# 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) \\

\]

Last updated date: 25th Mar 2023

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Answer

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308.1k+ views

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.

Complete step-by-step answer:

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 $.

Complete step-by-step answer:

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 $.

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