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

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

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.