# If \[\vec a.(\hat i) = \vec a.(\hat i + \hat j) = \vec a.(\hat i + \hat j + \hat k) = 1\] then \[\vec a = \]

\[

{\text{A}}{\text{.}}\hat i + \hat j \\

{\text{B}}{\text{.}}\hat i - \hat k \\

{\text{C}}{\text{.}}\hat i \\

{\text{D}}{\text{.}}\hat i + \hat j - \hat k \\

\]

Answer

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Hint – Assume the vector \[\vec a\] and operate the dot product. Dot product of two unit vectors is always one.

Let $\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......({\text{i}})$

We know, $\hat i.\hat i = 1,{\text{ }}\hat j.\hat j = 1,{\text{ \& }}\hat k.\hat k = {\text{1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......{\text{(ii)}}$

Given,

$\vec a.\hat i = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( {{\text{iii}}} \right)$

$

({a_1}\hat i + {a_2}\hat j + {a_3}\hat k).\hat i = 1\,\,\,\,\,\,\,\,\,\,\, \\

{a_1} = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,({\text{from (i),(ii),(iii))}} \\

\\

{\text{also, }}\vec a.(\hat i + \hat j) = 1 \\

({a_1}\hat i + {a_2}\hat j + {a_3}\hat k\,).(\hat i + \hat j) = 1\,\,\,\,\,\,\,\,\,\,({\text{iv}}) \\

{a_1} + {a_2} = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,({\text{from (i),(ii),(iv))}} \\

{a_2} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\,\because \,{a_1} = 1} \right) \\

$

Also,

$\vec a.(\hat i + \hat j + \hat k) = 1$

Then,

$

({a_1}\hat i + {a_2}\hat j + {a_3}\hat k\,).(\hat i + \hat j + \hat k) = 1\,\,\,\,\,\,\,\,\,\,({\text{v}}) \\

{a_1} + {a_2} + {a_3} = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,({\text{from (i),(ii),(v))}} \\

{a_3} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{a_1} = 1,{a_2}{\text{ = 0, above}}} \right) \\

$

We come to know ,

${a_1} = 1,{a_2}{\text{ = 0,}}{a_3} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,({\text{vi}})$

Then, ${\text{ }}\vec a = \hat i{\text{ (from (i))}}$

Hence the correct option is C.

Note – In these types of questions of vectors we have to use the concept that the dot product of two vectors is one and with different unit vectors is zero. Then we can get the value of the asked value by solving the obtained equations.

Let $\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......({\text{i}})$

We know, $\hat i.\hat i = 1,{\text{ }}\hat j.\hat j = 1,{\text{ \& }}\hat k.\hat k = {\text{1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......{\text{(ii)}}$

Given,

$\vec a.\hat i = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( {{\text{iii}}} \right)$

$

({a_1}\hat i + {a_2}\hat j + {a_3}\hat k).\hat i = 1\,\,\,\,\,\,\,\,\,\,\, \\

{a_1} = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,({\text{from (i),(ii),(iii))}} \\

\\

{\text{also, }}\vec a.(\hat i + \hat j) = 1 \\

({a_1}\hat i + {a_2}\hat j + {a_3}\hat k\,).(\hat i + \hat j) = 1\,\,\,\,\,\,\,\,\,\,({\text{iv}}) \\

{a_1} + {a_2} = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,({\text{from (i),(ii),(iv))}} \\

{a_2} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\,\because \,{a_1} = 1} \right) \\

$

Also,

$\vec a.(\hat i + \hat j + \hat k) = 1$

Then,

$

({a_1}\hat i + {a_2}\hat j + {a_3}\hat k\,).(\hat i + \hat j + \hat k) = 1\,\,\,\,\,\,\,\,\,\,({\text{v}}) \\

{a_1} + {a_2} + {a_3} = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,({\text{from (i),(ii),(v))}} \\

{a_3} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{a_1} = 1,{a_2}{\text{ = 0, above}}} \right) \\

$

We come to know ,

${a_1} = 1,{a_2}{\text{ = 0,}}{a_3} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,({\text{vi}})$

Then, ${\text{ }}\vec a = \hat i{\text{ (from (i))}}$

Hence the correct option is C.

Note – In these types of questions of vectors we have to use the concept that the dot product of two vectors is one and with different unit vectors is zero. Then we can get the value of the asked value by solving the obtained equations.

Last updated date: 26th Sep 2023

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