12 + 9 = 0$ $\therefore \left[ {\overrightarrow a \,\,\overrightarrow b \,\overrightarrow c } \right]=0$, vectors ${\overrightarrow a, \,\,\overrightarrow b, \,\overrightarrow c }$are coplanar

Question (3)

Find λ if the vectors $\widehat i - \widehat j + \widehat k$, $3\widehat i + \widehat j + 2\widehat k$, and $\widehat i + \lambda \widehat j - 3\widehat k$ are coplanar.

Solution

Since vector are coplaner $\left[ {\overrightarrow a \,\,\overrightarrow b \,\overrightarrow c } \right] = 0$
$\left| {\begin{array}{*{20}{c}}1&{ - 1}&1\\3&1&2\\1&\lambda &{ - 3}\end{array}} \right| = 0$ $1\left( { - 3 - 2\lambda } \right) + 1\left( { - 9 - 2} \right) + 1\left( {3\lambda - 1} \right) = 0$ $- 3 - 2\lambda - 11 + 3\lambda - 1 = 0$ $\lambda - 15 = 0$ $\lambda = 15$

Question (4)

Let $\overrightarrow a = \widehat i + \widehat j + \widehat k$ $\overrightarrow b = \widehat i$ and $\overrightarrow c = {c_1}\widehat i + {c_2}\widehat j + {c_3}\widehat k$ then
(a) If c2 = 1 and c2 = 2, find c3 which makes $\overrightarrow a$, $\overrightarrow b$, and $\overrightarrow c$, coplaner
(b) If c2 = -1 and c3 = 1, show that no value of c1 can make $\overrightarrow a ,\overrightarrow b \,and\overrightarrow {\,c}$ coplanar

Solution

(a) c1 = 1, c2 = 2, c3 = ---?
Since vector are coplaner $\left[ {\overrightarrow a \,\,\overrightarrow b \,\overrightarrow c } \right] = 0$
$\left| {\begin{array}{*{20}{c}}1&1&1\\1&0&0\\1&2&{{c_3}}\end{array}} \right| = 0$ $1\left( {0 - 0} \right) - 1\left( {{c_3} - 0} \right) + 1\left( {2 - 0} \right) = 0$ $- {c_3} + 2 = 0$ $\Rightarrow {c_3} = 2$
(b) c2 = -1, c3 = 1 , c1= ---?
Since vector are coplaner $\left[ {\overrightarrow a \,\,\overrightarrow b \,\overrightarrow c } \right] = 0$
$\left| {\begin{array}{*{20}{c}}1&1&1\\1&0&0\\{{c_1}}&{ - 1}&1\end{array}} \right| = 0$ $1\left( {0 - 0} \right) - 1\left( {1 - 0} \right) + 1\left( { - 1} \right) = 0$ $- 1 - 1 = 0 \quad \text{which is not possible}$ So for any value of c1, vectors will not be coplaner

Question (5)

Show that the four points with position vectors
$4\widehat i + 8\widehat j + 12\widehat k$, $2\widehat i + 4\widehat j + 6\widehat k$, and $3\widehat i + 5\widehat j + 4\widehat k$ are coplanar

Solution

$A\left( {4\widehat i + 8\widehat j + 12\widehat k} \right)$,
$B\left( {2\widehat i + 4\widehat j + 6\widehat k} \right)$,
$C\left( {3\widehat i + 5\widehat j + 4\widehat k} \right)$
$D\left( {5\widehat i + 8\widehat j + 5\widehat k} \right)$
$\overrightarrow {AB} = \overrightarrow b - \overrightarrow a = - 2\widehat i - 4\widehat j - 6\widehat k$ $\overrightarrow {AC} = \overrightarrow c - \overrightarrow a = - \widehat i - 3\widehat j - 8\widehat k$ $\overrightarrow {AD} = \overrightarrow d - \overrightarrow a = \widehat i + 0\widehat j - 7\widehat k$ $\left[ {\overrightarrow {AB} \,\,\overrightarrow {AC} \,\overrightarrow {AD} } \right] = \left| {\begin{array}{*{20}{c}}{ - 2}&{ - 4}&{ - 6}\\{ - 1}&{ - 3}&{ - 8}\\1&0&{ - 7}\end{array}} \right|$ $\left[ {\overrightarrow {AB} \,\,\overrightarrow {AC} \,\overrightarrow {AD} } \right] = - 2\left( {21} \right) + 4\left( {7 + 8} \right) - 6\left( {0 + 3} \right)$ $\left[ {\overrightarrow {AB} \,\,\overrightarrow {AC} \,\overrightarrow {AD} } \right] = - 42 + 60 - 18 = 0$ $\text{since} \quad \left[ {\overrightarrow {AB} \,\,\overrightarrow {AC} \,\overrightarrow {AD} } \right] = 0$ vectors ${\overrightarrow {AB} }$, ${\overrightarrow {AC} }$ and ${\overrightarrow {AD} }$ are coplaner vectors
^there4; A, B, C, D are coplaner points

Question (6)

Find x such that the four points A (3, 2, 1) B (4, x, 5), C (4, 2, –2) and D (6, 5, –1) are coplanar.

