Let the vectors u 1 = i ^ + j ^ + a k ^ , u 2 = i ^ + b j ^ + k ^ and u 3 = c i ^ + j ^ + k ^ be coplanar. If the vectors v ? 1 = ( a + b ) i ^ + c j ^ + c k ^ , v 2 = a i ^ + ( b + c ) j ^ + a k ^ and v 3 = b i ^ + b j ^ + ( c + a ) k ^ are

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Solution

Q.
Let the vectors ${\vec{u}}_{1} = \hat{i} + \hat{j} + a \hat{k},$ ${\vec{u}}_{2} = \hat{i} + b \hat{j} + \hat{k}$ and ${\vec{u}}_{3} = c \hat{i} + \hat{j} + \hat{k}$ be coplanar. If the vectors ${\vec{v}}_{1} = (a + b) \hat{i} + c \hat{j} + c \hat{k},$ ${\vec{v}}_{2} = a \hat{i} + (b + c) \hat{j} + a \hat{k}$ and ${\vec{v}}_{3} = b \hat{i} + b \hat{j} + (c + a) \hat{k}$ are also coplanar, then $6 (a + b + c)$ is equal to [2023]

1 4

2 0

3 12

4 6

Ans.
(3)

Given that ${\vec{u}}_{1}, {\vec{u}}_{2}$ and ${\vec{u}}_{3}$ are coplanar.

$∴$ $[{\vec{u}}_{1} {\vec{u}}_{2} {\vec{u}}_{3}] = 0 \Rightarrow$ $| \begin{matrix} 1 & 1 & a \\ 1 & b & 1 \\ c & 1 & 1 \end{matrix} |$ $= 0$

$\Rightarrow a + b + c - a b c = 2$ $...(i)$

Now, $[{\vec{v}}_{1} {\vec{v}}_{2} {\vec{v}}_{3}] =$ $| \begin{matrix} a + b & c & c \\ a & b + c & a \\ b & b & c + a \end{matrix} |$ $= 0 \Rightarrow a b c = 0$ $...(ii)$

From (i) and (ii), we get $a + b + c = 2$

So, $6 (a + b + c) = 12$

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