Let a - - - k- c - - - - - k- then b satisfying a-b - c - 0 and a- b - 3 is

Given: c = a×b c = nbsp;b×a Taking dot product of b both sides b. c = b.(b×a) = [b b a] = 0 Hence, b. c = nbsp; 0 ⋯ (i) Let b = b_1 î+b_2 ĵ+b_3 k̂ putti ...

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Byju's Answer

Standard XII

Physics

Work Done as Dot Product

Let a = ĵ - k...

Question

Let $\to a =^j -^k, \to c =^i -^j -^k$ then $\to b$ satisfying $\to a \times \to b + \to c = 0$ and $\to a . \to b = 3$ is

-^i +^j - 2^k

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^i +^j - 2^k

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2^i -^j + 2^k

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Solution

The correct option is A $-^i +^j - 2^k$
Given: $\to c = - \to a \times \to b$
$\to c = \to b \times \to a$
Taking dot product of $\to b$ both sides
$\to b . \to c = \to b . (\to b \times \to a) = [\to b \to b \to a] = 0$
Hence, $\to b . \to c = 0 \dots (i)$
Let $\to b = b_{1}^i + b_{2}^j + b_{3}^k$
putting values in equation $(i)$
$\Rightarrow (b_{1}^i + b_{2}^j + b_{3}^k) . (^i -^j -^k) = 0$
$\Rightarrow b_{1} - b_{2} - b_{3} = 0 \dots (i i)$
And given $\to a . \to b = 3$
$\Rightarrow (^j -^k) . (b_{1}^i + b_{2}^j + b_{3}^k) = 3$
$\Rightarrow b_{2} - b_{3} = 3 \dots (i i i)$
From $(i i)$ and $(i i i)$
$b_{1} = b_{2} + b_{3} = (3 + b_{3}) + b_{3}$
$b_{1} = 3 + 2 b_{3}$
Hence, $\to b = (3 + 2 b_{3})^i + (3 + b_{3})^j + b_{3}^k$
Now putting all values in $\to a \times \to b + \to c = 0$
$(^j -^k) \times ((3 + 2 b_{3})^i + (3 + b_{3})^j + b_{3}^k) + (^i -^j -^k) = 0$
$(2 b_{3} + 3)^i + (- 3 - 2 b_{3})^j + (- 3 - 2 b_{3})^k = (- 1)^i + (1)^j + (1)^k$
On comparing $2 b_{3} + 3 = - 1$
$2 b_{3} = - 4$
$b_{3} = - 2$
Hence $b_{1} = 3 + 2 (- 2) = - 1$
Hence $b_{2} = 3 + (- 2) = 1$
Therefore $\to b = -^i +^j - 2^k$

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Work Done as Dot Product

Standard XII Physics

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Let →a=^j−^k, →c=^i−^j−^k then →b satisfying →a×→b+→c=0 and →a.→b=3 is

Let $\to a =^j -^k, \to c =^i -^j -^k$ then $\to b$ satisfying $\to a \times \to b + \to c = 0$ and $\to a . \to b = 3$ is