Question

If a and b are two non-zero non-collinear vectors then \[2\left[ {a{\text{ }}b{\text{ i}}} \right]{\text{i }} + {\text{ }}2\left[ {a{\text{ }}b{\text{ j}}} \right]{\text{j }} + {\text{ }}2\left[ {a{\text{ }}b{\text{ k}}} \right]{\text{k }} + \left[ {a{\text{ }}b{\text{ }}a} \right]\] is equal to ?
\[\left( 1 \right)\] \[2(a \times b)\]
\[\left( 2 \right)\] \[a \times b\]
\[\left( 3 \right)\] \[a + b\]
\[\left( 4 \right)\] None of these

Answer 1

Hint: We have to find the value of the given expression of the vector quantities . We solve this question using the concept of cross - product of two vectors . We should also have the knowledge of non - collinear vectors and their properties . We should also have the knowledge of the concept of the scalar triple product of vectors . We should also have the idea of the dot product and cross product of two vectors . Using these properties and expanding the given vector expression , we find the required relation .

Complete answer:
Given :
We have to find the value of \[2\left[ {a{\text{ }}b{\text{ i}}} \right]{\text{i }} + {\text{ }}2\left[ {a{\text{ }}b{\text{ }}j} \right]{\text{j }} + {\text{ }}2\left[ {a{\text{ }}b{\text{ k}}} \right]{\text{k }} + \left[ {a{\text{ }}b{\text{ }}a} \right] - - - - (1)\]
Let us assume that two vectors \[a\] and \[b\] be such that \[a{\text{ }} = {\text{ }}{x_1}i{\text{ }} + {\text{ }}{y_1}j{\text{ }} + {\text{ }}{z_1}k\] and \[b{\text{ }} = {\text{ }}{x_2}i{\text{ }} + {\text{ }}{y_2}{\text{j }} + {\text{ }}{z_2}k\] .
Now , using the method of cross product of two vectors , we get
a \times b = \[\left| {{\text{ }}\begin{array}{*{20}{c}}
  i&j&k \\
  {{x_1}}&{{y_1}}&{{z_1}} \\
  {{x_2}}&{{y_2}}&{{z_2}}
\end{array}} \right|\]
On solving the determinant we obtain a vector equation , as
\[a \times b = ({y_1}{z_2} - {y_2}{z_1})i - ({x_1}{z_2} - {x_2}{z_1})j + ({x_1}{y_2} - {x_2}{y_1})k\]
We also know that \[i.i = j.j = k.k = 1\] and \[i.j = j.k = k.i = 0\]
Taking the dot product of the vector \[a \times b\] with each vector \[i\], \[j\], \[k\] separately , we get
\[(a \times b).i = ({y_1}{z_2} - {y_2}{z_1})i.i - ({x_1}{z_2} - {x_2}{z_1})j.i + ({x_1}{y_2} - {x_2}{y_1})k.i\]
\[(a \times b).i = ({y_1}{z_2} - {y_2}{z_1}) - - - - - (2)\]
\[(a \times b).j = ({y_1}{z_2} - {y_2}{z_1})i.j - ({x_1}{z_2} - {x_2}{z_1})j.j + ({x_1}{y_2} - {x_2}{y_1})k.j\]
\[(a \times b).j = - ({x_1}{z_2} - {x_2}{z_1}) - - - - - (3)\]
\[(a \times b).k = ({y_1}{z_2} - {y_2}{z_1})i.k - ({x_1}{z_2} - {x_2}{z_1})j.k + ({x_1}{y_2} - {x_2}{y_1})k.k\]
\[(a \times b).k = ({x_1}{y_2} - {x_2}{y_1}) - - - - - (4)\]
Now , we also know that scalar triple product of vectors is represents as below :
\[\left[ {a{\text{ }}b{\text{ }}c} \right]{\text{ }} = {\text{ }}\left( {a \times b} \right).c{\text{ }} = {\text{ }}a.\left( {b \times c} \right){\text{ }} = {\text{ }}\left( {a \times c} \right).b\]
So , using the concept of scalar triple product of vectors the equations (2), (3) and (4) can be written as :
\[\left( {a \times b} \right).i{\text{ }} = {\text{ }}\left[ {a{\text{ }}b{\text{ }}i} \right]\]
\[\left( {a \times b} \right).j{\text{ }} = {\text{ }}\left[ {a{\text{ }}b{\text{ }}j} \right]\]
\[\left( {a \times b} \right).k{\text{ }} = {\text{ }}\left[ {a{\text{ }}b{\text{ }}k} \right]\]
Now , substituting the values of equations (2) , (3) and (4) in equation (1) , we get
\[2\left[ {a{\text{ }}b{\text{ }}i} \right]i{\text{ }} + {\text{ }}2\left[ {a{\text{ }}b{\text{ }}j} \right]j{\text{ }} + {\text{ }}2\left[ {a{\text{ }}b{\text{ }}k} \right]k{\text{ }} + \left[ {a{\text{ }}b{\text{ }}a} \right] = 2({y_1}{z_2} - {y_2}{z_1}) - 2({x_1}{z_2} - {x_2}{z_1}) + 2({x_1}{y_2} - {x_2}{y_1})\; + {\text{ }}\left[ {a{\text{ }}b{\text{ }}a} \right] - - - (5)\]
Now , we have to get the value of [a b a] . As we stated above the expansion of a scalar triple product of a vector , we get
\[\left[ {a{\text{ }}b{\text{ }}a} \right]{\text{ }} = {\text{ }}\left( {a \times a} \right).b\]
Also , we know that cross product of two same vectors is always zero .
Hence , we get the value of \[\left[ {a{\text{ }}b{\text{ }}a} \right]\]
\[\left[ {a{\text{ }}b{\text{ }}a} \right] = 0\]
Substituting the value of \[\left[ {a{\text{ }}b{\text{ }}a} \right]\] in equation \[(5)\] , we get
\[2\left[ {a{\text{ }}b{\text{ }}i} \right]i{\text{ }} + {\text{ }}2\left[ {a{\text{ }}b{\text{ }}j} \right]j{\text{ }} + {\text{ }}2\left[ {a{\text{ }}b{\text{ }}k} \right]k{\text{ }} + \left[ {a{\text{ }}b{\text{ }}a} \right] = 2({y_1}{z_2} - {y_2}{z_1}) - 2({x_1}{z_2} - {x_2}{z_1}) + 2({x_1}{y_2} - {x_2}{y_1})\;\]
Taking 2 common from the R.H.S. , we get
\[2\left[ {a{\text{ }}b{\text{ }}i} \right]i{\text{ }} + {\text{ }}2\left[ {a{\text{ }}b{\text{ }}j} \right]{\text{j }} + {\text{ }}2\left[ {a{\text{ }}b{\text{ k}}} \right]k{\text{ }} + \left[ {a{\text{ }}b{\text{ }}a} \right] = 2\left[ {({y_1}{z_2} - {y_2}{z_1}) - ({x_1}{z_2} - {x_2}{z_1}) + ({x_1}{y_2} - {x_2}{y_1})\;} \right] - - - (6)\]
Also , we know that as solved above the cross product of the two vectors is given as :
\[a \times b = ({y_1}{z_2} - {y_2}{z_1})i - ({x_1}{z_2} - {x_2}{z_1})j + ({x_1}{y_2} - {x_2}{y_1})k\]
So , equation (6) becomes
\[2\left[ {a{\text{ }}b{\text{ }}i} \right]i{\text{ }} + {\text{ }}2\left[ {a{\text{ }}b{\text{ }}j} \right]j{\text{ }} + {\text{ }}2\left[ {a{\text{ }}b{\text{ }}k} \right]k{\text{ }} + \left[ {a{\text{ }}b{\text{ }}a} \right] = 2(a \times b)\]
Thus , the value of the given expression of vectors is \[2(a \times b)\] .
Hence , the correct option is \[\left( 1 \right)\] .