Solution

A (3, 2, 1) B (4, x, 5), C (4, 2, –2) and D (6, 5, –1)
$\overrightarrow {AB} = \left( {1,x - 2,4} \right)$,  $\overrightarrow {AC} = \left( {1,0, - 3} \right)$,  $\overrightarrow {AD} = \left( {3,3, - 2} \right)$
aince points are coplaner vectors ${\overrightarrow {AB} \,\,\overrightarrow {AC} \,\overrightarrow {AD} }$ are coplaner
$\left[ {\overrightarrow {AB} \,\,\overrightarrow {AC} \,\overrightarrow {AD} } \right] = 0$ $\left| {\begin{array}{*{20}{c}}1&{x - 2}&4\\1&0&{ - 3}\\3&{ - 3}&{ - 2}\end{array}} \right| = 0$ $1\left( 9 \right) - \left( {x - 2} \right)\left( { - 2 + 9} \right) + 4\left( {3 - 0} \right) = 3$ $9 - 7x + 14 + 12 = 0$ $- 7x + 35 = 0$ $\Rightarrow x = \frac{{ 35}}{{ 7}}=5$

Question (7)

Show that the vectors $\overrightarrow a ,\overrightarrow b \,and\overrightarrow {\,c}$ coplanar if $\overrightarrow a + \overrightarrow b ,\,\overrightarrow {\,b} + \overrightarrow c \,\,and\,\,\overrightarrow c + \overrightarrow a$ are coplanar.

Solution

$\overrightarrow a + \overrightarrow b ,\;\overrightarrow b + \overrightarrow c ,\;\overrightarrow c + \overrightarrow a$ are coplaners
$\left[ {\overrightarrow a + \overrightarrow b \;\;\overrightarrow b + \overrightarrow c \;\;\overrightarrow c + \overrightarrow a } \right] = 0$ $\left( {\overrightarrow a + \overrightarrow b } \right) \cdot \left[ {\left( {\overrightarrow b + \overrightarrow c } \right) \times \left( {\overrightarrow c + \overrightarrow a } \right)} \right] = 0$ $\therefore \left( {\overrightarrow a + \overrightarrow b } \right) \cdot \left[ {\overrightarrow b \times \overrightarrow c + \overrightarrow b \times \overrightarrow a + \overrightarrow c \times \overrightarrow c + \overrightarrow c \times \overrightarrow a } \right] = 0$ As $\overrightarrow c \times \overrightarrow c = 0$ $\left( {\overrightarrow a + \overrightarrow b } \right) \cdot \left[ {\overrightarrow b \times \overrightarrow c + \overrightarrow b \times \overrightarrow a + \overrightarrow 0 + \overrightarrow c \times \overrightarrow a } \right] = 0$ $\overrightarrow a \cdot \left( {\overrightarrow b \times \overrightarrow c } \right) + \overrightarrow a \cdot \left( {\overrightarrow b \times \overrightarrow a } \right) + \overrightarrow a \cdot \left( {\overrightarrow c \times \overrightarrow a } \right) + \overrightarrow b \cdot \left( {\overrightarrow b \times \overrightarrow c } \right) + \overrightarrow b \cdot \left( {\overrightarrow b \times \overrightarrow a } \right) + \overrightarrow b \cdot \left( {\overrightarrow c \times \overrightarrow a } \right) = 0$
$\left[ {\overrightarrow a \;\;\overrightarrow {b\,} \;\overrightarrow c } \right] + \left[ {\overrightarrow a \;\;\overrightarrow {b\,} \;\overrightarrow a } \right] + \left[ {\overrightarrow a \;\;\overrightarrow {c\,} \;\overrightarrow a } \right] + \left[ {\overrightarrow b \;\;\overrightarrow {b\,} \;\overrightarrow c } \right] + \left[ {\overrightarrow b \;\;\overrightarrow {b\,} \;\overrightarrow a } \right] + \left[ {\overrightarrow b \;\;\overrightarrow {c\,} \;\overrightarrow a } \right] = 0$
$\therefore \left[ {\overrightarrow a \;\;\overrightarrow {b\,} \;\overrightarrow c } \right] + 0 + 0 + 0 + 0 - \left[ {\overrightarrow a \;\;\overrightarrow {c\,} \;\overrightarrow b } \right] = 0$ $\therefore \left[ {\overrightarrow a \;\;\overrightarrow {b\,} \;\overrightarrow c } \right] + \left[ {\overrightarrow a \;\;\overrightarrow {b\,} \;\overrightarrow c } \right] = 0$ $\therefore 2\left[ {\overrightarrow a \;\;\overrightarrow {b\,} \;\overrightarrow c } \right] = 0$ $\left[ {\overrightarrow a \;\;\overrightarrow {b\,} \;\overrightarrow c } \right] = 0$ since $\left[ {\overrightarrow a \;\;\overrightarrow {b\,} \;\overrightarrow c } \right] = 0$ vectors ${\overrightarrow a \;\;\overrightarrow {b\,} \;\overrightarrow c }$ are coplaners