Question

The vector \[\vec{b} = 3 \hat { i }+ 4 \hat {k }\] is to be written as the sum of a vector \[\vec{\alpha}\] parallel to \[\vec{a} = \hat {i} + \hat {j}\] and a vector \[\vec{\beta}\] perpendicular to \[\vec{a}\]. Then \[\vec{\alpha} =\]

विकल्प

• \[\frac{3}{2}\left( \hat { i} + \hat {j} \right)\]

• \[\frac{2}{3}\left( \hat {i} + \hat {j} \right)\]

• \[\frac{1}{2}\left(\hat { i} + \hat {j} \right)\]

• \[\frac{1}{3}\left( \hat { i} + \hat {j} \right)\]

Answer 1

\[\frac{3}{2}\left( \hat { i} + \hat {j} \right)\]

Let:

\[ \vec{\alpha} = a_1 \hat {i} + a_2 \hat {j} + a_3 \hat {k} \]

\[ \vec{\beta} = b_1 \hat {i} + b_2 \hat {j} + b_3 \hat {k} \]

Now,

\[ \vec{b} =3 \hat {i} +4 \hat {k} = \vec{\alpha} + \vec{\beta} \text { (Given) }\]

\[ \Rightarrow 3\hat {  i} + 0 \hat {j} + 4 \hat {k} = \left( a_1 + b_1 \right) \hat {i} + \left( a_2 + b_2 \right) \hat{j} + \left( a_3 + b_3 \right) \hat {k} \]

\[ \Rightarrow a_1 + b_1 = 3; a_2 + b_2 = 0; a_3 + b_3 = 4\]

\[ \Rightarrow a_1 + b_1 = 3; a_2 = - b_2 ; a_3 + b_3 = 4 . . . (1)\]

\[ \vec{a} = \hat {i} + \hat {j} \text {(Given)} \]

\[\text { Also,} \vec{\alpha} \text { is parallel to } \vec{a} .\]

\[ \Rightarrow \vec{\alpha} \times \vec{a} = \vec{0} \]

\[ \Rightarrow \begin{vmatrix}\text { i} & \hat { j } & \hat {k} \\ a_1 & a_2 & a_3 \\ 1 & 1 & 0\end{vmatrix} = \vec{0} \]

\[ \Rightarrow - a_3 \hat {i} + a_3 \hat { j} + \left( a_1 - a_2 \right) \hat {k} = 0 \hat { i} + 0 \hat { j } + 0 \hat {k} \]

\[ \Rightarrow a_3 = 0; a_1 - a_2 = 0\]

\[ \Rightarrow a_3 = 0; a_1 = a_2 . . . (2)\]

\[\text { Since } \vec{\beta}\text {  is perpendicular to } \vec{a} ,\text {  we get }\]

\[ \Rightarrow \vec{\beta} . \vec{a} = 0\]

\[ \Rightarrow \left( b_1\hat {i} + b_2 \hat {j} + b_3 \hat { k} \right) . \left(\hat { i } + \hat {j} \right) = 0\]

\[ \Rightarrow b_1 + b_2 = 0\]

\[ \Rightarrow b_1 = - b_2 . . . (3)\]

Solving (1), (2) and (3), we get

\[ a_1 = \frac{3}{2}; a_2 = \frac{3}{2}; a_3 = 0\]

\[\therefore \vec{\alpha} = a_1\hat{  i } + a_2 \hat { j } + a_3 \hat { k } \]

\[ = \frac{3}{2} \hat { i } + \frac{3}{2} \hat { j } + 0 \hat { k } \]

\[ = \frac{3}{2} \left( \hat { i }+ \hat { j } \right)\]