Note:
Cross product of two vectors \[a{\text{ }} = {\text{ }}{x_1}i{\text{ }} + {\text{ }}{y_1}j{\text{ }} + {\text{ }}{z_1}k\] , \[b{\text{ }} = {\text{ }}{x_2}i{\text{ }} + {\text{ }}{y_2}{\text{j }} + {\text{ }}{z_2}k\] is given by solving the determinant

\[\left| {{\text{ }}\begin{array}{*{20}{c}}
  i&j&k \\
  {{x_1}}&{{y_1}}&{{z_1}} \\
  {{x_2}}&{{y_2}}&{{z_2}}
\end{array}} \right|\]
$ = \left( {{y_1}{z_2} - {y_2}{z_1}} \right)i - \left( {{x_1}{z_2} - {x_2}{z_1}} \right)j + \left( {{x_1}{y_2} - {y_2}{x_1}} \right)k$

Dot product of two vectors : let \[a{\text{ }} = {\text{ }}{x_1}i{\text{ }} + {\text{ }}{y_1}j{\text{ }} + {\text{ }}{z_1}k\] , \[b{\text{ }} = {\text{ }}{x_2}i{\text{ }} + {\text{ }}{y_2}{\text{j }} + {\text{ }}{z_2}k\] is given as
\[a.b = ({x_1}i{\text{ }} + {\text{ }}{y_1}j{\text{ }} + {\text{ }}{z_1}k{\text{) }}{\text{. (}}{x_2}i{\text{ }} + {\text{ }}{y_2}j + {\text{ }}{z_2}k)\]

\[a.b = ({x_1}{x_2}{\text{ }} + {\text{ }}{y_1}{y_2}{\text{ }} + {\text{ }}{z_1}{z_2})\]

The product \[i.i = j.j = k.k = 1\]
Two vectors are said to be collinear if the cross product of the two vectors results in a zero vector and one vector can be written as multiple another vector.