Question

If $r = \alpha b \times c + \beta c \times a + \gamma a \times b$ and $\left[ {abc} \right] = 2$ , then $\alpha + \beta + \gamma $ is equal to
A. $r\left[ {b \times c + c \times a + a \times b} \right]$
B. $\dfrac{1}{2}r\left( {a + b + c} \right)$
C. $2r(a + b + c)$
D. $4$

Answer 1

Hint: In this question we have been given that
$r = \alpha b \times c + \beta c \times a + \gamma a \times b$ .
So we will multiply both the left hand side and right hand side of the equation with $a$ .
We will take the constant out of the bracket in the right hand side and then simplify it. Similarly we will again multiply the original equation one by one with $b$ and $c$ . We should note the formula that we will be using in this question that if we have two terms similar such as $\left[ {xxy} \right]$ then its value will be zero i.e. $\left[ {xxy} \right] = 0$ .

Complete step-by-step solution:
Here we have
 $r = \alpha b \times c + \beta c \times a + \gamma a \times b$ and,
 $\left[ {abc} \right] = 2$ .
Let us multiply the given equation with $a$ , then we can write it as
\[a \cdot r = \alpha (a \cdot b \times c) + \beta (a \cdot c \times a) + \gamma (a \cdot a \times b)\]
We can see that we have two terms similar, i.e., \[\beta (a \cdot c \times a)\] and
\[\gamma (a \cdot a \times b)\] .
Now according to the formula we know that their values will be zero i.e.
\[\beta (a \cdot c \times a) = 0\] and,
\[\gamma (a \cdot a \times b) = 0\] .
By putting the values back in the equation we have:
\[a \cdot r = \alpha (a \cdot b \times c) + 0 + 0\]
So it gives us
$ar = \alpha (abc)$
Now we will multiply the equation with
$b$.
We can write it as:
\[b \cdot r = \alpha (b \cdot b \times c) + \beta (b \cdot c \times a) + \gamma (b \cdot a \times b)\]
Here we have the value of
\[\alpha (b \cdot b \times c) = 0\] and
\[\gamma (b \cdot a \times b) = 0\] .
We can substitute these values in the equation and we have :
\[b \cdot r = 0 + \beta (b \cdot c \times a) + 0\]
So it gives us value
 \[b \cdot r = \beta (b \cdot c \times a)\]
Again we will multiply the equation with
$c$ .
On multiplying we can write it as:
\[c \cdot r = \alpha (c \cdot b \times c) + \beta (c \cdot c \times a) + \gamma (c \cdot a \times b)\]
So it will give us value
\[c \cdot r = \gamma (c \cdot b \times a)\]
We will add all the three new terms i.e.
$ar + br + cr = \alpha \left[ {abc} \right] + \beta \left[ {abc} \right] + \gamma \left[ {abc} \right]$
We can also write this as
\[ar + br + cr = \left[ {abc} \right]\left( {\alpha + \beta + \gamma } \right)\]
We have been given that
 $\left[ {abc} \right] = 2$
So by putting this we have
\[(a + b + c)r = 2\left( {\alpha + \beta + \gamma } \right)\]
Now we can write this as
\[\dfrac{{r(a + b + c)}}{2} = \left( {\alpha + \beta + \gamma } \right)\]
Hence the correct option is (b) $\dfrac{1}{2}r(a + b + c)$

Note: We should note that in the above solution, the value of
\[\alpha (c \cdot b \times c) = 0\]
And,
 \[\beta (c \cdot a \times a) = 0\] .
By substituting these values in the equation and we have :
\[c \cdot r = 0 + 0 + \gamma (c \cdot a \times b)\] .
So it will be gives
$cr = \gamma (abc)$