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Work Done as Dot Product

Let a⃗=-î-k...

Question

Let $\to a = -^i -^k, \to b = -^i +^j$ and $\to c =^i + 2^j + 3^k$ be three given vectors. If $\to r$ is a vector such that $\to r \times \to b = \to c \times \to b$ and $\to r \cdot \to a = 0$ , then the value of $\to r \cdot \to b$ is?

A

8

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B

9

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C

6

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D

None of these

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Solution

The correct option is B $9$
Given $\to a = -^i -^k, \to b = -^i +^j, \to c =^i + 2^j + 3^k$
$\to r \times \to b = \to c \times \to b$ and $\to r . \to a = 0$
Cross product with $\to a$ on both sides.
$\Rightarrow (\to r \times \to b) \times \to a = (\to c \times \to b) \to a$
$\Rightarrow (\to r . \to a) \to b - (\to b . \to a) \to r = (\to c . \to a) \to b - (\to b . \to a) \to c$
$\to a . \to b = (- 1) (- 1) + (1) (0) + (- 1) (0) = 1$
$\to b . \to C = (- 1) (1) + (1) (2) + (0) (3) = 1$
$\to c . \to a = (- 1) (1) + (0) (2) + (- 1) (3) = - 4$
$\Rightarrow - \to r = - 4 \to b - \to c$
$\Rightarrow \to r = 4 \to b + \to c$
$= - 4^i + 4^j +^i + 2^j + 3^k$
$= - 3^i + 6^i + 3^k$
$∴ \to r . \to b (- 3^i + 6^j + 3^k) . (-^i +^j)$
$= 3 + 6$
$= 9$

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