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Theorems for Continuity

Let |a| = 1...

Question

Let $| \to a | = 1, | \to b | = 2$ and $| \to c | = 3$ and $\to a, \to b, \to c$ be three non-coplanar vectors. If $| \to d | = 4,$ then
$∣ ∣ ∣ ∣ ∣ ∣ \frac{[\to d \to b \to c] \to a + [\to d \to c \to a] \to b + [\to d \to a \to b] \to c}{[\to a \to b \to c]} ∣ ∣ ∣ ∣ ∣ ∣$ is equal to

A

2

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B

4

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C

1

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D

None of these

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Solution

The correct option is B 4
If $\to a, \to b, \to c$ are three non-coplanar vectors, then any vector $\to d$ can be written as
$\to d = x \to a + y \to b + z \to c$

First taking dot product with $\to b \times \to c$ , we get
$[\begin{matrix} \to d & \to b & \to c \end{matrix}] = x [\begin{matrix} \to a & \to b & \to c \end{matrix}] + y [\begin{matrix} \to b & \to b & \to c \end{matrix}] + z [\begin{matrix} \to c & \to b & \to c \end{matrix}]$

$\Rightarrow [\begin{matrix} \to d & \to b & \to c \end{matrix}] = x [\begin{matrix} \to a & \to b & \to c \end{matrix}] . . . . . . . . . (∵ [\begin{matrix} \to b & \to b & \to c \end{matrix}] = [\begin{matrix} \to c & \to b & \to c \end{matrix}] = 0)$

$\Rightarrow x = \frac{[\begin{matrix} \to d & \to b & \to c \end{matrix}]}{[\begin{matrix} \to a & \to b & \to c \end{matrix}]}$

Similarly, taking dot product with $\to c \times \to a$ and $\to a \times \to b$ we get,
$y = \frac{[\begin{matrix} \to d & \to c & \to a \end{matrix}]}{[\begin{matrix} \to b & \to c & \to a \end{matrix}]}$

$z = \frac{[\begin{matrix} \to d & \to a & \to b \end{matrix}]}{[\begin{matrix} \to c & \to a & \to b \end{matrix}]}$

Hence $\to d = \frac{[\begin{matrix} \to d & \to b & \to c \end{matrix}]}{[\begin{matrix} \to a & \to b & \to c \end{matrix}]} \to a + \frac{[\begin{matrix} \to d & \to c & \to a \end{matrix}]}{[\begin{matrix} \to b & \to c & \to a \end{matrix}]} \to b + \frac{[\begin{matrix} \to d & \to a & \to b \end{matrix}]}{[\begin{matrix} \to c & \to a & \to b \end{matrix}]} \to c$

$\to d = \frac{[\begin{matrix} \to d & \to b & \to c \end{matrix}] \to a + [\begin{matrix} \to d & \to c & \to a \end{matrix}] \to b + [\begin{matrix} \to d & \to a & \to b \end{matrix}] \to c}{[\begin{matrix} \to a & \to b & \to c \end{matrix}]}$

$\Rightarrow ∣ ∣ ∣ ∣ ∣ \frac{[\begin{matrix} \to d & \to b & \to c \end{matrix}] \to a + [\begin{matrix} \to d & \to c & \to a \end{matrix}] \to b + [\begin{matrix} \to d & \to a & \to b \end{matrix}] \to c}{[\begin{matrix} \to a & \to b & \to c \end{matrix}]} ∣ ∣ ∣ ∣ ∣ = ∣ ∣ \to d ∣ ∣ = 4$
Hence, option 'B' is correct.

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