Let a = i + j + k , b = i and c = c_ 1 i + c_ 2 j + c_ 3 k , then… - Vidyadip

Question

$\text{Let } \vec{\text a} = \hat{\text{i}} + \hat{\text{j}} + \hat{\text{k}}, \vec{\text{b}} = \hat{\text{i}} \text{ and } \vec{\text{c}} = \text{c}_{1} \hat{\text{i}} + \text{c}_{2} \hat{\text{j}} + \text{c}_{3} \hat{\text{k}}, \text{then}$

Let $c_1 = 1$ and $c_2 = 2$, find $c_3$ which makes $\vec{\text{a}}, \vec{\text{b}} \text{ and }\vec{\text{c}} \text{ coplanar.}$
If $c_2 = –1$ and $c_3 = 1$, show that no value of $c_1$ can make $\vec{\text{a}}, \vec{\text{b}} \text{ and } \vec{\text{c}} \text{ coplanar}.$

✓

Answer

$[\vec{\text{a}} \text{ }\vec{\text{b}} \text{ } \vec{\text{c}}] = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ \text{c}_{1} & \text{c}_{2} & \text{c}_{3} \end{vmatrix} = \text{c}_{2} - \text{c}_{3}$

$\text{c}_{1} = 1, \text{ c}_{2} = 2$

$[\vec{\text{a}} \text{ }\vec{\text{b}} \text{ } \vec{\text{c}}] = 2 - \text{c}_{3}$
$\therefore \vec{\text{a}}, \text{ }\vec{\text{b}}, \text{ } \vec{\text{c}} \text{ are coplanar} [\vec{\text{a}} \text{ }\vec{\text{b}} \text{ } \vec{\text{c}}] = 0 \Rightarrow \text{c}_{3} = 2$

$\text{c}_{2} = -1, \text{c}_{3} = 1$

$[\vec{\text{a}} \text{ }\vec{\text{b}} \text{ } \vec{\text{c}}] = \text{c}_{2} - \text{c}_{3} = -2 \neq 0$
$\Rightarrow \text{No value of }\text{c}_{1} \text{can make }\vec{\text{a}}, \text{ }\vec{\text{b}}, \text{ } \vec{\text{c}} \text{ coplanar}.$

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