Question

Let \[{x_0}\] be the point of local maxima of \[f\left( x \right) = \overrightarrow a \cdot \left( {\overrightarrow b \times \overrightarrow c } \right)\], where \[\overrightarrow a = x\widehat i - 2\widehat j + 3\widehat k\] , \[\overrightarrow b = - 2\widehat i + x\widehat j - \widehat k\] and \[\overrightarrow c = 7\widehat i - 2\widehat j + x\widehat k\]. What is the value of \[\overrightarrow a \cdot \overrightarrow b + \overrightarrow b \cdot \overrightarrow c + \overrightarrow c \cdot \overrightarrow a \] at \[x = {x_0}\]?
A. -22
B. -4
C. -30
D. 14

Answer 1

Hint: First we will apply the triple product formula to calculate the value of \[\overrightarrow a \cdot \left( {\overrightarrow b \times \overrightarrow c } \right)\]. We find the local maxima by finding the differentiation of \[f\left( x \right)\]. The value of \[{x_0}\] is the maxima of \[f\left( x \right)\]. By using dot product formula to calculate \[\overrightarrow a \cdot \overrightarrow b + \overrightarrow b \cdot \overrightarrow c + \overrightarrow c \cdot \overrightarrow a \] then we will put \[x = {x_0}\] to get required answer.

Formula used:
Triple product: \[\overrightarrow a = {x_1}\widehat i + {y_1}\widehat j + {z_1}\widehat k\], \[\overrightarrow b = {x_2}\widehat i + {y_2}\widehat j + {z_2}\widehat k\], \[\overrightarrow c = {x_3}\widehat i + {y_3}\widehat j + {z_3}\widehat k\]
\[\overrightarrow a \cdot \left( {\overrightarrow b \times \overrightarrow c } \right) = \left| {\begin{array}{*{20}{c}}{{x_1}}&{{y_1}}&{{z_1}}\\{{x_2}}&{{y_2}}&{{z_2}}\\{{x_3}}&{{y_3}}&{{z_3}}\end{array}} \right|\]
The formula of dot product \[\overrightarrow a = {x_1}\widehat i + {y_1}\widehat j + {z_1}\widehat k\] and \[\overrightarrow b = {x_2}\widehat i + {y_2}\widehat j + {z_2}\widehat k\] is \[\overrightarrow a \cdot \overrightarrow b = {x_1}{x_2} + {y_1}{y_2} + {z_1}{z_2}\].
Formula of differentiate: \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\]

Complete step by step solution:
Given that \[\overrightarrow a = x\widehat i - 2\widehat j + 3\widehat k\] , \[\overrightarrow b = - 2\widehat i + x\widehat j - \widehat k\] and \[\overrightarrow c = 7\widehat i - 2\widehat j + x\widehat k\].
Applying the triple product to calculate \[\overrightarrow a \cdot \left( {\overrightarrow b \times \overrightarrow c } \right)\].
\[\overrightarrow a \cdot \left( {\overrightarrow b \times \overrightarrow c } \right) = \left| {\begin{array}{*{20}{c}}x&{ - 2}&3\\{ - 2}&x&{ - 1}\\7&{ - 2}&x\end{array}} \right|\]
\[ \Rightarrow \overrightarrow a \cdot \left( {\overrightarrow b \times \overrightarrow c } \right) = x\left( {{x^2} - 2} \right) + 2\left( { - 2x + 7} \right) + 3\left( {4 - 7x} \right)\]
\[ \Rightarrow \overrightarrow a \cdot \left( {\overrightarrow b \times \overrightarrow c } \right) = {x^3} - 2x - 4x + 14 + 12 - 21x\]
\[ \Rightarrow \overrightarrow a \cdot \left( {\overrightarrow b \times \overrightarrow c } \right) = {x^3} - 27x + 26\]
Now putting the value of \[\overrightarrow a \cdot \left( {\overrightarrow b \times \overrightarrow c } \right)\] in \[f\left( x \right) = \overrightarrow a \cdot \left( {\overrightarrow b \times \overrightarrow c } \right)\]
\[f\left( x \right) = {x^3} - 27x + 26\]
Differentiate with respect to \[x\]
\[f'\left( x \right) = 3{x^2} - 27\]
Again, differentiate with respect to \[x\].
\[f''\left( x \right) = 6x\]
To calculate the local minima or maxima we will equate \[f'\left( x \right)\] with 0.
\[0 = 3{x^2} - 27\]
Solve the above equation:
\[ \Rightarrow 3{x^2} = 27\]
Divide both sides by 3
\[ \Rightarrow {x^2} = 9\]
Taking square root on both sides
\[ \Rightarrow x = \pm 3\]
Now putting \[x = 3\] in \[f''\left( x \right) = 6x\]
\[f''\left( x \right) = 6 \cdot 3 = 18 > 0\]
Since \[f''\left( x \right) > 0\], so the function \[f\left( x \right)\] has minima at \[x = 3\].
Now putting \[x = - 3\] in \[f''\left( x \right) = 6x\]
\[f''\left( x \right) = 6 \cdot \left( { - 3} \right) = - 18 < 0\]
Since \[f''\left( x \right) < 0\], so the function \[f\left( x \right)\] has maxima at \[x = - 3\].
Thus, the value of \[{x_0}\] is -3.
Now applying the dot product in \[\overrightarrow a \cdot \overrightarrow b + \overrightarrow b \cdot \overrightarrow c + \overrightarrow c \cdot \overrightarrow a \]
\[\overrightarrow a \cdot \overrightarrow b + \overrightarrow b \cdot \overrightarrow c + \overrightarrow c \cdot \overrightarrow a \]
\[ = \left( {x\widehat i - 2\widehat j + 3\widehat k} \right) \cdot \left( { - 2\widehat i + x\widehat j - \widehat k} \right) + \left( { - 2\widehat i + x\widehat j - \widehat k} \right) \cdot \left( {7\widehat i - 2\widehat j + x\widehat k} \right) + \left( {7\widehat i - 2\widehat j + x\widehat k} \right) \cdot \left( {x\widehat i - 2\widehat j + 3\widehat k} \right)\]
\[ = \left( { - 2x - 2x - 3} \right) + \left( { - 14 - 2 - x} \right) + \left( {7x + 4 + 3x} \right)\]
\[ = - 2x - 2x - 3 - 14 - 2 - x + 7x + 4 + 3x\]
\[ = 5x - 15\]
Now putting \[x = -3\] in the above expression.
\[ = 5 \cdot \left( { - 3} \right) - 15\]
\[ = - 30\]

Hence option C is the correct option.

Note: Students often confused to calculate the maxima or minima. If \[f''\left( x \right) < 0\] at \[x = a\], then the function \[f\left( x \right)\] has maxima at \[x = a\]. If \[f''\left( x \right) > 0\] at \[x = a\], then the function \[f\left( x \right)\] has minima at \[x = a\].