Let a - - - 2- -3k- and b - 2- - 3- - 5k- If r-a - b-r- r- - 2- - k- - 3 and r-2- - 5- - -k- - -1- - then the value of - - -r-2 is equal to -

R× nbsp;a = nbsp;b× nbsp;r ⇒r× nbsp;a = r× nbsp;b ⇒r× (a + b) = 0, where a+b = 3î ĵ + 2k̂ ⇒r | | (a + b) r = λ(a + b) r· (2î+5ĵ αk̂) = 1 ⇒λ[3î ĵ+2k ...

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Byju's Answer

Standard XII

Physics

Subtraction of a Vector

Let a = î + 2...

Question

Let $\to a =^i + 2^j - 3^k$ and $\to b = 2^i - 3^j + 5^k .$ If $\to r \times \to a = \to b \times \to r,$ $\to r \cdot (α^i + 2^j +^k) = 3$ and $\to r \cdot (2^i + 5^j - α^k) = - 1,$ $α \in R,$ then the value of $α + | \to r |^{2}$ is equal to :

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Solution

The correct option is B $15$
$\to r \times \to a = \to b \times \to r$
$\Rightarrow \to r \times \to a = - \to r \times \to b$
$\Rightarrow \to r \times (\to a + \to b) = \to 0,$ where $\to a + \to b = 3^i -^j + 2^k$
$\Rightarrow \to r | | (\to a + \to b)$
$\to r = λ (\to a + \to b)$

$\to r \cdot (2^i + 5^j - α^k) = - 1$
$\Rightarrow λ [3^i -^j + 2^k] \cdot [2^i + 5^j - α^k] = - 1$
$\Rightarrow λ (6 - 5 - 2 α) = - 1$
$\Rightarrow λ (1 - 2 α) = - 1 . . . (1)$

$\to r \cdot (α^i + 2^j +^k) = 3$
$\Rightarrow λ (3^i -^j + 2^k) \cdot (α^i + 2^j +^k) = 3$
$\Rightarrow λ [3 α - 2 + 2] = 3$
$\Rightarrow λ α = 1 . . . (2)$

Solving $(1)$ and $(2),$
$λ (1 - \frac{2}{λ}) = - 1$
$\Rightarrow λ - 2 = - 1$
$\Rightarrow λ = 1$ and $α = 1$
$\to r = 3^i -^j + 2^k$
$α + | \to r |^{2} = 1 + 14 = 15$

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Similar questions

Q. Let

\to a =^i + 2^j - 3^k

and

\to b = 2^i - 3^j + 5^k .

\to r \times \to a = \to b \times \to r,

\to r \cdot (α^i + 2^j +^k) = 3

and

\to r \cdot (2^i + 5^j - α^k) = - 1,

α \in R,

then the value of

α + | \to r |^{2}

is equal to :

Q. If

\to r = 3^i + 2^j - 5^k, \to a = 2^i -^j +^k, \to b =^i + 3^j - 2^k

and

\to c = - 2^i +^j - 3^k

such that

\to r = x \to a + y \to b + z \to c

, then

If $\to r \times \to b = \to c \times \to b$ and $\to r . \to a = 0$ where $\to a = 2^i + 3^j -^k, \to b = 3^i -^j +^k$ and $\to c =^i +^j + 3^k$ , then $\to r$ is equal to