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# If r satisfies the equation $r \times \left( {i + 2j + k} \right) = i - k$, then for any scalar $\alpha$, r is equal to$A.{\text{ }}i + \alpha \left( {i + 2j + k} \right) \\ B.{\text{ j}} + \alpha \left( {i + 2j + k} \right) \\ C.{\text{ k}} + \alpha \left( {i + 2j + k} \right) \\ D.{\text{ }}i - k + \alpha \left( {i + 2j + k} \right) \\$ Verified
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Hint: In this question assume $r = \alpha i + \beta j + \lambda k$ and then apply a cross product to reach the solution of the problem.

Let $r = \alpha i + \beta j + \lambda k$…………….. (1)
Now given equation is
$r \times \left( {i + 2j + k} \right) = i - k$
So, first find out the value of
$r \times \left( {i + 2j + k} \right)$
$\Rightarrow r \times \left( {i + 2j + k} \right) = \left( {\alpha i + \beta j + \lambda k} \right) \times \left( {i + 2j + k} \right)$
Now, apply cross product property we have,
$\Rightarrow r \times \left( {i + 2j + k} \right) = \left| {\begin{array}{*{20}{c}} i&j&k \\ \alpha &\beta &\lambda \\ 1&2&1 \end{array}} \right| = i\left( {\beta - 2\lambda } \right) - j\left( {\alpha - \lambda } \right) + k\left( {2\alpha - \beta } \right)$
Now the above value is equal to
$i\left( {\beta - 2\lambda } \right) - j\left( {\alpha - \lambda } \right) + k\left( {2\alpha - \beta } \right) = i - k$
So on comparing the terms $i,{\text{ j, k}}$ we have
$\left( {\beta - 2\lambda } \right)$ = 1, $\left( {\alpha - \lambda } \right)$ = 0, $\left( {2\alpha - \beta } \right)$ = -1
$\Rightarrow \beta = 1 + 2\lambda ,{\text{ }}\alpha = \lambda ,{\text{ }}\beta = 1 + 2\alpha$
So, from equation (1) we have,
$r = \alpha i + \beta j + \lambda k$, substitute the value of $\beta$ and $\lambda$ in this equation we have,
$\Rightarrow r = \alpha i + \left( {1 + 2\alpha } \right)j + \alpha k$
$\Rightarrow r = j + \alpha \left( {i + 2j + k} \right)$, for any scalar $\alpha$.
Hence, option (b) is correct.

Note: In such types of questions first assume r as above then apply cross product as above, then compare the value of cross product with the given value and find out the values of $\alpha ,{\text{ }}\beta {\text{, }}\lambda$, then substitute the value of $\beta$ and $\lambda$in terms of $\alpha$ in r we will get the required r for any scaler $\alpha$.
Last updated date: 20th Sep 2023
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