
If \[a,b\] and \[{\rm{c}}\] are unit coplanar vectors then the scalar triple product \[[2a - b2b - c2c - a]\] is equal to
A. 0
B. \[1\]
C. \[ - \sqrt 3 \]
D. \[\sqrt 3 \]
Answer
164.4k+ views
Hint: A vector is a quantity with both magnitude and direction. A unit vector is one with a magnitude of one. It is also referred to as the Direction Vector. The symbol (∧) sometimes known as a cap or hat. It is used to represent the Unit Vector. It is provided by \[\hat a = \frac{a}{{|a|}}\]
Formula Used:The dot product of two vectors is
\[{\bf{a}}.\left( {{\bf{b}} + {\bf{c}}} \right) = {\bf{a}}.{\bf{b}} + {\bf{a}}.{\bf{c}}\]
Complete step by step solution:There are two approaches to define the dot product: algebraically and geometrically. The dot product is defined algebraically as the sum of the products of the corresponding elements of the two number sequences.
We have been given that If \[a,b,c\] and \[c\] are unit coplanar vectors.
The scalar triple product
\[[2a - b,2b - c,2c - a]\]
Given that “\[a,{\rm{ }}b,{\rm{ }}c\]” are unit coplanar vectors
Therefore, that implies
\[\left[ {abc} \right] = 0\]-- (1)
Because \[[abc] = a.\{ b \times c\} \] and since \[{\rm{b}}\] and \[{\rm{c}}\] are unit coplanar vectors\[b \times c = 1\]
Therefore now \[[2a - b2b - c2c - a]\]
Now according to the data, the equation becomes,
\[ = (2a - b) \cdot \{ (2b - c) \times (2c - a)\} \]
Now, on computing the cross product, we obtain
\[ = (2a - b) \cdot \{ 4(b \times c) - 2(b \times a) + (c \times a)\} \]
Now, we have to simplify the above equation, we get
\[ = 8(a \cdot (b \times c)) - (b \cdot (c \times a))\]
Let’s further simplify the above equation to make it less complicated:
\[ = 8[abc] - [abc]\]
Now, we have to subtract the obtained equation, we get
\[ = 7[abc]\]
As “\[a,{\rm{ }}b,{\rm{ }}c\]” are unit coplanar vectors, it becomes
\[ = 0\]
Therefore, the scalar triple product \[[2a - b2b - c2c - a]\] is equal to \[0\].
Option ‘A’ is correct
Note: Vector is a quantity with both magnitude and direction. A unit vector is one with a magnitude of one. Dot product and cross product is not the same thing. Concentrate your efforts on the calculation part. Remember that the magnitude of a unit vector is always one.
Formula Used:The dot product of two vectors is
\[{\bf{a}}.\left( {{\bf{b}} + {\bf{c}}} \right) = {\bf{a}}.{\bf{b}} + {\bf{a}}.{\bf{c}}\]
Complete step by step solution:There are two approaches to define the dot product: algebraically and geometrically. The dot product is defined algebraically as the sum of the products of the corresponding elements of the two number sequences.
We have been given that If \[a,b,c\] and \[c\] are unit coplanar vectors.
The scalar triple product
\[[2a - b,2b - c,2c - a]\]
Given that “\[a,{\rm{ }}b,{\rm{ }}c\]” are unit coplanar vectors
Therefore, that implies
\[\left[ {abc} \right] = 0\]-- (1)
Because \[[abc] = a.\{ b \times c\} \] and since \[{\rm{b}}\] and \[{\rm{c}}\] are unit coplanar vectors\[b \times c = 1\]
Therefore now \[[2a - b2b - c2c - a]\]
Now according to the data, the equation becomes,
\[ = (2a - b) \cdot \{ (2b - c) \times (2c - a)\} \]
Now, on computing the cross product, we obtain
\[ = (2a - b) \cdot \{ 4(b \times c) - 2(b \times a) + (c \times a)\} \]
Now, we have to simplify the above equation, we get
\[ = 8(a \cdot (b \times c)) - (b \cdot (c \times a))\]
Let’s further simplify the above equation to make it less complicated:
\[ = 8[abc] - [abc]\]
Now, we have to subtract the obtained equation, we get
\[ = 7[abc]\]
As “\[a,{\rm{ }}b,{\rm{ }}c\]” are unit coplanar vectors, it becomes
\[ = 0\]
Therefore, the scalar triple product \[[2a - b2b - c2c - a]\] is equal to \[0\].
Option ‘A’ is correct
Note: Vector is a quantity with both magnitude and direction. A unit vector is one with a magnitude of one. Dot product and cross product is not the same thing. Concentrate your efforts on the calculation part. Remember that the magnitude of a unit vector is always one.
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